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The regular three-dimensional arrangement of atoms or ions in a crystal is usually described in terms of a space lattice and a unit cell. To see what these two terms mean, let us first consider the two-dimensional patterns shown in Figure $1$. We can think of each of these three structures as a large number of repetitions in two directions of the parallel-sided figure shown immediately below each pattern.
This parallel-sided figure is the unit cell. It represents the simplest, smallest shape from which the overall structure can be constructed. The pattern of points made by the comers of the unit cells when they are packed together is called the space lattice (Figure $2$). The lines joining the points of the space lattice are shown in color. Without some experience, it is quite easy to pick the wrong unit cell for a given structure. Some incorrect choices are shown immediately below the correct choice in the figure. Note in particular that the unit cell for structure b, in which each circle is surrounded by six others at the comers of a hexagon, is not a hexagon, but a parallelogram of equal sides (a rhombus) with angles of 60 and 120°.
Figure $2$ illustrates the space lattice and the unit cell for a real three-dimensional crystal structure—that of sodium chloride.
This is the same structure that was shown for lithium hydride, except that the sizes of the ions are different. A unit cell for this structure is a cube whose comers are all occupied by sodium ions. Alternatively, the unit cell could be chosen with chloride ions at the comers.
The unit cell of sodium chloride contains four sodium ions and four chloride ions. In arriving at such an answer we must bear in mind that many of the ions are shared by several adjacent cells (part c of Figure $2$ shows this well). Specifically, the sodium ions at the centers of the square faces of the cell are shared by two cells, so that only half of each lies within the unit cell. Since there are six faces to a cube, this makes a total of three sodium ions. In the middle of each edge of the unit cell is a chloride ion which is shared by four adjacent cells and so counts one-quarter. Since there are twelve edges, this makes three chloride ions. At each comer of the cube, a sodium ion is shared by eight other cells. Since there are eight comers, this totals to one more sodium ion. Finally, there is a chloride ion in the body of the cube unshared by any other cell. The grand total is thus four sodium and four chloride ions.
A general formula can be derived from the arguments just presented for counting N, the number of atoms or ions in a unit cell. It is:
$N=N_{\text{body}}\text{ + }\frac{N_{\text{face}}}{\text{2}}\text{ + }\frac{N_{\text{edge}}}{\text{4}}\text{ + }\frac{N_{\text{corner}}}{\text{8}} \nonumber$
10.04: Crystal Systems
Unit cells need not be cubes, but they must be parallel-sided, three-dimensional figures. A general example is shown in Figure \(1\). Such a cell can be described in terms of the lengths of three adjacent edges, a, b, and c, and the angles between them, α, β, and γ.
Crystals are usually classified as belonging to one of seven crystal systems, depending on the shape of the unit cell. These seven systems are shown in the image below.
The simplest is the cubic system, in which all edges of the unit cell are equal and all angles are 90°. The tetragonal and orthorhombic classes also feature rectangular cells, but the edges are not all equal. In the remaining classes, some or all of the angles are not 90°. The least symmetrical is the triclinic, in which no edges are equal and no angles are equal to each other or to 90°. Special note should be made of the hexagonal system whose unit cell is shown in Figure \(2\). It is related to the two-dimensional cell encountered previously as the second example of a 2D crystal lattice structure, in that two edges of the cell equal and subtend an angle of 120°. Hexagonal crystals are quite common among simple compounds, like quartz, seen here below. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.03%3A_Lattices_and_Unit_Cells.txt |
An important class of crystal structures is found in many metals and also in the solidified noble gases where the atoms (which are all the same) are packed together as closely as possible. Most of us are familiar with the process of packing spheres together, either from playing with marbles or BB’s as children or from trying to stack oranges or other round fruit into a pyramid. On a level surface we can easily arrange a collection of spheres of the same size into a very compact hexagonal layer in which each sphere is touching six of its fellows, as seen in Figure $1$.
Then we can add a second layer so that each added sphere snuggles into a depression between three spheres in the layer below. Within this second layer each sphere also contacts six neighbors, and the layer is identical to the first one. It appears that we can add layer after layer indefinitely, or until we run out of spheres. Each sphere will be touching twelve of its fellows since it is surrounded by six in the same plane and nestles among three in the plane above and three in the plane below. We say that each sphere has a coordination number of 12. It is impossible to make any other structure with a larger coordination number, that is, to pack more spheres within a given volume. Accordingly the structure just described is often referred to as a closest-packed structure. Nickel, shown below, is an example of a metal with a closest-packed structure.
In part a of Figure $2$ the first layer of spheres has been labeled A and the B to indicate that spheres in the second layer are not directly above those in the fist. The third layer is directly above the first, and so it is labeled A. If we continue in the fashion shown, adding alternately A, then B, then A layers, we obtain a structure whose unit cell (shown in part a) has two equal sides with an angle of 120° between them. Other angles are 90°, and so the cell belongs to the hexagonal crystal system. Hence this structure is called hexagonal closest packed (hpc). An example of a metal with an HPC structure is Magnesium, which is shown below.
Also shown in Figure $3$ is the unit cell of a structure called body-centered cubic (bcc). This is similar to the fcc structure except that, instead of spheres in the faces, there is a single sphere in the center of the cube. This central sphere is surrounded by eight neighbors at the corners of the unit cell, giving a coordination number of 8. Hence the bcc structure is not as compact as the closest-packed structures which had a coordination number 12. Nevertheless, some metals are found to have bcc structures.
Example $1$: Unit Cells
Count the number of spheres in the unit cell of (a) a face-centered cubic structure, and (b) a body-centered cubic structure.
Solution
Referring to the last figure and using the equation:
$N=N_{\text{body}}\text{ + }\frac{N_{\text{face}}}{\text{2}}\text{ + }\frac{N_{\text{edge}}}{\text{4}}\text{ + }\frac{N_{\text{corner}}}{\text{8}} \nonumber$
we find
a) $N=N_{\text{body}}\text{ + }\frac{N_{\text{face}}}{\text{2}}\text{ + }\frac{N_{\text{edge}}}{\text{4}}\text{ + }\frac{N_{\text{corner}}}{\text{8}}=\text{0 + }\frac{\text{6}}{\text{2}}\text{ + 0 + }\frac{\text{8}}{\text{8}}=\text{4} \nonumber$
b) $N=\text{1 + 0 + 0 + }\frac{\text{8}}{\text{8}}=\text{2} \nonumber$
Example $2$: Crystal Forms
Silicon has the same crystal structure as diamond. Techniques are now available for growing crystals of this element which are virtually flawless. Analysis on some of these perfect crystals found the side of the unit cell to be 543.102064 pm long. The unit cell is a cube containing eight Si atoms, but is ont one of the simple cubic cells discussed already. From the isotope make up, molar mass and density of the crystals, it was determined that one mole of Si in this crystal form has a volume of 12.0588349×10-6 m3. Determine NA from this data.
Solution This problem uses knowledge of silicon crystal structure to determine NA. From the edge length, we can obtain the volume of the cubic unit cell. We know that the unit contains eight atoms, and since we know the volume of one mole, we can calculate NA, with the Avogadro constant defined as the number of particles per unit amount of substance.
$N_{A}= \frac{N*V_{\text{m}}}{V_{\text{unit cell}}}=\frac{8\times{12.0588349}\times{10}^{-6}\text{m}^{3}}{({ 543.102064}\times{10}^{-12}\text{m})^{3}}={6.02214179}\times{10}^{23} \nonumber$
The values used to determine this value were taken from crystals using X Ray Crystal Density(XRCD), to determine side length. These values were used in the most recent analysis published by the Committee on Data for Science and Technology(CODATA)[1], which standardizes definitions of important scientific constants and units. The value you just calculated is therefore the most accurate determination of Avogadro's constant at this time.
1. Mohr, P.J., Taylor, B.N., and D. B. Newell. "CODATA Recommended Values of the Fundamental Physical Constants:fckLR2006." National Institute of Standards and Technology. December 28, 2007. Constants [physics.nist.gov] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.05%3A_Closest-Packed_Structures.txt |
When a crystalline solid melts, it loses its rigid form and adopts the shape of its container. At the same time there is usually an increase in volume of a few percent. On the molecular level we can interpret this as a breakdown in the regular structure of the solid. As temperature rises toward the melting point, the molecules vibrate more and more strongly. Above the melting point, these vibrations are so energetic that they overcome the forces holding the molecules in the crystal lattice.
In the two animations below from the Concord Consortium, one can compare solids and liquids. The animation on the left is that of a solid, compact and held tightly together by intermolecular forces. Pressing the play button on the bottom of the screen has little effect. On the other hand, pressing the play button on the liquid animation has a huge effect. The molecules in the liquid begin to move around rapidly, in stark contrast to the stability of the solid.
The molecules no longer vibrate around an average position but begin to slide past each other. The regular arrangement of the crystal disappears, but the molecules have not escaped each other’s attractive influence (as can be seen by the dotted lines representing attractive forces in the animation).
The very small volume change which occurs on melting shows that the molecules have moved apart to only a very limited extent and that there can be only a few gaps caused by the less-regular packing. This view is confirmed by the experimental fact that liquids, as opposed to gases, are very difficult to compress. Even at the bottom of the deepest oceans, under pressures of thousands of atmospheres, the density of water is only minutely larger than at the surface.
10.7.01: Lecture Demonstration
Because its molecules can slide around each other, a liquid has the ability to flow. The resistance to such flow is called the viscosity. Liquids which flow very slowly, like glycerin or honey, have high viscosities. Those like ether or gasoline which flow very readily have low viscosities.
Viscosity is governed by the strength of intermolecular forces and especially by the shapes of the molecules of a liquid. Liquids whose molecules are polar or can form hydrogen bonds are usually more viscous than similar nonpolar substances. Honey, mostly glucose and fructose (see image below) is a good example of a liquid which owes its viscosity to hydrogen bonding.
Liquids containing long molecules are invariably very viscous. This is because the molecular chains get tangled up in each other like spaghetti—in order for the liquid to flow, the molecules must first unravel. Fuel oil, lubricating grease, and other long-chain alkane molecules are quite viscous for this reason. Glycerol, CH2OHCHOHCH2OH, is viscous partly because of the length of the chain but also because of the extensive possibilities for hydrogen bonding between the molecules. The video below shows several different long chained oils, each progressively more viscous.
The viscosity of a liquid always decreases as temperature increases. As the molecules acquire more energy, they can escape from their mutual traction more readily. Long-chain molecules can also wriggle around more freely at a higher temperature and hence disentangle more quickly. Below is a video that demonstrates this effect with a household liquid: honey. As a warning, the video has loud background music.
10.07: Viscosity
Viscosity
A group of liquids can be compared relative to each other based on viscosity. This experiment is set up with the liquid in a buret. A stop watch is started and the time is recorded when the liquid passes the starting point and the ending point. These starting and ending points are chosen by the person running the experiment. Typically these points are not too close together and are not too far apart.
Look at the structures below. It may take several seconds for the molecule to show up on the screen. DO NOT SCROLL during this time.
Click on the molecules and hold down your mouse button. Move your mouse to rotate the molecule.
Can you predict how viscous a liquid will be by looking at their molecular structure?
Question 1) How are these five molecules different?
Question 2) If applicable, how are these five molecules similar?
Question 3) What does the red sphere represent?
water hexane (C6H14) octane (C8H18)
water hexane (C6H14) octane (C8H18)
branched (C8H18) pentanol
branched (C8H18) pentanol
BEFORE moving on write down the answer to these two questions
Question 4) Choose the molecule that you think will be the most viscous. Write down your answer on a piece of paper and briefly write why you chose that molecule.
Question 5) Choose the molecule that you think will be the least viscous. Write down your answer on a piece of paper and briefly write why you chose that molecule.
Now watch the videos below and keep these two questions in mind.
Question 6) How does the red sphere affect how fast the liquid falls?
Question 7) How does the time the liquid falls relate to viscosity?
water hexane (C6H14) octane (C8H18)
water hexane (C6H14) octane (C8H18)
Videos are from JCE Web Software, Chemistry Comes Alive!. Copyright ACS Division of Chemical Education, Inc. Used with permission.
branched (C8H18) pentanol
branched (C8H18) pentanol
Videos are from JCE Web Software, Chemistry Comes Alive!. Copyright ACS Division of Chemical Education, Inc.
Used with permission. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.06%3A_Liquids.txt |
Some liquids become extremely viscous as the temperature falls toward their freezing points, often because they consist of macromolecules. An example is quartz, SiO2, seen .
When quartz melts (at 1610°C), a few Si—O bonds break, but most remain intact. The liquid contains large covalently bonded fragments of the original structure and is highly viscous. When the liquid is cooled, the macromolecular fragments cannot readily slide past one another to attain the regular solid structure of quartz. Instead, a collection of interconnected, randomly oriented tetrahedrons of oxygen atoms surrounding silicon atoms is formed, as shown in the figure below. The material having this structure is known as fused silica.
Fused silica is an example of an amorphous material or glass. It is highly rigid at room temperature, but it does not have the long-range microscopic regularity of a solid crystal lattice. Consequently it cannot be made to cleave along a plane. Instead, like ordinary window glass, it shatters into irregular fragments when struck sharply. (Window glass is primarily silica, but oxides of sodium and calcium are added to lower the melting point.) Since the microscopic structure of a glass is random, like that of a liquid, scientific purists describe glasses as highly viscous liquids, not as solids.
To the left is an example of fused silica, which has many industry applications due to its high purity, high melting temperature, and high radiation resistance.
10.09: Phase Transitions
We have now looked at the physical properties which chemists use to define the solid, liquid, and gas phases. In a solid, atoms, ions or molecules, are locked into an organized, long range lattice structure, unable to move beyond an average position due to intermolecular forces. In a liquid, this structure breaks down, molecules can slip past each other, but they are still held together by attractive forces. In a gas, these attractive forces are overcome, and the substance expands to fill space, each particle having gained mobility to break free of the others. Below, all 3 phases are shown at the submicroscopic level in animations. Notice how the movement and freedom of molecules steadily increases as attractive forces decrease from solid to liquid to gas phase.
Substances can be transformed from one phase into another. Solids melt into liquids and liquids boil to form vapors at temperatures which depend on their molecular properties, so chemists are interested in these transitions between phases. We are all familiar with the changes in macroscopic properties that accompany these transitions. YouTube has time lapse movies of ice melting on a small scale, or of the more environmentally critical arctic ice melt from 1979 to 2007.
This is a familiar process. As the solid melts, the resulting liquid is able to flow and conforms to the shape of the container. Heat from a flame is needed to bring about this transition. On a microscopic level melting involves breaking the intermolecular interactions between molecules. This requires an increase in the potential energy of the molecules, and the necessary energy is supplied by the Bunsen burner. Melting (or freezing) can, in some cases, be caused by changing just the pressure.
Boiling is equally familiar. Under specific temperature and pressure conditions, liquids start to bubble, and are converted to a gaseous form. A YouTube video "Boiling water with ice" shows that water boils at low temperatures if the pressure is reduced. Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy.
In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to boil at higher temperatures. It should be noted as well, that because there is a distribution in the kinetic energies of molecules, an equilibrium between gas and liquid phase is established at temperatures other than the boiling point, and this behavior is another aspect of phase transitions that chemists study.
For phase transitions from solid to liquid, liquid to gas, or solid to gas, energy is required because they involve separation of particles which attract one another. Further, we can predict under which conditions of temperature and pressure such transitions will occur. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.08%3A_Amorphous_Materials-_Glasses.txt |
When heat is supplied to a solid (like ice) at a steady rate by means of an electrical heating coil, we find that the temperature climbs steadily until the melting point is reached and the first signs of liquid formation become evident, as can be seen on the graph below. Thereafter, even though we are still supplying heat to the system, the temperature remains constant as long as both liquid and solid are present. On the graph below, this is represented by the flat line, where energy is being added to the ice, but no change is occurring in the temperature. All energy added to the system at this stage is used to convert solid ice into liquid water.
Image Credits: By Cawang via Wikimedia Commons
This macroscopic behavior demonstrates quite clearly that energy must be supplied to a solid in order to melt it. On a microscopic level melting involves separating molecules which attract each other. This requires an increase in the potential energy of the molecules, and the necessary energy is supplied by the heating coil. The kinetic energy of the molecules (rotation, vibration, and limited translation) remains constant during phase changes, because the temperature does not change.
The heat which a solid absorbs when it melts is called the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”) When 1 mol of ice, for example, is melted, we find from experiment that 6.01 kJ are needed. The molar enthalpy of fusion of ice is thus +6.01 kJ mol–1, and we can write
$\ce{ H2O(s) ->[0^{o} \text{C}] H2O(l)} \nonumber$
$\triangle H_m = 6.01 \frac {\text{kJ}}{\text{mol}} \nonumber$
Selected molar enthalpies of fusion are tabulated in Table $1$. Solids like ice which have strong intermolecular forces have much higher values than those like CH4 with weak ones. Note that the enthalpies of fusion and vaporization change with temperature.
When a liquid is boiled, the variation of temperature with the heat supplied is similar to that found for melting. When heat is supplied at a steady rate to a liquid at atmospheric pressure, the temperature rises until the boiling point is attained. After this the temperature remains constant until the enthalpy of vaporization (ΔHm) has been supplied. Once all the liquid has been converted to vapor, the temperature again rises. In the case of water the molar enthalpy of vaporization is 40.67 kJ mol–1. In other words
$\ce{ H2O(l) ->[100^{o} \text{C}] H2O(g)} \nonumber$
$\triangle H_m = 40.67 \frac {\text{kJ}}{\text{mol}} \nonumber$
heat is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy. In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the table bears this out.
Table $1$: Molar Enthalpies of Fusion and Vaporization of Selected Substances.
Substance Formula ΔH(fusion)
/ kJ mol-1
Melting Point / K ΔH(vaporization) / kJ mol-1 Boiling Point / K (ΔHv/Tb)
/ JK-1 mol-1
Neon Ne 0.33 24 1.80 27 67
Oxygen O2 0.44 54 6.82 90.2 76
Methane CH4 0.94 90.7 8.18 112 73
Ethane C2H6 2.85 90.0 14.72 184 80
Chlorine Cl2 6.40 172.2 20.41 239 85
Carbon tetrachloride CCl4 2.67 250.0 30.00 350 86
Water* H2O 6.00678 at 0°C, 101kPa
6.354 at 81.6 °C, 2.50 MPa
273.1 40.657 at 100 °C,
45.051 at 0 °C,
46.567 at -33 °C
373.1 109
n-Nonane C9H20 19.3 353 40.5 491 82
Mercury Hg 2.30 234 58.6 630 91
Sodium Na 2.60 371 98 1158 85
Aluminum Al 10.9 933 284 2600 109
Lead Pb 4.77 601 178 2022 88
*www1.lsbu.ac.uk/water/data.html
Two other features of the table deserve mention. One is the fact that the enthalpy of vaporization of a substance is always higher than its enthalpy of fusion. When a solid melts, the molecules are not separated from each other to nearly the same extent as when a liquid boils. Second, there is a close correlation between the enthalpy of vaporization and the boiling point measured on the thermodynamic scale of temperature. Periodic trends in boiling point closely follow periodic trends in heat of vaporization. If we divide the one by the other, we find that the result is often in the range of 75 to 90 J K–1 mol–1. To a first approximation therefore the enthalpy of vaporization of a liquid is proportional to the thermodynamic temperature at which the liquid boils. This interesting result is called Trouton’s rule. An equivalent rule does not hold for fusion. The energy required to melt a solid and the temperature at which this occurs depend on the structure of the crystal as well as on the magnitude of the intermolecular forces.
Example $1$: heat
Compare the heat required to vaporize 100 g of lead to the energy required (1)to melt 100 g of lead; (2) to melt 100 g water; and (3) to vaporize 100 g of water.
Solution
To vaporize 100 g of lead:
$\text{Pb}(l) \rightarrow\text{Pb}(g)\;\;\;\;\;\; (1749^{o} \text{C}) \;\;\;\;\triangle H_m = 178 \dfrac {\text{kJ}}{\text{mol}} \nonumber$
$100 g ~ \times ~\dfrac {1 \text{mol Pb}}{207.2 ~\text{g}~ \text{Pb}}~ \times ~\dfrac{178~ \text{kJ}}{\text{mol}} = 85.9~ \text{kJ} \nonumber$
(1) To melt 100 g of lead:
$\text{Pb}(s) \rightarrow\text{Pb}(l) \nonumber$
$(328^{o}\text{C})$
$\triangle H_m = 4.77 \frac {\text{kJ}}{\text{mol}} \nonumber$
$100 \text{g} ~\times~ \frac {1 \text{mol Pb}}{207.2 ~\text{g}~ \text{Pb}} ~\times ~\frac{4.77~ \text{kJ}}{\text{mol}} = 2.30~ \text{kJ} \nonumber$
(2) To melt 100 g of water:
$100 \text{g} ~\times~ \frac {1 \text{mol}}{18.0 ~\text{g}}~ \times~ \frac{6.01~ \text{kJ}}{\text{mol}} = 33.4~ \text{kJ} \nonumber$
(3) To vaporize 100 g of water:
$100 \text{g} ~\times ~\frac {1 \text{mol}}{18.0~ \text{g}} ~\times ~\frac{40.657~ \text{kJ}}{\text{mol}} = 226~ \text{kJ} \nonumber$
It might be surprising that the heat required to melt or vaporize 100 g of lead is so much less than that require to melt or vaporized water. First, the temperature at which the substance melts has nothing to do with the enthalpy of fusion, although in practice we would have to add more heat to get lead to the melting point. The molar enthalpy of fusion is actually smaller for lead, because of smaller bonding energies between particles. The molar enthalpy of vaporization of lead is larger than that of water, but this problem reminds us that in some cases a mass-based result can be of practical value, showing that less heat is required to vaporize an equal mass of lead.
10.10: Enthalpy of Fusion and Enthalpy of Vaporization
Sunrise Sol gif [chemwiki.ucdavis.edu]
This sequence of nine images taken by NASA's Phoenix Mars Lander shows the sun rising on the morning of the lander's 101st Martian day after landing. [1]
Mars, with an atmosphere of 95.32% CO2 and 0.03% water [2] and temperatures (measured at 1.5 meters above the surface) ranging from + 1° F, ( -17.2° C) to -178° F (-107° C), is not at all Earth-like, but the two planets have a weather phenomenon in common:
A "virga" is a rain or snow shower that evaporates before it reaches the surface, and it appears that virga snowstorms occur on Mars[3]. We can use thermodynamic parameters developed on Earth to understand weather phenomena on Mars, and conditions there are right for snow to evaporate directly into gaseous water (a process called "sublimation"). There are also dry ice storms on Mars that are analogous to thunderstorms on Earth, where the precipitation is solid dry ice.
Virgas on Earth can be disastrous to aviation and cause strong damaging winds on the ground because of their heat effects. When a rain virga occurs, the raindrops absorb heat from the air, creating a "microburst" of cold air that descends rapidly, and the heat removed by sublimation of snow showers is much greater.
Although there is some putative evidence for liquid water on Mars, most water exists as ice, mixed in with dry ice (frozen carbon dioxide). Solar radiation may raise equatorial soil temperature to +81°F (27° C) briefly, but the polar cap surface temperatures are always much colder. But even at the highest surface temperatures, most ice appears to sublime. This is because the mean surface level atmospheric pressure is only 600 Pa (0.6 kPa, 0.6% of Earth's 101.3 kPa). The solid carbon dioxide (dry ice) on the surface of Mars sublimes, just as dry ice sublimes on earth.
We can understand these effects in the Earth and Martian atmospheres in terms of the enthalpy of fusion and enthalpy of vaporization of water, described in more familiar terms:
When heat energy is supplied to a solid, like ice, at a steady rate we find that the temperature climbs steadily until the melting point is reached and the first signs of liquid formation become evident. Thereafter, even though we are still supplying heat energy to the system, the temperature remains constant as long as both liquid and solid are present. Only when the last vestiges of the solid have disappeared does the temperature start to climb again.
A large quantity of energy must be supplied to a solid in order to melt it. On a microscopic level melting involves separating molecules which attract each other. This requires an increase in the potential energy of the molecules, and the necessary energy is supplied by the heating coil. The kinetic energy of the molecules (rotation, vibration, and limited translation) remains constant during phase changes, because the temperature does not change.
The heat energy which a solid absorbs when it melts is called the enthalpy of fusion or heat of fusion and is usually quoted on a molar basis. (The word fusion means the same thing as “melting.”) When 1 mol of ice, for example, is melted, we find from experiment that 6.01 kJ are needed. The molar enthalpy of fusion of ice is thus +6.01 kJ mol–1, and we can write
$\ce{H2O (s) -> H2O (l)} \nonumber$
(0°C) ΔHm = 6.01 kJ mol–1
If we continue adding heat after the solid melts, the temperature again gradually increases until the liquid begins to boil. At this point, the temperature remains constant until the enthalpy of vaporization has been supplied. Once all the liquid has been converted to vapor, the temperature again rises. In the case of water the molar enthalpy of vaporization is 40.67 kJ mol–1. In other words
$\ce{H2O (l) -> H2O (g)} \nonumber$
(100°C) ΔHm = 40.67 kJ mol–1
The enthalpies of fusion and vaporization both depend on temperature and pressure, as you can see in the table of selected molar enthalpies below. Solids like ice which have strong intermolecular forces have much higher values than those like CO2 with weak ones. Even on Earth and Mars, CO2 sublimes rather than melting, then evaporating.
If snow in a virga evaporates at, say 0°C, in removes heat from the surroundings equal to the sum of the heat of fusion and vaporization, which at 0°C is 45.051 + 6.007= 51.058 kJ/mol.[6]
$\ce{H2O (s) -> H2O (g)} \nonumber$
(0°C) ΔHsub = 51.058 kJ mol–1 or 2.83 kJ/g
While the temperature of the water does not change during sublimation, the surrounding atmosphere is cooled dramatically. This can be noted as snow on the ground disappears without melting as it sublimes, or as frozen clothes on a clothesline dry without melting.
The enthalpy of sublimation for CO2 at atmospheric pressure, and −78.5 °C (−109.3°F). is
$\ce{CO2 (s) -> CO2 (g)} \nonumber$
ΔHsub = 25.2 kJ mol–1 or 0.57 kJ/g. Note that much less heat is required to sublime CO2 (s) than H2O(s).
Figure $3$ Sublimation of dry ice when placed on the surface of water at room temperature[7]
Table $1$ Molar Enthalpies of Fusion and Vaporization of Selected Substances.
Molar Enthalpies of Fusion and Vaporization of Selected Substances.
Substance Formula ΔH(fusion)
/ kJ mol1
Melting Point / K ΔH(vaporization) / kJ mol-1 Boiling Point / K (ΔHv/Tb)
/ JK-1 mol-1
Neon Ne 0.33 24 1.80 27 67
Oxygen O2 0.44 54 6.82 90.2 76
Methane CH4 0.94 90.7 8.18 112 73
Ethane C2H6 2.85 90.0 14.72 184 80
Chlorine Cl2 6.40 172.2 20.41 239 85
Carbon tetrachloride CCl4 2.67 250.0 30.00 350 86
Water* H2O 6.00678 at 0°C, 101kPa
6.354 at 81.6 °C, 2.50 MPa
273.1 40.657 at 100 °C,
45.051 at 0 °C,
46.567 at -33 °C
373.1 109
n-Nonane C9H20 19.3 353 40.5 491 82
Mercury Hg 2.30 234 58.6 630 91
Sodium Na 2.60 371 98 1158 85
Aluminum Al 10.9 933 284 2600 109
Lead Pb 4.77 601 178 2022 88
*www1.lsbu.ac.uk/water/data.html
Heat energy is absorbed when a liquid boils because molecules which are held together by mutual attraction in the liquid are jostled free of each other as the gas is formed. Such a separation requires energy. In general the energy needed differs from one liquid to another depending on the magnitude of the intermolecular forces. We can thus expect liquids with strong intermolecular forces to have larger enthalpies of vaporization. The list of enthalpies of vaporization given in the table bears this out.
Two other features of the table deserve mention. One is the fact that the enthalpy of vaporization of a substance is always higher than its enthalpy of fusion. When a solid melts, the molecules are not separated from each other to nearly the same extent as when a liquid boils. Second, there is a close correlation between the enthalpy of vaporization and the boiling point measured on the thermodynamic scale of temperature. Periodic trends in boiling point closely follow periodic trends in heat of vaporization. If we divide the one by the other, we find that the result is often in the range of 75 to 90 J K–1 mol–1. To a first approximation therefore the enthalpy of vaporization of a liquid is proportional to the thermodynamic temperature at which the liquid boils. This interesting result is called Trouton’s rule. An equivalent rule does not hold for fusion. The energy required to melt a solid and the temperature at which this occurs depend on the structure of the crystal as well as on the magnitude of the intermolecular forces.
From ChemPRIME: 10.9: Enthalpy of Fusion and Enthalpy of Vaporization
10.10.02: Lecture Demonstrations
The q/T Paradox: Which "Contains More Heat", a Cup of Coffee at 95 °C or a Liter of Ice water?[1]
A small mass of water at 0oC is added to a measured mass of liquid nitrogen, and the amount that evaporates is compared to the mass that evaporates when a larger mass of water at 95oC is added to liquid nitrogen. This demonstration requires knowledge of both specific heat and heat capacity.
Determination of the Enthalpy of Fusion of Water
Calculate Enthalpy of Fusion of Ice (assuming heat capacity prerequisite)
Dip a computer-interfaced thermistor probe in 100g of water in a styrofoam cup calorimeter. Add 3-5 g of ice to a paper towel on a balance, and record the total mass. Start temperature acquisition 1 sample/second, 3 minutes total, and after a few readings, remove ~2 g of ice from the balance and add it to the calorimeter. Record the final mass on the balance and calculate the mass of ice. Display the T vs. time plot [2]. Record the final temperature.
q (cal) + q (water ) = q (water from ice) + q (ice)
14.4 J/oC (20.97-22.70oC) + 100 (4.18)(20.97-22.70oC) = -(q + 1.90 (4.18)(20.97-0oC))
q = 581 J
ΔH = 581J/1.90 g x (1kg/1000 J) x (18 g/mol) = 5.5 kJ/mol (6.07 kJ/mol true value)
References
1. ↑ J. Chem. Educ., 2005, 82 (6), p 856
2. ↑ We use Vernier LoggerPro(R)software | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.10%3A_Enthalpy_of_Fusion_and_Enthalpy_of_Vaporization/10.10.01%3A_Astronomy-_Water_on_Mars.txt |
When a liquid such as water or alcohol is exposed to air in an open container, the liquid evaporates. This happens because the distribution of speeds (and hence kinetic energies) among molecules in a liquid is similar to that illustrated for gases, shown again below.
At any given instant a small fraction of the molecules in the liquid phase will be moving quite fast. If one of these is close to the surface and is traveling upward, it can escape the attraction of its fellow molecules entirely and pass into the gas phase. As the higher energy molecules depart, the average energy of the molecules in the liquid decreases and the temperature of the liquid falls. Heat energy will be absorbed from the surroundings, an effect which you can feel if you let water or alcohol evaporate from your skin. Absorption of heat maintains the average molecular speed in the liquid, so that, given enough time, all the liquid can evaporate. The heat absorbed during the entire process corresponds to the enthalpy of vaporization.
If the liquid is placed in a closed, rather than an open, container, we no longer find that it evaporates completely. Once a certain partial pressure of gas has been built up by the evaporation of liquid, no more change occurs, and the amount of liquid remains constant. The partial pressure attained in this way is called the vapor pressure of the liquid. It is different for different liquids and increases with temperature for a given liquid. So long as some liquid is present, the vapor pressure is always the same, regardless of the size of the container or the quantity of liquid.
For example, we find that any size sample of water held at 25°C will produce a vapor pressure of 23.8 mmHg (3.168 kPa) in any closed container, provided only that all the water does not evaporate. On the macroscopic level, once the vapor pressure has been attained in a closed container, evaporation appears to stop, as seen in the water bottle below.
On the microscopic level, though, molecules are still escaping from the liquid surface into the vapor above, as shown in the accompanying figure. The amount of vapor remains the same only because molecules are reentering the liquid just as fast as they are escaping from it. The molecules of the vapor behave like any other gas: they bounce around colliding with each other and the walls of the container. However, one of these “walls” is the surface of the liquid. In most cases a molecule colliding with the liquid surface will enter the body of the liquid, not have enough energy to escape, and be recaptured.
When the liquid is first introduced into the container, there are very few molecules of vapor and the rate of recapture will be quite low, but as more and more molecules evaporate, the chances of a recapture will become proportionately larger. Eventually the vapor pressure will be attained, and the rate of recapture will exactly balance the rate of escape. There will then be no net evaporation of liquid or condensation of gas.
Once the vapor-liquid system has attained this state, it will appear on the macroscopic level not to be undergoing any change in its properties. The amount, the volume, the pressure, the temperature, the density, etc., of both liquid and gas will all remain constant with time. When this happens to a system, it is said to be in an equilibrium state or to have attained equilibrium.
Later, we will encounter many other quite different examples of equilibrium, but they all have one property in common. The lack of change on the macroscopic level is always the result of two opposing microscopic processes whose rates are equal. The effect of each process is to nu1lify the effect of the other. Since both microscopic processes are still in motion, such a situation is often referred to as dynamic equilibrium.
The magnitude of the vapor pressure of a liquid depends mainly on two factors: the strength of the forces holding the molecules together and the temperature. It is easy to see that if the intermolecular forces are weak, the vapor pressure will be high. Weak intermolecular forces will permit molecules to escape relatively easily from the liquid. The rate at which molecules escape will thus be high. Quite a large concentration of molecules will have to build up in the gas phase before the rate of reentry can balance the escape rate. Consequently the vapor pressure will be large.
By contrast, strong intermolecular forces result in a low escape rate, and only a small concentration of molecules in the vapor is needed to balance it. The vapor pressure of a liquid is quite a sensitive indicator of small differences in intermolecular forces. In the case of the alkanes, for example, we find that the vapor pressure of normal pentane at 25°C is 512.3 mmHg (68.3 kPa) while that of normal hexane is 150.0 mmHg (20.0 kPa) and that of normal heptane only 45.7 mmHg (6.1 kPa), despite the fact that the intermolecular forces for the three substances differ by less than 10 percent. The following video compares the vapor pressures of pentane, hexane, and heptane, by injecting each into a column of mercury. If you look closely, you can see the small amount of each liquid on top of the mercury. This serves to demonstrate the described difference in vapor pressure: stronger intermolecular forces result in lower vapor pressure.
The other major factor governing the magnitude of the vapor pressure of a liquid is temperature. At a low temperature only a minute fraction of the molecules have enough energy to escape from the liquid. As the temperature is raised, this fraction increases very rapidly and the vapor pressure increases with it, which makes sense given our previous discussion on temperature and gases. Moreover, the higher the temperature, the more rapid the rate of increase of the energetic fraction of molecules. The result is a variation of vapor pressure with temperature like that shown for four liquids alkanes. Note from this figure how the vapor-pressure increase for a 10°C increase in temperature is larger at higher temperatures.
The following video also showcases the effect of heat on vapor pressure. The subject is one of the novelty "drinking bird" devices. These birds operate due to vapor pressure. There are two distinct spaces above the liquid, one in the base of the bird, and one in the head of the bird which is connected to the liquid by a long glass tube. By heating the head of the bird, the temperature of the vapor there increase, thus the vapor pressure increases, and it pushes the liquid back down into the base. When the base is heated, the gas in the base increases in vapor pressure compared to the head, and thus pushes liquid back up the neck of the bird. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.11%3A_Vapor-Liquid_Equilibrium.txt |
When we heat a liquid until it boils, the bubbles that form inside the liquid consist of pure vapor. If the liquid is well stirred while boiling occurs, the vapor in the bubbles will be in equilibrium with the liquid and will have a pressure equal to the vapor pressure at the boiling temperature. However, the pressure inside the bubbles must also be equal to the external pressure above the liquid. If this were not so, the bubbles would either suddenly collapse or suddenly expand. It follows therefore that when a liquid boils, the vapor pressure of the liquid is equal to the external pressure.
Normally when we boil a liquid, we do so at atmospheric pressure. If this pressure is the standard pressure of 1 atm (101.3 kPa), then the temperature at which the liquid boils is referred to as its normal boiling point. This is the boiling point which is usually quoted in chemical literature. Not everyone lives at sea level, though. Denver, Colorado, for example, is about a mile high, and the average atmospheric pressure there is only 630 mmHg (84 kPa). Liquids attain a vapor pressure of 630 mmHg at a somewhat lower temperature than is required to produce 760 mmHg (1 atm). Consequently liquids in Denver boil some 4 to 5°C lower than the normal boiling point. Since the boiling point is often used to identify a liquid, chemists living at high altitudes must be careful to allow for this difference.
The dependence of the boiling point on the external pressure can often be very useful. Chemists often purify liquids by boiling them and collecting the vapor, a process known as distillation. Some liquids have such high normal boiling points that they begin to decompose before distillation can be carried out. Such a liquid can often be distilled at reduced pressure. The temperature of boiling is then much lower, and the risk of decomposition considerably less. The reverse procedure is used in a pressure cooker. The pressure inside the sealed cooker builds up until it is larger than atmospheric, and so the water used for cooking boils at a temperature above its normal boiling point. Therefore the cooking proceeds more rapidly.
The following video highlights the idea that boiling point is dependent upon both temperature and pressure. In the video, water is boiled in a flask, which is then stoppered and removed from the heat source. When cold water is poured over the top of the flask, it cools the gas above the liquid water. This decreases the vapor pressure above the water. The lower vapor pressure corresponds to a lower boiling point, and therefore the water boils again. Note that if cooling had been applied to the liquid on the bottom, these subsequent boilings would not occur.
Video \(1\): So you thought you could only boil water with heat? Watch as we flip that upside down and use ice cubes. Keep in mind there is a little Science involved and all the items you need are already in your home. We take the glass bottle fill it 1/3 with water set on your stove top until a nice boil 100 ºC remove with oven mit let cool for 5-10 seconds then put cap back on top. Invert bottle into a cold bowl of ice with ice cubes on the bottom and watch as the water starts to re boil in front of your eyes.
Example \(1\): Boiling Points
From the figure displaying boiling points of four alkanes, estimate the boiling points of the four alkanes when the pressure is reduced to 600 mmHg.
The vapor pressure of four liquid alkanes.
Solution
Reading along the 600-mmHg line in the graph, we find that it meets the vapor-pressure curve for pentane at about 29°C. Accordingly this is the boiling point of pentane at 600 mmHg. Similarly we find the boiling point of hexane to be 61°C, and of heptane to be 90°C. The boiling point of octane is above 100°C and cannot be estimated from the graph. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.12%3A_Boiling_Point.txt |
Suppose we seal a pure liquid and its vapor in a strong glass tube and heat it to a very high temperature. As we increase the temperature, the vapor pressure will rise. (It is not a good idea to heat a liquid this way unless you are sure the container can withstand the increased pressure.) The rising vapor pressure corresponds to a greater number of molecules in the limited volume of the vapor phase. In other words, the vapor becomes considerably denser. Eventually we reach a temperature at which the density of the vapor becomes the same as that of the liquid. Since liquids are usually distinguished from gases on the basis of density, at this point both have become identical. The temperature at which this occurs is called the critical temperature, and the pressure is called the critical pressure.
The accompanying videos illustrate what happens experimentally in the case of Chlorine. In the first video, as the temperature nears the critical temperature, liquid and vapor become very similar in appearance and the meniscus between them becomes difficult to distinguish. Finally, at the critical temperature the meniscus disappears completely. Above the critical temperature the sample is quite uniform and it is difficult to know whether to call it a liquid or a gas.
In the second half of the video, the flask is brought back below the critical temperature. The speed of gas molecules decreases to a point where intermolecular forces can cause a liquid phase to condense out. The meniscus reappears, and the Chlorine separates back again into a liquid and vapor phase.
Once a gas is above its critical temperature, it is impossible to get it to separate into a liquid layer below and a vapor layer above no matter how great a pressure is applied, as can be seen in the graph below. On the graph, once the temperature is higher than 300 K, it is not possible to revert to liquid form. Increasing the pressure only leads to the transition from gas to supercritical fluid.
Oxygen, for instance, is well above its critical temperature at room temperature. If we increase the pressure on it to a few thousand atmospheres, its density becomes so high that we are forced to classify it as a liquid. Nevertheless as we increase the pressure, there is no point at which drops of liquid suddenly appear in the gas. Instead the oxygen gradually changes from something which is obviously a gas to something which is obviously a liquid (a supercritical fluid). Conversely, if we gradually relax the pressure, there is no point at which the oxygen will start to boil.
The following table lists the critical temperatures and critical pressures for some well-known gases and liquids. Such data are often quite useful. Many gases are sold commercially in strong steel cylinders at high pressures. The behavior of the gas in such a cylinder depends on whether it is above or below the critical temperature. The critical temperature of propane, for instance, is 97°C, well above room temperature. Thus propane in a high-pressure cylinder consists of a mixture of liquid and vapor, and you can sometimes hear the liquid sloshing about inside.
Table \(1\) Critical Temperatures and Pressures of Some Simple Substances.
Critical Temperatures and Pressures of Some Simple Substances
Name Critical Temperature (K) Critical Pressure (MPa) Critical Pressure (atm)
Hydrogen (H) 33.2 1.30 12.8
Neon (Ne) 44.5 2.7 26.9
Nitrogen (N) 126.0 3.39 33.5
Carbon dioxide (CO2) 304.2 7.39 73.0
Propane (C3H8) 370 4.23 41.8
Ammonia (NH3) 405.5 11.29 111.5
Water (H2O) 647.1 22.03 217.5
The pressure of the gas in such a cylinder will be the vapor pressure of propane at namely, 9.53 atm (965.4 kPa).As long as there is some liquid left in the cylinder, the pressure will remain at 9.53 atm. Only when all the liquid has evaporated the pressure begin to drop. At that point the cylinder will be virtually empty. A very different behavior is found in the case of a cylinder of oxygen. Since oxygen is above its critical temperature at 20°C, the cylinder will contain a uniform fluid rather than a liquid-vapor mixture. As we use up the oxygen, the pressure will gradually decrease to 1 atm, at which point no more O2 will escape from the cylinder.
The principles discussed in the preceding paragraph apply to the aerosol sprays most of us encounter every day. Such spray cans contain a small quantity of the active ingredient—hair conditioner, deodorant, shaving cream, and the like—and a large quantity of propellant. The propellant is a substance, such as propane, whose critical temperature is well above room temperature. Therefore it can be liquefied at the high pressure in the spray can.
When the valve is opened, the vapor pressure of the liquid propellant causes the active ingredient and the propellant to spray out of the can. As long as liquid propellant remains, the pressure inside the can will be constant (it will be the vapor pressure), and the spray will be reproducible. It should be obvious why such cans always bear a warning against throwing them in a fire--vapor pressure increases more rapidly at higher temperatures, and so heating an enclosed liquid is far more likely to produce an explosion than heating a gas alone. (The latter case was described in an earlier example.) | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure.txt |
Solids as well as liquids exhibit vapor pressure. If you hang out a wet cloth in winter, the cloth first freezes as hard as a board, but after sufficient time all the ice evaporates and the cloth becomes soft and dry. Another solid which exhibits evidence of a vapor pressure is para-dichlorobenzene, C6H4Cl2, which is used for mothballs. The fact that you can smell this solid from across the room means that some of its molecules must have evaporated into the air and entered your nose. Indeed, given a year or so, moth crystals will evaporate completely. Still another example is dry ice, CO2(s), whose vapor pressure reaches atmospheric pressure at –78.5°C. Consequently CO2(s) sublimes, forming vapor without passing through the liquid state. Unlike ordinary ice, it remains dry, making it very convenient as a refrigerant.
Figure \(1\) Phase diagram of water. The scale has been distorted for purposes of diagrammatic clarity. Point T is the, triple point, C the critical point, B the boiling point, and F the freezing point.
Like the vapor pressure of a liquid, the vapor pressure of a solid increases with temperature. This variation is usually presented in combination with other curves in a phase diagram, such as that for water in accompanying figure. The term phase in this context is used to distinguish solid, liquid, and gas, each of which has its own distinctive properties, such as density, viscosity, etc. Each phase is separated from the others by a boundary.
In the phase diagram for water, the variation of the vapor pressure of ice with temperature is shown by the line AT. As might be expected, the vapor pressure of ice is quite small, never rising above 0.006 atm (0.61 kPa). The vapor pressure of liquid water is usually much higher, as is shown by the curve TC. The point C on this curve corresponds to the critical point, and the vapor pressure is the critical pressure of water, 218.3 atm (22 MPa). The curve stops at this point since any distinction between liquid and gas ceases to exist above the critical temperature (this state is called a supercritical fluid).
The other end of this vapor-pressure curve, point T, is of particular interest. At the temperature of point T (273.16 K or 0.01°C) both ice and water have the same vapor pressure, 0.006 atm. Since the same vapor is in equilibrium with both liquid and solid, it follows that all three phases, ice, water, and vapor, are in equilibrium at point T. This point where all 3 phases exist at the same time is therefore referred to as the triple point.
Line TD remains to be discussed. This line describes how the melting point of ice varies with pressure. In other words, it includes temperatures and pressures at which solid and liquid phases are in equilibrium. Note that the line is not exactly vertical. Point D corresponds to a slightly lower temperature than point T. This means that, as we increase the pressure on ice, its freezing point decreases. Again we can draw on everyday experience for an example of this behavior. A person on a pair of skates standing on ice exerts his or her whole weight on the ice through the thin skate blade. The very high pressure immediately under the blade causes some ice to melt, affording lubrication which enables the skater to glide smoothly over the ice.
It should be emphasized that point T is not the normal freezing point of water. When we measure the normal freezing point of water (or any other liquid), we do so at standard atmospheric pressure in a container open to the air (part a of the next figure). In this container we have not only ice and water in equilibrium with each other but also air saturated with water vapor.
The total pressure on the contents of this container is 1 atm (101.3 kPa) and its temperature is exactly 273.15 K (0.00°C). As far as liquid and solid are concerned, this corresponds to point F in the phase diagram.
If we could now pump all the air out of the container so that only pure water in its three phases was left, we would have the situation shown in part b. Ice, water, and pure water vapor, but no air, now occupy the container. The pressure has dropped from the 1.00 atm of the atmosphere to the vapor pressure (0.006 atm). Because of this decrease in pressure, the melting point of ice increases slightly and the new equilibrium temperature is 0.01°C. This corresponds to the triple point of water, point T in the phase diagram.
Because the triple point is more readily attained (one need not be at sea level) and also more reproducible, it is now used as a primary reference temperature for the international thermodynamic scale of temperature. In SI units the temperature of the triple point of water is defined as 273.16 K. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.14%3A_Phase_Diagrams.txt |
Up to this point we have discussed only the properties of pure solids and liquids. Of more importance to a chemist, though, are the properties of solutions. Very few chemical reactions involve only pure substances―almost all involve a solution of some sort.
We defined a solution as a homogeneous mixture of two or more substances previously, that is, a mixture which appears to be uniform throughout. Under this definition we would refer to sugar or salt dissolved in water as solutions, but we would not apply the term to muddy water or to milk. A close inspection of muddy water reveals that it is not uniform in appearance but consists of small solid particles dispersed in water. We refer to such a mixture as a suspension. Under the microscope, milk can also be seen to be nonuniform, It consists of small drops of milk fat dispersed throughout an aqueous phase. A more obvious example of a suspension is chia seeds in water, seen below. The chia seeds are randomly suspended in the cup of water, a perfect example of a suspension.
Our definition of a solution in terms of the homogeneity of a mixture is somewhat unsatisfactory since it does not tell us where to draw the line. Field-emission microscopes and electron microscopes have now been developed which can just about ”see” a single atom. With such a microscope virtually all matter looks nonuniform and hence not homogeneous. If our definition extends to such microscopes, then true solutions do not exist. In practice we draw the line somewhere around the 5 nanometer (nm) mark, even though some molecules are larger than this.
On the molecular level a solution corresponds to the random arrangement of one kind of molecule or ion around another. In the accompanying figure, the illustration on the left corresponds to a solution since each black molecule is randomly surrounded by black and red molecules, and vice versa. The illustration on the right is a suspension. The distribution is not random, and most red molecules have red neighbors, while most black molecules have black neighbors.
Strictly speaking, the term solution applies to any homogeneous mixture, but we will concentrate our discussion on those solutions which involve liquids since these are the most common. It should be realized, though, that other types of solutions also exist. Air is a solution of a large number of gases (oxygen is the most concentrated) in another gas (nitrogen). A 5-cent coin is made from an alloy in which one solid (nickel) is dissolved in another (copper). Solutions of hydrogen gas in solid palladium and some other metals are also possible.
As mentioned in the brief discussion of solutions earlier, it is sometimes difficult to decide which component of a solution is the solute and which is the solvent. Usually the amount of solvent is much larger than that of the solute. If the pure components were initially in separate phases (a gas and a solid, for example), the phase corresponding to the state of the solution is taken to be the solvent. In the case of H2(g) and Pd(s) mentioned above, for example, Pd would be the solvent because the solution is a solid phase.
10.16: Saturated and Supersaturated Solutions
We often find that there is a limit to the quantity of solute which will dissolve in a given quantity of solvent. This is especially true when solids dissolve in liquids. For example, if 36 g KCl crystals is shaken with 100 g H2O at 25°C only 35.5 g of the solid dissolves. If we raise the temperature somewhat, all the KCl will dissolve, but on cooling to 25°C again, the extra 0.5 g KC1 will precipitate, leaving exactly 35.5 g of the salt dissolved. We describe this phenomenon by saying that at 25°C the solubility of KCl in H2O is 35.5 g KC1 per 100 g H2O. A solution of this composition is also described as a saturated solution since it can accommodate no more KCl.
Under some circumstances it is possible to prepare a solution which behaves anomalously and contains more solute than a saturated solution. Such a solution is said to be supersaturated. A good example of supersaturation is provided by Na2S2O3, sodium thiosulfate, whose solubility at 25°C is 50 g Na2S2O3 per 100 g H2O. If 70 g Na2S2O3 crystals is dissolved in 100 g hot H2O and the solution cooled to room temperature, the extra 20 g Na2S2O3 usually does not precipitate. The resulting solution is supersaturated; consequently it is also unstable. It can be “seeded” by adding a crystal of Na2S2O3, whereupon the excess salt suddenly crystallizes and heat is given off. After the crystals have settled and the temperature has returned to 25°C, the solution above the crystals is a saturated solution—it contains 50 g Na2S2O3
Another example of crystallizing salt out of a supersaturated solution can be seen in the following video. In this case, a supersaturated solution of sodium acetate is poured over a crystal of sodium acetate. These crystals provide the lattice structure "seed" which causes the sodium acetate ions in solution to crystallize out.
The salt begins to crystallize, forming a large sodium acetate structure from the precipitation of the ions out of solution. When the sodium acetate crystallizes, the oppositely charged ions are brought closer together by the crystal structure. Since formation of a crystal lattice lowers potential energy by placing like charges close together, the system releases the excess energy in the crystallization process. Thus, the structure ends up being warm to the touch from this excess energy. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.15%3A_Solutions.txt |
When a solid dissolves in a liquid, we very seldom find that the liquid has any tendency to dissolve in the solid. In a saturated solution of potassium chloride, for example, essentially no water dissolves in the potassium chloride crystals. With liquids the situation is usually different. If equal quantities of 1-butanol and water are shaken together, the mixture slowly separates into two layers. The bottom layer is a saturated solution of 1-butanol in water—it contains about 8% 1-butanol by weight. The top layer is not pure 1-butanol but a saturated solution of water in 1-butanol. It contains about 32% water by weight. A pair of liquids, like 1-butanol and water, which separates into two layers, but still partially mix, is said to be partially miscible. On a molecular level, a partially miscible mixture resembles the image below, with the two liquids (1 butanol in light blue and water in dark blue) clearly separated with slight mixing occurring.
By contrast with the solubilities of solids in liquids, a great many liquid pairs are completely miscible. That is, regardless of the proportions in which the two liquids are mixed, each will dissolve completely in the other. There will be no phase boundary as in the case of partially miscible liquids like 1-butanol and water. Ethanol and water provide a good example of two liquids which are completely miscible. If you have a source of pure ethanol, it is possible to mix a drink in any proportions you like-even up to 200 proof—without forming two separate liquid phases.
A well known example of two immiscible liquids are oil and water. Check out the below video to get a visual representation for what immiscibility looks like. Once you are done watching, take a guess, what would a mix of water and 1-butanol look like?
10.17: Miscibility
Figure \(1\). Exxon Valdez Oil Carrier
Oil has been a staple of our every day lives for thousands of years. In ancient times it was used as a fuel source, but more widely as a medicine. In the late 19th century oil was found in the Texas area, and from then on the world would never be the same. Oil was used for cars, planes, fuel, and helped to spur the industrial revolution. With the profitability of oil rising, many oil companies and drilling areas evolved around the world. The production and extraction of oil became one of the most profitable business in the world. However, a major problem with the oil industry was that it was all around the world. In order to get oil to the largest consumers the oil had to be shipped. Some of the largest oil reserves in the world are in the middle east, Alaska, and in the deep ocean. Each of these places has great access to shipping lanes, and therefore have been very profitable. However, a major problem is that along with the shipping lanes comes huge bodies of water, and oil and water definitely don't mix.
In 1989 an Exxon Valdez tanker was taking oil from Alaska to California. It ran aground and spilled 32 million gallons of oil all over the Alaskan coastline. The spill took place in Prince William Sound, which is in Southern Alaska by the Kenai Peninsula. The oil spread to over 1,300 miles of coastline and 11,000 square miles in the ocean. 250,000 animals were killed in the spill, including endangered species like the Bald Eagle. The reason that the oil is so deadly is that it sits on top of the water, because the two are not miscible, or do not mix. The sea life or birds come to the surface and get caked in oil. When covered in oil the birds wings do not work, and gills of fish are covered and they can't breathe. No animal is safe from the oil. The Exxon Valdez spill was extremely destructive, and was the United States' largest oil spill until the 2010 sinking of the oil rig Deepwater Horizon off the coast of Louisiana. The Louisiana oil spill released over 90 million gallons of oil into the ocean. However, these two oil spills are minute compared to some of the biggest in the world. These include the Kuwait oil spill (336 million gallons) and the Ixtoc, Mexico oil well collapse (140 million gallons). All oil spills no matter how big have devastating effects on the environments for many years because of the difficulty of cleaning up oil.
It could be seen that oil would be easy to clean up because it is not miscible with water. It seems easy to simply take off the top layer of oil and that would be it. However, the problem is that there is no great solvent to dissolve the oil. Many different techniques are used to clean up oil. One of these techniques is called using a dispersant. The dispersant takes large clumps of oil and breaks them down to very small particles. Since there is so much water available, and the particles are so small, the oil starts to be dissolved into the water column. The problem though is that with excessive amounts of oil, the water column can become toxic because of the large amounts of oil being dissolved. Another technique is to use skimmers. The skimmers take the oil off the water and get rid of it. The skimmers are prone to clogging though, especially with thick oil. One option that many people ask about is burning the oil. This is done to some extent, but burning oil creates many bad after effects for the atmosphere and people to breathe in. There really is no silver bullet in treating an oil spill, and because of this all of the above are done to clean up an oil spill.
Why Oil and Water don't Mix
In basic chemistry classes everyone learns that in solvent chemistry like dissolves like. In a more in depth sense, polar solutions dissolve polar solutes, and non-polar solvents dissolve non-polar solutes. In the case of oil and water, one is polar (water), and the other is non-polar (oil). Water is polar because of the bonds between oxygen and hydrogen are polar. The electronegativity difference between these two elements is about 1.24, meaning it is very polar. The water molecule is in the bent shape, which is important because the two polar bonds don't cancel each other out. For oil on the other hand, none of the bonds are polar. Oil is made of alkanes, or long carbon chains. Gas fuel for cars is made up of carbon chains between 5-8 carbons long and diesel and jet fuel is made up of carbon chains of 9-16 carbons long. These are the kinds of oil that the Exxon Valdez was carrying, and are components of every oil spill. Alkanes are simply made up of carbons and hydrogens, and the electronegativity difference between these elements is only .35. This means that the bonds are not polar. Since alkanes are not polar and water is polar, the two do not mix (not miscible), which creates the many problems with oil spills.
From ChemPRIME: 10.16: Miscibility | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.17%3A_Miscibility/10.17.01%3A_Cultural_Connections-_Oil_and_Water.txt |
When speaking of solubility or miscibility, or when doing quantitative experiments involving solutions, it is necessary to know the exact composition of a solution. This is invariably given in terms of a ratio telling us how much solute is dissolved in a unit quantity of solvent or solution. The ratio can be a ratio of masses, of amounts of substances, or of volumes, or it can be some combination of these. For example, concentration was defined before as the amount of solute per unit volume of solution:
$c_{\text{solute}}=\dfrac{n_{\text{solute}}}{V_{\text{solute}}} \nonumber$
The two simplest measures of the composition are the mass fraction w, which is the ratio of the mass of solute to the total mass of solution, and the mole fraction x, which is the ratio of the amount of solute to the total amount of substance in the solution. If we indicate the solute by A and the solvent by B, the mass fraction and the mole fraction are defined by
$w_{A}=\dfrac{m_{A}}{m_{A}\text{ + }m_{B}}\text{ and }x_{A}=\dfrac{n_{A}}{n_{A}\text{ + }n_{B}} \nonumber$
Example $1$: Fractions
A solution is prepared by dissolving 18.65 g naphthalene, C10H8 in 89.32 g benzene, C6H6. Find (a) the mass fraction and (b) the mole fraction of the naphthalene.
Solution
a) The mass fraction is easily calculated from the masses:
$w_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{m_{\text{C}_{\text{10}}\text{H}_{\text{8}}}}{m_{\text{C}_{\text{10}}\text{H}_{\text{8}}}\text{ + }m_{\text{C}_{\text{6}}\text{H}_{\text{6}}}}=\dfrac{\text{18}\text{.65 g}}{\text{18}\text{.65 g + 89}\text{.32 g}}=\text{0}\text{.1727} \nonumber$
It is sometimes useful to distinguish mass of solute and mass of solution for purposes of calculation. In such a case we can write
wC10H8 = 0.1727 g C10H8 per g solution
b) In order to calculate the mole fraction, we must first calculate the amount of each substance. Since
$M_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\text{128}\text{.18 g mol}^{-\text{1}}\text{ and }M_{\text{C}_{\text{6}}\text{H}_{\text{6}}}=\text{78}\text{.11 g mol}^{-\text{1}}\text{ } \nonumber$
we find $n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{\text{18}\text{.65 g}}{\text{128}\text{.18 g mol}^{-\text{1}}}=\text{0}\text{.1455} \nonumber$
$n_{\text{C}_{\text{6}}\text{H}_{\text{6}}}=\dfrac{\text{89}\text{.32 g}}{\text{78}\text{.11 g mol}^{-\text{1}}}=\text{1}\text{.144} \nonumber$
Thus
$x_{\text{C}_{\text{10}}\text{H}_{\text{8}}}=\dfrac{n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}}{n_{\text{C}_{\text{10}}\text{H}_{\text{8}}}\text{ + }n_{\text{C}_{\text{6}}\text{H}_{\text{6}}}}=\dfrac{\text{0}\text{.1455 mol}}{\text{0}\text{.1455 mol + 0}\text{.144 mol}}=\text{0}\text{.1455}=\text{0}\text{.1128} \nonumber$
or 0.1128 mol C10H8 per mol solution
The mass fraction is useful because it does not require that we know the exact chemical nature of both solute and solvent. Thus if we dissolve 10 g crude oil in 10 g gasoline, we can calculate wcrude oil= 0.5 even though the solute and solvent are both mixtures of alkanes and have no definite molar mass. By contrast, the mole fraction is useful when we want to know the nature of the solution on the microscopic level. In the above example, for instance, we know that for every 100 mol of solution, 11.28 mol is naphthalene. On the molecular level this means that out of each 100 molecules in the solution, 11.28 will, on the average, be naphthalene molecules.
The mass fraction of a solution is often encountered in other disguises. The weight percentage (strictly speaking, the mass percentage) of a solution is often defined by the formula:
$\text{Weight percentage of }A=\dfrac{m_{A}}{m_{A}\text{ + }m_{B}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ } \nonumber$
This definition is really the same as that of the mass fraction because the percent sign means “divided by 100.” Thus 100% is merely a synonym for 100/100, that is, the number 1, and we can write:
$w_{\text{C}_{10} \text{H}_8} = 0.1727 \times 1 = 0.1727 \times 100\% = 17.27\% \nonumber$
When the mass fraction is very small, it is often expressed in parts per million (ppm) or parts per billion (ppb). These symbols can be handled in much the same way as a percentage if you remember how they are related to unity: 1 = 100% = 106 ppm = 109 ppb
In other words:
\begin{align} 1\% = \tfrac{1}{100} & = 10^{-2} \ \text{1 ppm} & = {10}^{-6} \ \text{1 ppb} & = {10}^{-9} \end{align}
Example $2$: Mass Fraction
A 1-kg sample of water from Lake Powell, Utah, is found to contain 10 ng mercury. Walleyed pike caught in the lake contain 0.427 ppm mercury. (a) What is the mass fraction of mercury (in ppb) in water from Lake Powell? (b) If you ate 2 lb of walleyed pike caught from the lake, what mass of mercury would you ingest?
Solution
a) The mass fraction of mercury is by definition
$w_{\text{Hg}}=\dfrac{m_{\text{Hg}}}{m_{\text{solution}}} \nonumber$
Therefore
$w_{\text{Hg}}=\dfrac{\text{10 ng}}{\text{1 kg}}=\dfrac{\text{10 }\times \text{ 10}^{-\text{9}}\text{ g}}{\text{1 }\times \text{ 10}^{\text{3}}\text{ g}}=\text{1 }\times \text{ 10}^{-\text{11}} \nonumber$
In ppb
$w_{\text{Hg}}=\text{1 }\times \text{ 10}^{-\text{11}}\text{ }\times \text{ 1}=\text{1 }\times \text{ 10}^{-\text{11}}\text{ }\times \text{ }10^{\text{9}}\text{ ppb}=\text{0}\text{0.01 ppb} \nonumber$
b) Assuming the mercury to be uniformly distributed throughout the walleyed pike, we have
$w_{\text{Hg}} = 0.427 \text{ ppm} = 0.427 \times 10^{-7} \nonumber$
By definition and
$w_{\text{Hg}}=\dfrac{m_{\text{Hg}}}{m_{\text{walleyed pike}}} \nonumber$
and
$m_{\text{Hg}}=w_{\text{Hg}}\text{ }\times \text{ }m_{\text{walleyed pike}}=\text{4}\text{0.27 }\times \text{ 10}^{-\text{7}}\text{ }\times \text{ 2}\text{.0 lb} \nonumber$
$=\text{4}\text{.27 }\times \text{ 10}^{-\text{7}}\text{ }\times \text{ 2}\text{.0 lb }\times \dfrac{\text{1 kg}}{\text{2}\text{.2 lb}}\text{ }\times \text{ }\dfrac{\text{10}^{\text{3}}\text{ g}}{\text{1 kg}}=\text{3}\text{.9 }\times \text{ 10 g}=\text{390 }\mu \text{g} \nonumber$
From this example you can see that from nearly the same mass of fish, almost 40 000 times as much mercury would be obtained as from the water. Indeed, if you ate 2 lb of walleyed pike every day, you would exceed the minimum dosage (300 μg for a 70-kg human) at which symptoms of mercury poisoning can appear. Fortunately, most of us do not eat fish every day, nor are we as gluttonous as the example suggests. Nevertheless, the much higher mass fraction of mercury in fish than in water shows that very small quantities of mercury in the environment can be magnified many times in living systems. This process of bioamplification will be discussed more thoroughly in the sections on Biochemistry. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.18%3A_Measuring_the_Composition_of_a_Solution.txt |
Chemical theory has not reached the point where it can predict exactly how much of one substance will dissolve in another. The best we can do is to indicate in general terms the relationships between solubility and the microscopic structures of solute and solvent.
To begin with, moving particles of any kind tend to become more randomly distributed as time passes. If you put a layer of red marbles in the bottom of a can and cover it with a second layer of white marbles, shaking the can for a short time will produce a nearly random distribution. The same principle applies on the microscopic level. Moving molecules tend to become randomly distributed among one another, unless something holds them back. Thus gases, whose molecules are far apart and exert negligible forces on one another, are all completely miscible with other gases.
In liquid solutions, the molecules are much closer together and the characteristics of different types of molecules are much more important. In particular, if the solute molecules exert large intermolecular forces on each other but do not attract solvent molecules strongly, the solute molecules will tend to group together. This forms a separate phase and leaves the solvent as a second phase. Conversely, if the solvent molecules attract each other strongly but have little affinity for solute molecules, solvent molecules will segregate, and two phases will form.
A classic example of the second situation described in the previous paragraph is the well-known fact that oil and water do not mix—or if they do, they do not stay mixed for long. The reason is that oil consists of alkanes and other nonpolar molecules, while water molecules are polar and can form strong hydrogen bonds with each other. Suppose that alkane or other nonpolar molecules are randomly dispersed among water molecules, as shown in part a of the figure.The constant jostling of both kinds of molecules will soon bring two water molecules together. Dipole forces and hydrogen bonding will tend to hold the water molecules together, but there are only weak London forces between water and nonpolar molecules. Before long, clusters of water molecules like those in part b will have formed.
These clusters will be stable at room temperature because the energy of interaction between the water molecules will be larger than the average energy of molecular motion. Only an occasional molecular collision will be energetic enough to bump two water molecules apart, especially if they are hydrogen bonded. Given enough time, this process of aggregation will continue until the polar molecules are all collected together. If the nonpolar substance is a liquid, this process corresponds on the macroscopic level to the liquids separating from each other and forming two layers.
If instead of mixing substances like oil and water, in which there are quite different kinds of intermolecular attractions, we mix two polar substances or two nonpolar substances, there will be a much smaller tendency for one type of molecule to segregate from the other. Thus two alkanes like n-heptane, C7H16, and n-hexane, C6H14, are completely miscible in all proportions. The C7H16 and C6H14 molecules are so similar (recall the projection formulas of alkanes) that there are only negligible differences in intermolecular forces. Thus the molecules remain randomly mixed as they jostle among one another.
For a similar reason, methanol, CH3OH, is completely miscible with water. In this case both molecules are polar and can form hydrogen bonds among themselves, and so there are strong intermolecular attractions within each liquid. However, CH3OH dipoles can align with H2O dipoles, and CH3OH molecules can hydrogen bond to H2O molecules, and so the attractions among unlike molecules in the solution are similar to those among like molecules in each pure liquid. Again there is little tendency for one type of molecule to become segregated from the other.
All the cases just discussed are examples of the general rule that like dissolves like. Two substances whose molecules have very similar structures and consequently similar intermolecular forces will usually be soluble in each other. Two substances whose molecules are quite different will not mix randomly on the microscopic level. In general, polar substances will dissolve other polar substances, while nonpolar materials will dissolve other nonpolar materials. The greater the difference in molecular structure (and hence in intermolecular attractions), the lower the mutual solubility. The following video succinctly showcases this principle.
In the video a number of mixing events occur. A nonpolar, colored solid is added to CCl4. Since CCl4 is also nonpolar, like dissolves like, and the solid is dissolved. Next, water is added. Since water is polar, it does not mix with the CCl4 solution, even after vigorous shaking. Two layers remain, with the less dense water on top. Finally, hexane is added. Since it is nonpolar and less dense than water, it forms a third layer, on top of the water. When the test tube is shaken, however, two layers remain. Since hexane is nonpolar, it is miscible with CCl4, and so both form a single layer below the water.
Example $1$: Solubility in Water
Predict which of the following compounds will be most soluble in water:
1. $\underset{\text{Ethanol}}{\mathop{\text{CH}_{\text{3}}\text{CH}_{\text{2}}\text{OH}}}\,$
2. $\underset{\text{Hexanol}}{\mathop{\text{CH}_{\text{3}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{CH}_{\text{2}}\text{OH}}}\,$
Solution
Since ethanol contains an OH group, it can hydrogen bond to water. Although the same is true of hexanol, the OH group is found only at one end of a fairly large molecule. The rest of the molecule can be expected to behave much as though it were a nonpolar alkane. This substance should thus be much less soluble than the first. Experimentally we find that ethanol is completely miscible with water, while only 0.6 g hexanol dissolves in 100 g water.
Our discussion of solubility in terms of microscopic structure concludes with one more point. The solubility of other substances in solids are usually small. The constituent particles in a solid crystal lattice are packed tightly together in a very specific geometric arrangement. For one particle to replace another in such a structure is very difficult, unless the particles are almost identical. The most common solid solutions are alloys, in which one essentially spherical metal atom replaces another. Thus alloys are easily made by melting two metals and cooling the liquid solution. In many other cases, however, completely miscible liquids separate when a solid phase forms. A good example of this is benzene and naphthalene:
A naphthalene molecule is almost twice as big as a benzene molecule and cannot fit in the benzene lattice. Therefore, even though the liquids are miscible, the solids are not due to the molecular structures of benzene and naphthalene. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.19%3A_Solubility_and_Molecular_Structure.txt |
When a pure substance is mixed with another pure substance in which it is soluble, the substances become completely interspersed at the molecular level. In thinking about making solutions at the molecular level an analogy to a can of marbles may be useful. In the analogy, a layer of red marbles is placed in the bottom of a can and covered with a second layer of white marbles. After shaking the can for a short time, the marbles are mixed randomly.
Now let us imagine that you want to collect all the red marbles again. If you simply shake the can, it is unlikely that you will ever divide the marbles into two layers, each with only one kind of marble. Similarly, if two miscible liquids are combined, a chemist cannot simply un-mix the liquids into pure components.
Continuing the analogy, what if a few green marbles and blue marbles are placed into the can? Given enough red and white marbles, it may be difficult to determine that the green marbles and blue marbles are actually there. Similarly, when chemists have a multi-component solution which may contain traces of important chemical species, they are faced with the challenge of detecting whether these chemicals are present in solution.
To deal with these difficulties, chemists employ different methods to separate solutions into their components. Two essential techniques are distillation and chromatography
10.22: Distillation
It is immediately apparent from the cartoon in the section on Raoult’s law ( seen below ) that the composition of the vapor in equilibrium with a solution and the composition of the solution itself must be different. While only 50 percent of the molecules in the liquid phase are benzene (red) molecules, in the vapor phase they represent 80 percent of the molecules. This difference in composition between the two phases is the reason we can separate components of a liquid mixture by distillation.
In the laboratory, distillation is usually carried out in an apparatus similar to that shown in figure 1. The liquid mixture to be distilled is heated in a round-bottom flask until it boils. The emerging vapor travels into the condenser where it is cooled sufficiently to return to the liquid state. The condensed liquid can then be collected in a suitable container. In practice this container is usually changed several times during the course of the distillation. Each sample collected in this way is called a fraction, and the whole process is called fractional distillation. A thermometer may be used to determine when vapor reaches the condenser, and to determine the progression of the distillation.
This process can also be viewed in the following video. Here distillation follows the same process already described for the figure.
Suppose we use the distillation apparatus shown in the figure to distill an equimolar mixture of benzene and toluene. We would first have to raise the temperature of the mixture to 92.4°C, at which point the total vapor pressure would be 1 atm (101.3 kPa) and the liquid would boil. At 92.4°C the vapor in equilibrium with an equimolar mixture has a mole fraction of benzene equal to 0.71 (slightly different from that in the cartoon on the page discussing Raoult’s law).
Accordingly, when our mixture is distilled, vapor of this composition will be condensed and the first drops in the collection vessel will contain 71 percent benzene molecules and 29 percent toluene molecules. We will now have a sample which is much richer in benzene than the original. In theory we could go on repeating this process indefinitely. If we redistilled a liquid mixture having xb = 0.71, it would boil at 87°C and yield a condensed vapor whose mole fraction of benzene would be 0.87. A third distillation would yield benzene with xb = 0.95, and a fourth would yield a product with xb = 0.99. If we went on long enough, we could obtain a sample of benzene of any desired purity.
In practice repeated distillation takes up a lot of time, and we would lose part of the sample each time. Fortunately there is a way out of this difficulty. We insert a glass column filled with round beads, or with specially designed packing, between the distillation flask and the condenser. When a vapor reaches the bottom-most glass beads en route to the condenser, part of it will condense. This condensate will be richer in the less volatile component, toluene, and so the vapor which passes on will be richer in benzene than before. As it passes through the column, the vapor will undergo this process several times. In effect the fractionating column allows us to perform several distillations in one operation. A well-designed laboratory column makes it possible to effect several hundred distillations in one pass. By using such a column, it would be possible to distill virtually pure benzene from the equimolar mixture, leaving virtually pure toluene behind.
Fractional distillation is used on a very large scale in the refining of petroleum. Some of the large towers seen in petroleum refineries are gigantic fractionating columns. These columns enable the oil company to separate the petroleum into five or six fractions called cuts. The more-volatile cuts can be used directly for gasoline, kerosene, etc. The less-volatile cuts contain the very long-chain alkanes and are usually converted into shorter-chain alkanes by cracking, as described in the section on unsaturated hydrocarbons.
While fractional distillation can be used to separate benzene and toluene and most mixtures of alkanes which show only small deviations from ideal behavior, it cannot be used to separate mixtures which exhibit a maximum or a minimum in the vapor-pressure versus mole-fraction curve. Thus any attempt to separate benzene and methanol (seen in the graph on the page discussing Raoult’s law) by fractional distillation does not result in methanol distilling over initially but rather results in a mixture with xCH3OH = 0.614. A mixture of this composition is not only more volatile than methanol or benzene but also has a higher vapor pressure than any other mixture, as can be seen from that graph. This is why it distills first. Such a mixture is called a minimum boiling azeotrope. Conversely, for a mixture such as chloroform and acetone which exhibits large enough negative deviations from Raoult’s law, a maximum boiling azeotrope of low volatility distills at the end of a fractional distillation, preventing complete separation of the pure components. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.21%3A_The_Separation_of_Mixtures.txt |
Another useful set of techniques for separating mixtures is called chromatography. Perhaps the simplest of these techniques to describe is paper chromatography, an example of which is shown in the video below.
Video: Simple paper chromatography. A simple demonstration on paper chromatography using marker ink and water
Three substances are applied to a strip of chromatography-grade paper (the stationary phase of this experiment). As the liquid level rises and meets the spots, the sample partially dissolves in the liquid (the mobile phase because it is moving) and travels up the plate within the solution. Different substances will travel different distances along the plate. The distance that a substance will travel depends on how strongly it adheres to the stationary phase (a process called adsorption) versus how much time it spends dissolved in the mobile phase.
The more a substance adsorbs, the less it dissolves and the less it moves along the plate. The pink and blue spots at the end of the video are examples of substances highly adsorbed to the stationary phase. The less a substance adsorbs, the more it dissolves and the farther it travels, such as the yellow on the far left. The process is continued until a good separation is created. In this manner, a mixture of substances may be separated such as the middle sample, which was originally green but separated into blue and yellow. Notice that while it was not initially obvious that the middle spot contained both substances, this fact is clear after performing paper chromatography.
All forms of chromatography work on the same general principle as paper chromatography. There is always a stationary phase which does not move and a mobile phase which does. The various components in the mixture being chromatographed separate from each other because they are more strongly held by one phase or the other. Those which have the greatest affinity for the mobile phase move along the fastest.
The most important form of chromatography is gas chromatography or vapor-phase chromatography. A long column is packed with a finely divided solid whose surface has been coated with an inert liquid. This liquid forms the stationary phase. The mobile phase is provided by an inert carrier gas, such as He or N2, which passes continuously through the column (seen below), among the solid particles. A liquid sample can be injected into the gas stream at the sample injector and vaporized just before it enters the tube. As this sample is carried through the column by the slow stream of gas, those components which are most soluble in the inert liquid are held up, while the less-soluble components move on more rapidly. The components thus emerge one by one from the end of the tube, into the detector. In this way it is possible to separate and analyze mixtures of liquids which it would be impossible to deal with by distillation or any other technique.
The development of chromatography is one of the major revolutions in technique in the history of chemistry, comparable to that which followed the development of an accurate balance. Separations which were previously considered impossible are now easily achieved, sometimes with quite simple apparatus. This technique is particular essential to the science of biochemistry, in which complex mixtures are almost always encountered. In the field of environmental chemistry, chromatography has helped us separate and detect very low concentrations of contaminants like DDT or PCB (polychlorinated biphenyls). The major drawback to chromatography is that it does not lend itself to large-scale operation. As a result it remains largely a laboratory, rather than an industrial, technique for separating mixtures.
10.24: Colligative Properties of Solutions
The colligative properties of a solution are those which depend on the number of particles (and hence the amount) of solute dissolved in a given quantity of solvent, irrespective of the chemical nature of those particles. We have already seen from Raoult’s law that the vapor pressure of a solution depends on the mole fraction of solute (amount of solute), and now we are in a position to see how this affects several other properties of solutions.
In the image below, there are two solutions. Both have a dark blue solvent and the same number of solute particles dissolved in said solvent, light blue particles on the left and green particles on the right. A colligative property is a property that would be the same for both of the solutions below, even though they contain different solutes. Since they contain the same number of solute particles, any colligative properties of the solutions would be identical. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.23%3A_Chromatography.txt |
We often encounter solutions in which the solute has such a low vapor pressure as to be negligible. In such cases the vapor above the solution consists only of solvent molecules and the vapor pressure is always lower than that of the pure solvent. Consider, for example, the solution obtained by dissolving 0.020 mol sucrose (C12H22O11) in 0.980 mol H2O at 100°C. The sucrose will contribute nothing to the vapor pressure, while we can expect the water vapor, by Raoult’s law, to contribute
$P_{\text{H}_2 \text{O}}=P^*_{\text{H}_2\text{O}} * x_{ \text{H}_2 \text{O}} = 760 \text{mmHg} * 0.980 = 744.8 \text{mmHg} \nonumber$
We would thus expect the vapor pressure to be 744.8 mmHg, in reasonable agreement with the observed value of 743.3 mmHg.
A direct consequence of the lowering of the vapor pressure by a nonvolatile solute is an increase in the boiling point of the solution relative to that of the solvent. We can see why this is so by again using the sucrose solution as an example. At 100°C this solution has a vapor pressure which is lower than atmospheric pressure, and therefore it will not boil. In order to increase the vapor pressure from 743.3 to 760 mmHg so that boiling will occur, we need to raise the temperature. Experimentally we find that the temperature must be raised to 100.56°C. We say that the boiling-point elevation ΔTb is 0.56 K.
The image below demonstrates a solute dissolved in water. As you can see in the image, the presence of the solute interferes with the ability of the water to "escape" into the gaseous phase. This interference can be physical (as seen in the oversimplified image above) or it can be due to more complex factors, like intermolecular forces. Therefore, the solution has a lower vapor pressure for any given temperature. Thus, the mixture will require a higher temperature overall than a pure solution in order to boil, which is evident as a higher boiling boiling point.
A second result of the lowering of the vapor pressure is a depression of the freezing point of the solution. Any aqueous solution of a nonvolatile solute, for example, will have a vapor pressure at 0°C which is less than the vapor pressure of ice (0.006 atm or 4.6 mmHg) at this temperature. Accordingly, ice and the aqueous solution will not be in equilibrium. If the temperature is lowered, though, the vapor pressure of the ice decreases more rapidly than that of the solution and a temperature is soon reached when both ice and the solution have the same vapor pressure. Since both phases are now in equilibrium, this lower temperature is also the freezing point of the solution. In the case of the sucrose solution of mole fraction 0.02 described above, we find experimentally that the freezing point is –2.02°C.
We say that the freezing-point depression ΔTf is 2.02 K.
The image below demonstrates how this freezing point depression works on a very simple level. The salt (purple) in the water prevents the rigid, ordered arrangement of a solid from being achieved. The particles of solute block the crystallization of water, at least up until a certain point, causing a depression in the freezing point of the water.
The depression of the freezing point of water by a solute explains why the sea does not freeze at 0°C. Because of its high salt content the sea has a freezing point of –2.2°C. If the sea froze at 0°C, larger stretches of ocean would turn into ice and the climate of the earth would be very different. We can now also understand why we add ethylene glycol, CH2OHCH2OH, to water in a car radiator in winter. Without any additive the water would freeze at 0°C and the resulting increase in volume would crack the radiator. Since ethylene glycol is very soluble in water, it can form a solution with a freezing point low enough to prevent freezing even on the coldest winter day. Both the freezing-point depression and the boiling-point elevation of a solution were once important methods for determining the molar mass of a newly prepared compound. Nowadays a mass spectrometer is usually used for this purpose, often on an assembly-line basis. Many chemists send samples of newly prepared compounds to a laboratory specializing in these determinations in much the same way as a medical doctor will send a sample of your blood to a laboratory for analysis.
Note
The reason we can use the boiling-point elevation and the freezing-point depression to determine the molar mass of the solute is that both properties are proportional to the mole fraction and independent of the nature of the solute. The actual relationship, which we will not derive, is
$x_{A}=\frac{\Delta H_{m}}{RT^{\text{2}}}\text{ }\Delta T \nonumber$
where xA is the mole fraction of the solute and ΔT is the boiling-point elevation or freezing-point depression. T indicates either the boiling point or freezing point of the pure solvent, and ΔHm is the molar enthalpy of vaporization or fusion, whichever is appropriate. This relationship tells us that we can measure the mole fraction of the solute in a solution merely by finding its boiling point or freezing point of the solution.
Example $1$: Mole Fraction
A solution of sucrose in water boils at 100.56°C and freezes at –2.02°C. Calculate the mole fraction of the solution from each temperature.
Solution
a) For boiling we have, from the Table of Molar Enthalpies of Fusion and Vaporization,
$\triangle H_m = 40.7 \dfrac{\text{kJ}}{\text{mol}} \nonumber$ and $T=373.15 \text{K}$
As in previous examples the units of R should be compatible with the other units appearing in the equation. In this case since ΔHm is given in units of kJ mol–1, R = 8.314 J K–1 mol–1 is most appropriate. Since ΔT = 0.56 K, we have
$x_{\text{sucrose}}=\frac{\Delta H_{m}}{RT^{\text{2}}}\Delta T=\frac{\text{40}\text{.7 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 0}\text{.56 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ (373}\text{.15)}^{\text{2}}}=\text{0}\text{.0197} \nonumber$
b) Similarly for freezing we have
$\triangle H_m = 6.01 \dfrac{kJ}{mol} \nonumber$ $T=273.15 \text{K}$ $\triangle T = 2.02 \text{K}$
so that
$x_{\text{sucrose}}=\dfrac{\text{6}\text{.01 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}\text{ }\times \text{ 2}\text{.02 K}}{\text{8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ }\times \text{ (273}\text{.15 K)}^{\text{2}}}=\text{0}\text{.0196} \nonumber$
The two values are in reasonable agreement.
Once we know the mole fraction of the solute, its molar mass is easily calculated from the mass composition of the solution.
Example $2$: Molar Mass
33.07 g sucrose is dissolved in 85.27 g H2O. The resulting solution freezes at –2.02°C. Calculate the molar mass of sucrose.
Solution
Since the freezing point of the solution is the same as in part b of Example 1, the mole fraction must be the same. Thus
$x_{\text{sucrose}}=\text{0}\text{.0196}=\dfrac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + }n_{\text{H}_{\text{2}}\text{O}}} \nonumber$
Furthermore $\text{ }n_{\text{H}_{\text{2}}\text{O}}=\dfrac{\text{85}\text{.27 g}}{\text{18}\text{.02 g mol}^{-\text{1}}}=\text{4}\text{.732 mol} \nonumber$
Thus $\text{0}\text{.0196}=\dfrac{n_{\text{sucrose}}}{n_{\text{sucrose}}\text{ + 4}\text{.732 mol}} \nonumber$
so that \begin{align} & \text{4}\text{.732 }\times \text{ 0}\text{.0196 mol}=n_{\text{sucrose}}-\text{0}\text{.0196}n_{\text{sucrose}} \ & \text{ 0}\text{.092 75 mol}=\text{0}\text{.9804}n_{\text{sucrose}} \ \end{align} \nonumber or $\text{ }n_{\text{sucrose}}=\dfrac{\text{0}\text{.092 75}}{\text{0}\text{.9804}}\text{mol}=\text{0}\text{.0946 mol} \nonumber$
From which $\text{ }M_{\text{sucrose}}=\dfrac{\text{33}\text{.07 g}}{\text{0}\text{.0946 mol}}=\text{350 g mol}^{-\text{1}} \nonumber$
Note: The correct molar mass is 342.3 g mol–1. Neither the freezing point nor the boiling point gives a very accurate value for the molar mass of the solute. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.25%3A_Boiling-Point_Elevation_and_Freezing-Point_Depression.txt |
Suppose we have a solution of sugar in water separated by a membrane from a sample of pure water. The membrane is porous, but the holes are not large enough to allow sucrose molecules to pass through from one side to the other while still being large enough to allow water molecules to pass freely through them. In such a situation, water molecules will be hitting one side of the membrane more often than the other. As a result, water molecules will move more often from right to left through the membrane in the animation than they will in the reverse direction. There will thus be a net flow of water from the compartment containing pure solvent, through the membrane, into the compartment containing the sucrose. This is another example of the tendency for moving molecules to become more thoroughly mixed together. The following movie shows this process, called osmosis.
Osmosis occurs when two solutions of different concentrations are separated by a membrane which will selectively allow some species through it but not others. Then, material flows from the less concentrated to the more concentrated side of the membrane. A membrane which is selective in the way just described is said to be semipermeable. Osmosis is of particular importance in living organisms, since most living tissue is semipermeable in one way or another. In the movie, we have depicted a membrane which is selective purely because of the pore size. In biological systems, the semipermeability relies on a set of solute transporters and channels. The cell membrane is formed of a lipid bilayer with polar head groups facing out, and nonpolar hydrocarbon tails in the middle of the membrane. The consequence is that charged and polar substances cannot cross the membrane. Aquaporins are membrane proteins which allow water, but no other molecule, not even H3O+ to pass through. For other solutes and ions, there exist specific transporters, some which allow a solute to diffuse down a natural gradient, and others which actively pump ions or other solutes in or out of the cell. These transporters, pumps and channels can be gated and regulated as well, allowing a cell to respond to varying osmotic conditions.[1]
A simple demonstration of osmosis is provided by the behavior of red blood cells. If these are immersed in water and observed under the microscope, they will be seen to gradually swell and to finally burst, as seen in the video below. Osmotic flow occurs from the surrounding water into the more concentrated solutions inside the cell. If the blood cells are now immersed in a saturated solution of NaC1, osmosis occurs in the opposite direction, since the solution inside the cell is not as concentrated as that outside. Under the microscope, the blood cells can be seen to shrink and shrivel. In medical practice, any solution which is to be introduced into the blood must take the possibility of osmosis into account. Normal saline, a solution of 0.16 M NaCl (0.16 mol NaCl per dm3 of solution) is always employed for intravenous feeding or injection, because it has the same concentration of salts as blood serum. The video below shows what happens when a red blood cell is immersed in water and osmosis occurs.
The tendency for osmotic flow to occur from a solvent to a solution is usually measured in terms of what is called the osmotic pressure of the solution, symbol Π. This osmotic pressure is not a pressure which the solution itself exerts but is rather the pressure which must be applied to the solution (but not the solvent) from outside in order to just prevent osmosis from occurring. A simple method for measuring the osmotic pressure is shown in Figure 2. The wider end of a funnel-shaped tube is covered with a membrane. The tube is filled with solution and placed in a container of the solvent. The height of the solution above the solvent increases until a maximum value is reached. The osmotic pressure is then the pressure exerted by the column of a solution of height h where ρ is the density of the solution and g is the gravitational acceleration.
$\Pi = \rho gh \nonumber$
Experimentally the osmotic pressure is found to obey a law similar in form to the ideal gas law and hence easy to remember:
$\Pi V = nRT \label{1}$
where n is the amount of solute in volume V of solution. In practice it is more useful to have Eq. (1) in terms of the concentration of solute n/V. Accordingly we rearrange it to read
$\Pi = n * V^{-1}RT \nonumber$ or $\Pi = cRT \label{2}$
Equation $\ref{2}$ is a very useful relationship since it means that we can find the concentration of any solution merely by measuring its osmotic pressure. This, in turn, allows us to find the molar mass of the solute. Suppose we have a solution in which a known mass of solute is dissolved in a known volume of solution. By measuring the osmotic pressure of this solution, we can find the concentration of solute and hence the amount of solute in the total volume. Since we already know the mass of solute, the molar mass follows immediately.
Example $1$: Molar Mass
A solution of 20.0 g of polyisobutylene in 1.00 dm3 of benzene was placed in an osmometer, similar to the one shown in Figure 2, at 25°C. After equilibrium had been obtained, the height h was found to be 24.45 mm of benzene. Find the average molar mass of the polymer. The density of the solution is 0.879 g cm–3.
Solution
We must first find the osmotic pressure from the height h with the formula $\Pi = \rho gh \nonumber$. In doing this, it is most convenient to convert everything to SI base units.
\begin{align} & \Pi =\rho gh=\text{0}\text{.879}\frac{\text{g}}{\text{cm}^{\text{3}}}\text{ }\times \text{ 9}\text{.807}\frac{\text{m}}{\text{s}^{\text{2}}}\text{ }\times \text{ 24}\text{.45 mm} \ & \text{ }=210.8\frac{\text{g}}{\text{cm}^{\text{3}}}\frac{\text{m}}{\text{s}^{\text{2}}}\text{mm }\times \text{ }\frac{\text{1 kg}}{\text{1000 g}}\text{ }\times \text{ }\left( \frac{\text{100 cm}}{\text{1 m}} \right)^{\text{3}}\text{ }\times \text{ }\frac{\text{1 m}}{\text{1000 mm}} \ & \text{ }=\text{210}\text{.8}\frac{\text{kg m}^{\text{2}}}{\text{m}^{\text{3}}\text{ s}^{\text{2}}}=\text{210}\text{.8}\frac{\text{kg m}}{\text{s}^{\text{2}}}\text{ }\times \text{ }\frac{\text{1}}{\text{m}^{\text{2}}}=\text{210}\text{.8 N m}^{-\text{2}}=\text{210}\text{.8 Pa} \ \end{align} \nonumber
Knowing Π, we can now calculate the concentration c by rewriting
$\Pi = cRT \nonumber$
as
$\text{c}=\frac{\Pi }{RT}=\frac{\text{210}\text{.8 }\times \text{ 10}^{-\text{3}}\text{ kPa}}{\text{8}\text{.3143 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{ }\times \text{ 298}\text{.15 K}}=\text{8}\text{.50 }\times \text{ 10}^{-5}\text{ mol dm}^{-\text{3}} \nonumber$
For 1 dm3 of solution therefore,
$n_{solute} = 8.50 × 10^{–5} \text{mol} \nonumber$
while
$m_{solute} = 20 \text{g} \nonumber$
Thus
$M_{\text{solute}}=\frac{\text{20 g}}{\text{8}\text{.50 }\times \text{ 10}^{-\text{5}}\text{ mol}}\text{2}\text{.35 }\times \text{ 10}^{\text{5}}\text{ g mol}^{-\text{1}} \nonumber$
Note: An average polyisobutylene molecule thus has a molecular weight of close to a quarter of a million! Such a molecule is made from units, each with a molar mass of 56.10 g mol-1.
An average polyisobutylene chain is thus $\frac{\text{2}\text{.35 }\times \text{ 10}^{\text{5}}\text{ g mol}^{-\text{1}}}{\text{56}\text{.10 g mol}^{-\text{1}}}=\text{4189} \nonumber$ units long.In other words the chain length is over 8000 carbon atoms in this sample of polymer.
1. Nelson, D.L., Cox, M.M. Lehninger Principles of Biochemistry(5th ed.). New York: W.H. Freeman and Company, 2008. pp. 52,389,405. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.26%3A_Osmotic_Pressure.txt |
Solutions are homogeneous. Dissolved molecules as large as 1000 pm never separate as a result of gravitational forces, even in an ultracentrifuge. When suspended particles reach µm size (106 pm), they separate readily under gravity, and we classify the mixture as definitely heterogeneous. Surprisingly, suspensions of particles between these sizes, in the range 5 000 - 200 000 pm, never settle under gravity or centrifugation, yet the mixtures are definitely heterogeneous because beams of light passing though them are visible. If a laser pointer beam passes through a solution, it is invisible, but if it passes through a colloidal suspension it is easily seen. Laser shows would be invisible if it weren't for the colloidal smoke or fog that renders the beams visible.
Colloids are stable suspensions of a dispersed medium, like the solid particles in smoke or fat particles in milk, in a continuous medium, like air or water. The particles and medium may be any combination of solid, liquid, and gas (except that they both can't be gases, because gasses are completely miscible), as the following table [1] shows:
For reference, the table is arranged so that the suspended particles are in the column while the medium the particles are suspended in are on the row. For example, solid foam's medium is solid (row) while the colloid particles are gaseous (column).
Medium Gas Liquid Solid
Gas None
(All gases are mutually miscible)
Liquid Aerosol
Examples: fog1, mist, hair spray
Solid Aerosol
Examples: smoke, cloud1
Liquid Foam
Example: Whipped cream
Emulsion
Examples: milk, mayonnaise, hand cream
Sol
Examples: pigmented ink, blood
Solid Solid Foam
Examples: aerogel, styrofoam, pumice
Gel
Examples: agar, gelatin, jelly, silica gel
Solid Sol
Example: red stained glass
1 Average water droplets in fogs and clouds are around 0.01 mm in diameter, so they are ten times as big as typical colloids. Clouds "float" mostly because of air updrafts, and the hardly noticeable (~0.3 cm/s) terminal falling velocity of small droplets[2].
Colloids are stable, that is, their particles don't aggregate and precipitate, because the particles all tend to adsorb ions of one charge from solution. The colloidal particles then repel one another electrostatically. The individual particles are kept in suspension by Brownian motion.
When white light passes through a colloidal suspension, light scattering is highly wavelength dependent, so colloids may appear to be brightly colored. This is the case with the gold colloid that is responsible for the deep red color in most stained glass. Colloidal gold is easily made in the laboratory by reducing a gold salt dissolved in water to give elemental gold clusters. As the clusters change in size, the color of the colloid may change through virtually every color of the rainbow.
Red sunsets are due to the selective reflectance of blue light by colloidal particles in the sky, as can be seen in a YouTube video where milk is added to a container of pure water while a beam of white light from a bright flashlight or projector is passed through the suspension and toward a white screen, resulting in the deep red of sunset.
References
1. http://en.Wikipedia.org/wiki/Colloid
2. www.suite101.com/article.cfm/...nce_sky/116567
3. http://en.Wikipedia.org/wiki/Sunset
4. The colloidal sulfur can also be prepared by dissolving sulfur in methanol and adding the solution dropwise to water: www.woodrow.org/teachers/ci/1986/exp14.html | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/10%3A_Solids_Liquids_and_Solutions/10.27%3A_Colloids.txt |
Water is by far the most important liquid solvent, partly because it is plentiful and partly because of its unique properties. In your body, in other living systems, and in the outside environment a tremendous number of reactions take place in aqueous solutions. Consequently this section, as well as significant portions of many subsequent sections, will be devoted to developing an understanding of reactions which occur in water. Since ionic compounds and polar covalent compounds constitute the main classes which are appreciably soluble in water, reactions in aqueous solutions usually involve these types of substances. There are three important classes of reactions which occur in aqueous solution: precipitation reactions, acid-base reactions, and redox reactions.
11: Reactions in Aqueous Solutions
In other sections we emphasized the importance of liquid solutions as a medium for chemical reactions. Water is by far the most important liquid solvent, partly because it is plentiful and partly because of its unique properties. In your body, in other living systems, and in the outside environment a tremendous number of reactions take place in aqueous solutions. Consequently this section, as well as significant portions of many subsequent sections, will be devoted to developing an understanding of reactions which occur in water. Since ionic compounds and polar covalent compounds constitute the main classes which are appreciably soluble in water, reactions in aqueous solutions usually involve these types of substances.
There are three important classes of reactions which occur in aqueous solution: precipitation reactions, acid-base reactions, and redox reactions.
• Precipitation reactions are useful for detecting the presence of various ions and for determining the concentrations of solutions.
• Acid-base reactions and redox reactions are similar in that something is being transferred from one species to another.
• Acid-base reactions involve proton transfers, while redox reactions involve electron transfers.
• Redox reactions are somewhat more complicated, though, because proton transfers and other bond-making and bond-breaking processes occur at the same time as electron transfer.
Below are demonstrations of each of the types of reactions we will be investigating throughout the course of this chapter.
The following video demonstrates precipitation reactions (carried out in an aqueous solution):
Next up is a demonstration of acids and bases reacting with Aluminum (in the form of a soda can):
Finally we have a video that shows a few demos of different redox reactions:
11.02: Ions in Solution (Electrolytes)
In Binary Ionic Compounds and Their Properties we point out that when an ionic compound dissolves in water, the positive and negative ions originally present in the crystal lattice persist in solution. Their ability to move nearly independently through the solution permits them to carry positive or negative electrical charges from one place to another. Hence the solution conducts an electrical current.
Electrolytes
Substances whose solutions conduct electricity are called electrolytes. All soluble ionic compounds are strong electrolytes. They conduct very well because they provide a plentiful supply of ions in solution. Some polar covalent compounds are also strong electrolytes. Common examples are HCl, HBr, HI and H2SO4, all of which react with H2O to form large concentrations of ions. A solution of HCl, for example, conducts even better than one of NaCl having the same concentration.
The effect of the concentration of ions on the electrical current flowing through a solution is illustrated in Figure $1$. Part a of the figure shows what happens when a battery is connected through an electrical meter to two inert metal strips (electrodes) dipping in ethanol. Each cubic decimeter of such a solution contains 0.10 mol NaCl (that is, 0.10 mol Na+ and 0.10 mol Cl). An electrical current is carried through the solution both by the Na+ ions moving toward the negative electrode and by the Cl- ions which are attracted toward the positive electrode. The dial on the meter indicates the quantity of current.
Figure 1b shows that if we replace the 0.10-M NaCl solution with a 0.05-M NaCl solution, the meter reading falls to about one-half its former value. Halving the concentration of NaCl halves the number of ions between the electrodes, and half as many ions can only carry half as much electrical charge. Therefore the current is half as great. Because it responds in such a direct way to the concentration of ions, conductivity of electrical current is a useful tool in the study of solutions.
Conductivity measurements reveal that most covalent compounds, if they dissolve in water at all, retain their original molecular structures. Neutral molecules cannot carry electrical charges through the solution, and so no current flows. A substance whose aqueous solution conducts no better than water itself is called a nonelectrolyte. Some examples are oxygen, O2, ethanol, C2H5OH, and sugar, C12H22O11.
Some covalent substances behave as weak electrolytes—their solutions allow only a small current flow, but it is greater than that of the pure solvent. An example is mercury(II) chloride (seen in the Figure above). For a 100-M HgCl2 solution the meter reading shows only about 0.2 percent as much current as for 0.10 M NaCl. A crystal of HgCl2 consists of discrete molecules, like those shown for HgBr2 in Figure $2$. When the solid dissolves, most of these molecules remain intact, but a few dissociate into ions according to the equation
$\underbrace{HgCl_2}_{99.8\%} \rightleftharpoons \underbrace{HgCl^+}_{0.2\%} + Cl^- \nonumber$
(The double arrows indicate that the ionization proceeds only to a limited extent and an equilibrium state is attained.) Since only 0.2 percent of the HgCl2 forms ions, the 0.10 M solution can conduct only about 0.2 percent as much current as 0.10 M NaCl.
Conductivity measurements can tell us more than whether a substance is a strong, a weak, or a nonelectrolyte. Consider, for instance, the data in Table $1$ which shows the electrical current conducted through various aqueous solutions under identical conditions. At the rather low concentration of 0.001 M, the strong electrolyte solutions conduct between 2500 and 10 000 times as much current as pure H2O and about 10 times as much as the weak electrolytes HC2H3O2 (acetic acid) and NH3 (ammonia).
Closer examination of the data for strong electrolytes reveals that some compounds which contain H or OH groups [such as HCl or Ba(OH)2] conduct unusually well. If these compounds are excluded, we find that 1:1 electrolytes (compounds which consist of equal numbers of +1 ions and –1 ions) usually conduct about half as much current as 2:2 electrolytes (+2 and -2 ions), 1:2 electrolytes (+1 and -2 ions), or 2:1 electrolytes (+2 and -1 ions).
TABLE $1$: Electrical Current Conducted Through Various 0.001 M Aqueous Solutions at 18°C.*
Substance Current /mA Substance Current /mA
Pure Water 1:2 Electrolytes
H2O 3.69 x 10-4 Na2SO4 2.134
Weak Electrolytes Na2CO3 2.24
HC2H3O2 0.41 K2CO3 2.660
NH3 0.28 2:1 Electrolytes
1:1 Electrolytes MgCl2 2.128
NaCl 1.065 CaCl2 2.239
NaI 1.069 SrCl2 2.290
KCl 1.273 BaCl2 2.312
KI 1,282 Ba(OH)2 4.14
AgNO3 1.131 2:2 Electrolytes
HCl 3.77 MgSO4 2.00
HNO3 3.75 CaSO4 2.086
NaOH 2.08 CuSO4 1.97
KOH 2.34 ZnSO4 1.97
* All measurements refer to a cell in which the distance between the electrodes is 1.0 mm and the area of each electrode is 1.0 cm². A potential difference of 1.0 V is applied to produce the tabulated currents.
There is a simple reason for this behavior. Under similar conditions, most ions move through water at comparable speeds. This means that ions like Mg2+ or SO42, which are doubly charged, will carry twice as much current through the solution as will singly charged ions like Na+ or Cl. Consequently, a 0.001 M solution of a 2:2 electrolyte like MgSO4 will conduct about twice as well as a 0.001 M solution of a 1:1 electrolyte like NaCl.
A similar argument applies to solutions of 1:2 and 2:1 electrolytes. A solution like 0.001 M Na2SO4 conducts about twice as well as 0.001 M NaCl partly because there are twice as many Na ions available to move when a battery is connected, but also because SO42 ions carry twice as much charge as Cl ions when moving at the same speed. These differences in conductivity between different types of strong electrolytes can sometimes be very useful in deciding what ions are actually present in a given electrolyte solution as the following example makes clear.
A second, slightly more subtle, conclusion can be drawn from the data in Table $1$. When an electrolyte dissolves, each type of ion makes an independent contribution to the current the solution conducts. This can be seen by comparing NaCl with KCl, and NaI with KI. In each case the compound containing K+ conducts about 0.2 mA more than the one containing Na+. If we apply this observation to Na2CO3 and K2CO3, each of which produces twice as many Na+ or K+ ions in solution, we find that the difference in current is also twice as great—about 0.4 mA.
Thus conductivity measurements confirm our statement that each ion exhibits its own characteristic properties in aqueous solutions, independent of the presence of other ions. One such characteristic property is the quantity of electrical current that a given concentration of a certain type of ion can carry.
Example $1$: Ions
At 18°C a 0.001-M aqueous solution of potassium hydrogen carbonate, KHCO3, conducts a current of 1.10 mA in a cell of the same design as that used to obtain the data in Table 11.1. What ions are present in solution?
Solution
Referring to Table 6.2 which lists possible polyatomic ions, we can arrive at three possibilities for the ions from which KHCO3 is made:
1. K+ and H+ and C4+ and three O2–
2. K+ and H+ and CO32
3. K+ and HCO3
Since the current conducted by the solution falls in the range of 1.0 to 1.3 mA characteristic of 1:1 electrolytes, possibility c is the only reasonable choice. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.01%3A_Prelude_to_Aqueous_Phase_Reactions.txt |
The independent behavior of each type of ion in solution was illustrated in Chemical Bonding by means of precipitation reactions. Precipitation is a process in which a solute separates from a supersaturated solution. In a chemical laboratory it usually refers to a solid crystallizing from a liquid solution, but in weather reports it applies to liquid or solid water separating from supersaturated air.
A typical precipitation reaction occurs when an aqueous solution of barium chloride is mixed with one containing sodium sulfate. The equation
$\ce{Ba^{2+}(aq) + Na2SO4(aq) -> BaSO4(s) + 2NaCl(aq)}\label{1}$
can be written to describe what happens, and such an equation is useful in making chemical calculations. However, Equation $\ref{1}$ does not really represent the microscopic particles (that is, the ions) present in the solution. Thus we might write
$\ce{Ba^{2+}(aq) + 2Cl^{-}(aq) + 2Na^{+}(aq) + SO4^{2-}(aq) -> BaSO4(s) + 2Na^{+} + 2Cl^{-}(aq)}\label{2}$
Equation $\ref{2}$ is rather cumbersome and includes so many different ions that it may be confusing. In any case, we are often interested in the independent behavior of ions, not the specific compound from which they came. A precipitate of BaSO4(s) will form when any solution containing Ba2+(aq) is mixed with any solution containing SO42(aq) (provided concentrations are not extremely small). This happens independently of the Cl(aq) and Na+(aq) ions in Eq. $\ref{2}$. These ions are called spectator ions because they do not participate in the reaction (see the figure above). When we want to emphasize the independent behavior of ions, a net ionic equation is written, omitting the spectator ions. For precipitation of BaSO4 the net ionic equation is
$\ce{Ba^{2+}(aq) + SO4^{2-}(aq) -> BaSO{4}(s)}\label{3}$
Example $1$: Precipitation Reaction
When a solution of AgNO3 is added to a solution of CaCl2, insoluble AgCl precipitates. Write three equations to describe this process.
Solution
Both AgNO3 and CaCl2 are soluble ionic compounds, and so they are strong electrolytes. The three equations are
$\ce{2AgNO3(aq) + CaCl2(aq) -> 2AgCl(s) + Ca(NO3)2(aq)}$
$\ce{2Ag^+(aq) + 2NO3^{-}(aq) + Ca^{2+}(aq) + Cl^{-}(aq) -> 2AgCl(s) + Ca^{2+}(aq) + 2NO3^{-}(aq)}$
$\ce{Ag^+(aq) + Cl^{-} (aq) -> AgCl(s)}$
The occurrence or nonoccurrence of precipitates can be used to detect the presence or absence of various species in solution. BaCl2 solution, for instance, is often used as a test for SO42(aq) ion. There are several insoluble salts of Ba, but they all dissolve in dilute acid except for BaSO4. Thus, if BaCl2 solution is added to an unknown solution which has previously been acidified, the occurrence of a white precipitate is proof of the presence of the SO42 ion. AgNO3 solution is often used in a similar way to test for halide ion.
If AgNO3 solution is added to an acidified unknown solution, a white precipitate indicates the presence of Cl ions, a cream-colored precipitate indicates the presence of Br ions, and a yellow precipitate indicates the presence of I ions. Further tests can then be made to see whether perhaps a mixture of these ions is present. When AgNO3 is added to tap water, a white precipitate is almost always formed. The Cl ions in tap water usually come from the Cl2 which is added to municipal water supplies to kill microorganisms.
Precipitates are also used for quantitative analysis of solutions, that is, to determine the amount of solute or the mass of solute in a given solution. For this purpose it is often convenient to use the first of the three types of equations described above. Then the rules of stoichiometry may be applied.
Example $2$: Concentration
When a solution of 0.1 M AgNO3is added to 50.0 cm3 of a CaCl2 solution of unknown concentration, 2.073 g AgCl precipitates. Calculate the concentration of the unknown solution.
Solution
We know the volume of the unknown solution, and so only the amount of solute is needed to calculate the concentration. This can be found using Eq. (2a) in Example 1. From the equation the stoichiometric ratio S(CaCl2/AgCl) may be obtained. A road map to the solution of the problem is
$m_{AgCl} \underset{\text{M_{AgCl}}}{\mathop{\rightarrow}}\, n_{AgCl} \underset{\text{M_{CaCL_2}}}{\mathop{\rightarrow}}\, n_{CaCl_2}$
$n_{CaCl-2} = 2.073 g AgCl * \frac{1mol AgCl}{143.32 g AgCl} * \frac{1 mol CaCl_2}{2 mol AgCl} = 7.23 * 10^{-3} \text{mol CaCl}_2$
$c_{CaCl_2} = \frac{n_{CaCl_2}}{V_{soln}} = \frac{7.23 * 10^{-3} mol CaCl_2}{50.0 cm^3} * \frac{10^3 cm^3}{1 dm^3} = 0.145 \frac{mol}{dm^3}$
Thus the concentration of the unknown solution is 0.145 M.
Because of the general utility of precipitates in chemistry, it is worth having at least a rough idea of which common classes of compounds can be precipitated from solution and which cannot. Table $1$ gives a list of rules which enable us to predict the solubility of the most commonly encountered substances. Use of this table is illustrated in the following example.
Table $1$: Solubility. Rules
Soluble in Water Important Exceptions (insoluble)
All Group IA and NH4+ salts CaSO4, BaSO4, SrSO4, PbSO4
All nitrates, chlorates, perchlorates and acetates AgX, Hg2X2, PbX2 (X= Cl, Br, or I)
All sulfates
All chlorides, bromides, and iodides
Slightly Soluble in Water Important Exceptions (soluble)
All carbonates and phosphates Group IA and NH4+ salts
All hydroxides Group IA and NH4+ salts; Ba2+, Sr2+, Ca2+ sparingly soluble
All sulfides Group IA, IIA and NH4+ salts; MgS, CaS, BaS sparingly soluble
All oxalates Group IA and NH4+ salts
The following electrolytes are of only moderate solubility in water: CH3COOAg, Ag2SO4, KClO4 They will precipitate only if rather concentrated solutions are used.
Example $3$: Net Ionic Equation
Write balanced net ionic equations to describe any reactions which occur when the following solutions are mixed:
1. 0.1 M Na2SO4 + 0.1 M NH4I
2. 0.1 M K2CO3 + 0.1 M SrCl2
3. 0.1 M FeSO4 + 0.1 M Ba(OH)2
Solution
a) If any precipitate forms, it will be either a combination of Na+ ions and I ions, namely, NaI, or a combination of ammonium ions, NH4+, and sulfate ions, SO42, namely, (NH4)2SO4. From Table 11.2 we find that NaI and (NH4)2SO4 are both soluble. Thus no precipitation reaction will occur, and there is no equation to write.
b) Possible precipitates are KCl and SrCO3. From Table 11.2 we find that SrCO3 is insoluble. Accordingly we write the net ionic equation as
$\ce{Sr^{2+}(aq) + CO3^{2-}(aq) -> SrCO3(s)}$
omitting the spectator ions K+ and Cl.
c) Possible precipitates are Fe(OH)2 and BaSO4. Both are insoluble. The net ionic equation is thus
$\ce{Fe^{2+}(aq) + SO4^{2-}(aq) + Ba^{2+}(aq) + 2OH^{-}(aq) -> Fe(OH)2(s) + BaSO4(s)}$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.03%3A_Precipitation_Reactions.txt |
We have already stated several times that solubility in water is a characteristic property of many ionic compounds, and Table 1 from Precipitation Reactions provides further confirmation of this fact. We have also presented experimental evidence that ions in solution are nearly independent of one another. This raises an important question, though, because we have also stated that attractive forces between oppositely charged ions in a crystal lattice are large. The high melting and boiling points of ionic compounds provide confirmation of the expected difficulty of separating oppositely charged ions. How, then, can ionic compounds dissolve at room temperature? Surely far more energy would be required for an ion to escape from the crystal lattice into solution than even the most energetic ions would possess.
The resolution of this apparent paradox lies in the interactions between ions and the molecules of water or other polar solvents. The negative (oxygen) side of a dipolar water molecule attracts and is attracted by any positive ion in solution. Because of this ion-dipole force, water molecules cluster around positive ions, as shown in Figure $1$ a. Similarly, the positive (hydrogen) ends of water molecules are attracted to negative ions. This process, in which either a positive or a negative ion attracts water molecules to its immediate vicinity, is called hydration.
When water molecules move closer to ions under the influence of their mutual attraction, there is a net lowering of the potential energy of the microscopic particles. This counteracts the increase in potential energy which occurs when ions are separated from a crystal lattice against their attractions for other ions.
Thus the process of dissolving an ionic solid may be divided into the two hypothetical steps shown in Figure $2$. First, the crystalline salt is separated into gaseous ions. The heat energy absorbed when the ions are separated this way is called the lattice enthalpy (or sometimes the lattice energy). Next, the separate ions are placed in solution; that is, water molecules are permitted to surround the ions. The enthalpy change for this process is called the hydration enthalpy.
Since there is a lowering of the potential energy of the ions and water molecules, heat energy is given off and hydration enthalpies are invariably negative. The heat energy absorbed when a solute dissolves (at a pressure of 1.00 atm) is called the enthalpy of solution. It can be calculated using Hess' law, provided the lattice enthalpy and hydration enthalpy are known.
Example $1$: Enthalpy
Using data given in Figure $2$, calculate the enthalpy of solution for NaCl(s).
Solution
According to the figure, the lattice enthalpy is 773 kJ mol–1. The hydration enthalpy is – 769 kJ mol–1. Thus we can write the thermo-chemical equations
$\text{NaCl}(s) \rightarrow \text{Na}^{+}(g) + \text{Cl}^{-}(g)$ $\triangle H_f = 773 \frac{kJ}{mol}$
$\underline{\text{Na}^{+}(g) + \text{Cl}^{-}(g) \rightarrow \text{Na}^{+}(aq) + \text{Cl}^-(aq)}$ $\triangle H_h = -769 \frac{kJ}{mol}$
$\text{NaCl}(s) \rightarrow \text{Na}^{+}(aq) + \text{Cl}^{-}(aq)$ $\triangle H_s=\triangle H_f + \triangle H_h$
$\triangle H_s = (773-769)\frac{kJ}{mol} = +4\frac{kJ}{mol}$
When NaCl(s) dissolves, 773 kJ is required to pull apart a mole of Na+ ions from a mole of Cl ions, but almost all of this requirement is provided by the 769 kJ released when the mole of Na+ and the mole of Cl becomes surrounded by water dipoles. Only 4 kJ of heat energy is absorbed from the surroundings when a mole of NaCl(s) dissolves. You can verify the small size of this enthalpy change by putting a few grains of salt on your moist tongue. The quantity of heat energy absorbed as the salt dissolves is so small that you will feel no cooling, even though your tongue is quite a sensitive indicator of temperature changes.
Few molecules are both small enough and polar to cluster around positive and negative ions in solution as water does. Consequently water is one of the few liquids which readily dissolves many ionic solids. Hydration of Na+, Cl and other ions in aqueous solution prevents them from attracting each other into a crystal lattice and precipitating. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.04%3A_Hydration_of_Ions.txt |
If you refer to Table 1 from Ions in Solution (Electrolytes), you will see that pure water does conduct some electrical current, albeit much less than-even the weak electrolytes listed there. This is because water itself is a very weak electrolyte. It ionizes to hydrogen ions and hydroxide ions to an extremely small extent:
$\text{H}_{2}\text{O}(l) \rightleftharpoons \text{H}^{+}(aq) + \text{OH}^{-}(aq)\label{1}$
Careful measurements show that at 25°C the concentrations of H+(aq) and OH(aq) are each 1.005 × 107 mol dm–3. At higher temperatures more H+(aq) and OH(aq) are produced while at lower temperatures less ionization of water occurs. Nevertheless, in pure water the concentration of H+(aq) always equals the concentration of OH(aq). Dissolving acids or bases in water can change the concentrations of both H+(aq) and OH(aq), causing them to differ from one another. The special case of a solution in which these two concentrations remain equal is called a neutral solution.
A hydrogen ion, H+, is a hydrogen atom which has lost its single electron; that is, a hydrogen ion is just a proton. Because a proton is only about one ten-thousandth as big as an average atom or ion, water dipoles can approach very close to a hydrogen ion in solution. Consequently the proton can exert a very strong attractive force on a lone pair of electrons in a water molecule—strong enough to form a coordinate covalent bond:
The H3O+ formed in this way is called a hydronium ion (on the left in the figure below). All three of its O―H bonds are exactly the same, and the ion has a pyramidal structure as predicted by VSEPR theory (1a). To emphasize the fact that a proton cannot exist by itself in aqueous solution, Eq. (1a) is often rewritten as
$\text{2H}_{2}\text{O}(l) \rightleftharpoons \text{H}_{3}\text{O}^{+}(aq) + \text{OH}^{-}(aq) \nonumber$
Like other ions in aqueous solution, both hydronium and hydroxide ions are hydrated. Moreover, hydrogen bonds are involved in attracting water molecules to hydronium and hydroxide ions. In both cases three water molecules appear to be rather tightly held, giving formulas H3O(H2O)3+ (or H9O4+) and HO(H2O)3 (or H7O4). Possible structures for the hydrated hydronium and hydroxide ions are shown in Figure $1$.
Hydrogen bonding of hydronium and hydroxide ions to water molecules accounts rather nicely for the unusually large electrical currents observed for some electrolytes containing H and OH. The case of the hydronium ion is illustrated in Figure $2$. When a hydronium ion collides with one end of a hydrogen-bonded chain of water molecules, a different hydronium ion can be released at the other end. Only a slight movement of six protons and a rearrangement of covalent and hydrogen bonds is needed. In effect a hydronium ion can almost instantaneously “jump” the length of several water molecules. It need not elbow its way through a crowd as other ions must. The same is true of aqueous hydroxide ions. Since both ions move faster, they can transfer more electrical charge per unit time, that is, more current.
11.06: Acid-Base Reactions
Early in the history of chemistry it was noted that aqueous solutions of a number of substances behaved very similarly, although the substances themselves did not at first seem to be related. Solutions were classified as acids if they had the following characteristics: sour taste; ability to dissolve metals such as Zn, Mg, or Fe; ability to release a gas from solid limestone (CaCO3) or other carbonates; ability to change the color of certain dyes (litmus paper turns red in the presence of acid). Another group of substances called bases or alkalies can also be distinguished by the properties of their aqueous solutions.
These are bitter taste, slippery or soapy feel, and the ability to change the color of certain dyes (litmus paper turns blue in base). Most important of all, acids and bases appear to be opposites. Any acid can counteract or neutralize the properties of a base. Similarly any base can neutralize an acid.
The video below demonstrates one of the ways acids and bases are determined (indicators - often colorful as seen below) and shows an example of a neutralization reaction.
$\text{HCl}(aq) + \text{NaOH}(aq) \rightleftharpoons \text{Na}^{+}(aq) + \text{Cl }^{-}(aq) + \text{H}_{2}\text{O}(l) \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.05%3A_Hydrogen_and_Hydroxide_Ions.txt |
A typical example of an acid is hydrogen chloride gas, HCl(g). When it dissolves in water, HCl reacts to form hydronium ions and chloride ions:
$\text{HCl}(g) + \text{H}_{2}\text{O}(l) \rightarrow \text{H}_{3}\text{O}^{+}(aq) + \text{Cl}^{-}(aq) \nonumber$
Thus the concentration of hydronium ions is increased above the value of 1.00 × 10–7 mol/L characteristic of pure water. Other acids, such as nitric acid, HNO3 behave in the same way:
$\text{HNO}_{3}(l) + \text{H}_{2}\text{O}(l) \rightarrow \text{H}_{3}\text{O}^{+}(aq) + \text{NO}_{3}^{-}(aq) \nonumber$
Thus the characteristic properties of solutions of acids are due to the presence of hydronium ions (or hydrogen ions). Whenever the concentration of hydronium ions exceeds 1.00 × 10–7 mol/L, an aqueous solution is said to be acidic. In 1884 a Swedish chemist, Svante Arrhenius (1859 to 1927), first recognized the importance of hydrogen ions. He defined an acid as any substance which increases the concentration of hydrogen (or hydronium) ions in aqueous solution.
The formation of a hydronium ion involves transfer of a proton from an acid molecule to a water molecule (as seen in the submicroscopic images above). This process is immediate―there are no free protons in solution which have left an acid molecule but have not yet attached themselves to a water molecule. To put it another way, a proton transfer is like a quarterback hand-off as opposed to a forward pass in foot- ball. The proton is always under the control and influence of one molecule or another. In the case of HCl we can indicate the transfer as:
As the molecule collides with an H2O molecule, a hydrogen bond forms between the H and O atoms: Cl—H---OH2. When it begins to bounce away from the H2O molecule, the Cl atom loses control of the proton, leaving it attached to the O atom. The Cl atom retains control over both electrons which were in the H—Cl bond and thus ends up as a Cl ion. The H2O molecule ends up with an extra proton, becoming H3O+.
Example $1$ : Proton Transfer
Write balanced equations to describe the proton transfer which occurs when each of the following acids is dissolved in H2O:
1. HClO4 (perchloric acid) and
2. HBr (hydrogen bromide, or hydrobromic acid).
Solution:
Although a free proton is never actually produced in solution, it is often convenient to break the proton-transfer process into two hypothetical steps: (1) the loss of a proton by the acid, and (2) the gain of a proton by H2O.
a) When HClO4 loses a proton, H+, the valence electron originally associated with the H atom is left behind, producing a negative ion, ClO4. The proton can then be added to a water molecule in the second hypothetical step. Summing the two steps gives the overall proton transfer:
$\text{HClO}_{4} \rightarrow \cancel{\text{H}^{+}} + \text{ClO}_{4}^{-}$ step 1
$\cancel{\text{H}^{+}} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+}$ step 2
__________________________
$\text{HClO}_{4} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{ClO}_{4}^{-}$ overall
b) Proceeding as in part a, we have
$\text{HBr} \rightarrow \cancel{\text{H}^{+}} + \text{Br}^{-}$ step 1
$\cancel{\text{H}^{+}} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+}$ step 2
______________________
$\text{HBr} +\text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{Br}^{-}$ overall
With practice, you should be able to write overall proton transfers without having to write steps 1 and 2 every time.
Another point to note about proton transfers is that in any equation involving ions, the sum of the ionic charges on the left side must equal the sum of the ionic charges on the right. For example, the last overall equation in Example 1 has HBr and H2O on the left. Neither is an ion, and so the sum of the ionic charges is zero. On the right we have H3O+ and Br, which satisfy the rule because +1 + (–1) = 0. An equation which does not satisfy this rule of charge balance will involve creation or destruction of one or more electrons and therefore cannot be valid. For example, the equation
$\text{2HBr} \rightarrow \text{2H}^{+} +\text{Br}_{2} \nonumber$
cannot describe a valid proton transfer because the charges sum to zero on the left but +2 (because 2H+ ions) on the right. Careful examination reveals that there are 16 valence electrons (two octets in 2HBr) on the left but only 14 valence electrons (none in 2H+ and 14 in ) on the right. Two electrons have been destroyed—something which cannot happen. Therefore the equation must be incorrect.
Because hydronium ions can be formed by transferring protons to water molecules, it is convenient when dealing with aqueous solutions to define an acid as a proton donor. This definition was first proposed in 1923 by the Danish chemist Johannes Brönsted (1879 to 1947) and the English chemist Thomas Martin Lowry (1874 to 1936). It is called the Brönsted-Lowry definition of an acid, and we will use it for the majority of this site. The Brönsted-Lowry definition has certain advantages over Arrhenius’ idea of an acid as a producer of H3O+(aq). This is especially true when acid strengths are compared, a subject we shall come to a bit later. Consequently, when we speak of an acid, we will mean a proton donor, unless some qualification, such as Arrhenius acid, is used. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.07%3A_Acids.txt |
Bases have characteristics opposite those of acids, and bases can be neutralized by acids. Therefore it is logical, in the Brönsted-Lowry scheme, to define a base as a proton acceptor, that is, a species which can incorporate an extra proton into its molecular or ionic structure. For example, when barium oxide, BaO, dissolves in water, oxide ions accept protons from water molecules according to the equation
$\text{BaO} + \text{H}_{2}\text{O} \rightarrow \text{Ba}^{2+} + \text{2OH}^{-} \nonumber$ $\text{O}^{2-} + \text{H}_{2}\text{O} \rightarrow \text{OH}^{-} + \text{OH}^{-} \nonumber$
The added proton transforms the oxide ion, into a hydroxide ion. Removal of a proton from a water molecule leaves behind a hydroxide ion also, accounting for the OH on the right side of the equation. Since it can accept protons, barium oxide (or, more specifically, oxide ion) serves as a base. When a base is added to water, its molecules or ions accept protons from water molecules, producing hydroxide ions.
Thus the general properties of solutions of bases are due to the presence of hydroxide ions [OH(aq)]. Any aqueous solution which contains a concentration of hydroxide ions greater than the 1.00 × 10–7 mol/L characteristic of pure water is said to be basic. Unlike the hydronium ion, which forms very few solid compounds, hydroxide ions are often present in solid crystal lattices (like NaOH, seen below). Therefore it is possible to raise the hydroxide-ion concentration above 1.00 × 10–7 mol/L by dissolving compounds such as NaOH, KOH, or Ba(OH)2.
Hydroxide ions can accept protons from water molecules, but of course such a proton transfer has no net effect because the hydroxide ion itself becomes a water molecule:
$\text{HOH} + \text{OH}^{-} \rightarrow \text{HO}^{-} + \text{HOH} \tag{1}$ Nevertheless, the hydroxide ion fits the Brönsted-Lowry definition of a base as a proton acceptor.
Example $1$ : Proton Transfer Equation
Write a balanced equation to describe the proton transfer which occurs when the base sodium hydride, NaH, is added to water.
Solution NaH consists of Na+ and H ions. Since positive ions repel protons, the H ion is the only likely base. Again it may be useful to use two hypothetical steps: (l) donation of a proton by an H2O molecule, and (2) acceptance of a proton by the base. As in Example 11.6, we can then sum the steps
$\text{HOH} \rightarrow \cancel{\text{H}^{+}}\ + \text{OH}^{-}$ step 1 $\text{H}^{-} + \cancel{\text{H}{+}}\ \rightarrow \text{H}_{2}$ step 2
_______________
$\text{H}^{-} + \text{H}_{2}\text{O} \rightarrow \text{H}_{2} + \text{OH}^{-}$ overall
Note that adding a proton to H balances the excess electron of that ion, producing a neutral H2 molecule. Note also that charges balance in all equations.
11.09: Strong Acids and Bases
The most straight-forward examples involving acids and bases deal with strong acids and bases. Strong acids, like HCl or HNO3, are such good proton donors that none of their own molecules can remain in aqueous solution. All HCl molecules, for example, transfer their protons to H2O molecules, and so the solution contains only H3O+(aq) and Cl(aq) ions. Similarly, the ions of strong bases, like BaO or NaH, are such good proton acceptors that they cannot remain in aqueous solution. All O2– ions, for example, are converted to OH ions by accepting protons from H2O molecules, and the H2O molecules are also converted to OH. Therefore a solution of BaO contains only Ba2+(aq) and OH(aq) ions.
Table \(1\) lists molecules and ions which act as strong acids and bases in aqueous solution. In addition to those which react completely with H2O to form H3O+ and OH, any compound which itself contains these ions will serve as a strong acid or base. Note that the strength of an acid refers only to its ability to donate protons to H2O molecules and the strength of a base to its ability to accept protons from H2O molecules. The acidity or basicity of a solution, on the other hand, depends on the concentration as well as the strength of the dissolved acid or base.
TABLE \(1\) Species Which Are Strong Acids and Bases in Aqueous Solution.
Strong Acids Strong Bases
H3O+ (Only a few compounds like H3OCl
and H3ONO3 are known to contain hydronium ions.)
OH [Only LiOH, NaOH, KOH, RbOH,
CsOH, Ca(OH)2, Sr(OH)2 and Ba(OH)2 are sufficiently soluble to produce large concentrations of OH(aq).]
HCl, HBr, HI
HClO4
O2– (Li2O, Na2O, K2O, Rb2O, Cs2O, CaO,
SrO, and BaO are soluble
HNO3, H2SO4, HClO3 H, S2–, NH2, N3–, P3–
Notice that the cations of the strong bases are also soluble in water, as seen in the table from 11.2 on solubility rules. For example, all the Group 1 cations form strong bases and all Group 1 cations are soluble in water. Therefore, you can use your prior understanding of solubility rules to guide your understanding of strong bases. As a general rules, if a cation is soluble in water, it can form a strong base.
As a general rule, strong proton donors are molecules in which a hydrogen is attached to a rather electronegative atom, such as oxygen or a halogen. Considerable electron density is shifted away from hydrogen in such a molecule, making it possible for hydrogen ions to depart without taking along any electrons. The strong acids in Table \(1\) fit this rule nicely. They are either hydrogen halides (HCl, HBr, HI) or oxyacids (whose general formula is HnXOm).
Below are the resonance structures for oxoacids after they donate a proton. The reason for the strength of the following acids is the stability of the anion, which is shown by the number of resonance structures and the distribution of the negative charge amongst all of the oxygen atoms. The distribution of the negative charge can be visualized in the 3D structure, with red being representing negative charge and blue representing positive charge. The 3D structure represents the average of the resonance structures shown to the left.
Name Lewis Structure (Resonance) 3D Structure
Perchloric Acid
Nitric Acid
The Lewis structures indicate a proton bonded to oxygen in each of the oxyacids, hence their general name. Note that for a strong oxyacid the number of oxygens is always larger by two or more than the number of hydrogens. That is, in the general formula HnXOm, mn + 2.
The strength of a base depends on its ability to attract and hold a proton. Therefore bases often have negative charges, and they invariably have at least one lone pair of electrons which can form a coordinate covalent bond to a proton. The strong bases in Table 1 might be thought of as being derived from neutral molecules by successive removal of protons. For example, OH can be obtained by removing H+ from H2O, and O2– can be obtained by removing H+ from OH. When the strong bases are considered this way, it is not surprising that they are good proton acceptors. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.08%3A_Bases.txt |
Not all molecules which contain hydrogen are capable of donating protons. For example, methane (CH4) and other hydrocarbons show no acidic properties at all. Carbon is not highly electronegative, and so electron density is fairly evenly shared in a C―H bond, and the hydrogen atom is unlikely to depart without at least one electron. Even when it is bonded to highly electronegative atoms like oxygen or fluorine, a hydrogen atom is not always strongly acidic.
Example $1$ : Proton Transfer
Acetic acid has the projection formula
Write an equation for transfer of a proton from acetic acid to water.
Solution
By the electronegativity rule given above, only the hydrogen attached to oxygen should be acidic. The equation is
$\text{CH}_{3}\text{COOH} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{CH}_{3}\text{COO}^{-} \label{1}$
Note: To emphasize that only one hydrogen atom is acidic, the formula acetic acid is often written HC2H3O2. You may also find the formulas HAc or HOAc for acetic acid, where Ac or OAc represents the acetate ion, CH3COO.
The equation for the proton transfer between acetic acid and water is written with a double arrow because it occurs to only a limited extent. Like HgCl2, acetic acid is a weak electrolyte. According to Table 1 in Ions in Solution (Electrolytes), 0.001 M HC2H3O2 conducts slightly more than one-tenth as much current as the same concentration of the strong acids HCl or HNO3. Therefore we can conclude that at a given instant only a little over 10 percent of the acetic acid molecules have donated protons to water molecules according to Eq. $\ref{1}$. Nearly 90 percent are in molecular form as CH3COOH (or HC2H3O2) and make no contribution to the current. Because acetic acid is not a strong enough proton donor to be entirely converted to hydronium ions in aqueous solution, it is called a weak acid. A given concentration of a weak acid produces fewer hydronium ions per unit volume and therefore less acidity than the same concentration of a strong acid.
For those of you in laboratory, you can identify a weak acid using litmus paper. Below is an image of litmus paper after the addition of the weak acid acetic acid (CH3COOH). Notice the red hue, which indicates a weak acid. If HCl of the same molarity were tested, the color would be darker, more purple than red.
There are a large number of weak acids, but fortunately they fall into a few well-defined categories:
Carboxylic acids
These compounds have the general formula RCOOH. All react with water in the same way as acetic acid [Eq. $\ref{1}$]. The strength of carboxylic acids is dependent on the electronegative strength of the atoms in the "R" group. Consider the compounds F3COOH and H3COOH. Fluorine is the most electronegative element, while hydrogen is comparable to carbon in electronegativity. Thus, the fluorines pull electron density away from the carboxyl group. This removes electron density from the acidic oxygen-hydrogen bond, which weakens it. This weaker bond means that the hydrogen can be removed more easily, which creates a stronger acid. This concept can be applied to any R group. The more electronegative the R group, the stronger the carboxylic acid will be.
Weak oxyacids
These have the same general formula HnXOm as strong oxyacids, but the number of hydrogens is equal to or one less than the number of oxygens. For a weak oxyacid, in other words, mn + 1. Some examples are:
Some of the weak oxyacids, H2CO3 for example, are very unstable and cannot be separated in pure form from aqueous solution. Other molecules containing acidic hydrogen atoms
Hydrogen fluoride (HF) has a very strong bond and does not donate its proton as readily as other hydrogen halides. Other molecules in this category are hydrogen sulfide (H2S) and hydrogen cyanide (HCN). In the latter case, even though H is bonded to C, the electronegative N atom pulls some electron density away, and the HCN molecule is a very weak proton donor.
Hydrated cations
Cations, especially those of charge +3 or more or of the transition metals, are surrounded closely by four to six water molecules in aqueous solution. An example is Cr(H2O)63+, shown in Figure $1$. The positive charge of the metal ion pulls electron density away from the surrounding water molecules, weakening the hold of the oxygen atoms for the hydrogen atoms. The latter can consequently be more easily donated as protons:
$\text{Cr}(\text{H}_{2}\text{O})_{6}^{3+} + \text{H}_{2}\text{O} \rightleftharpoons \text{Cr}(\text{H}_{2}\text{O})_{5}\text{OH}^{2+} + \text{H}_{3}\text{O}^{+} \nonumber$
Ions having acidic protons
Certain other ions can donate protons. One example is the ammonium ion, NH4+:
$\text{NH}_{4}^{+} + \text{H}_{2}\text{O} \rightleftharpoons \text{NH}_{3} + \text{H}_{3}\text{O}^{+} \nonumber$
Anions formed when some acids donate protons can lose yet another H+. An example of this is the hydrogen sulfate ion formed when sulfuric acid donates a proton:
$\text{H}_{2}\text{SO}_{4} + \text{H}_{2}\text{O} \rightarrow \text{H}_{3}\text{O}^{+} + \text{HSO}_{4}^{-} \nonumber$
$\text{HSO}_{4}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{SO}_{4}^{2-} \nonumber$
Although sulfuric acid is strong, the negative charge on the hydrogen sulfate ion holds the proton tighter, and so the ion is a considerably weaker acid. Acids such as H2SO4, H2S, H2SO3, and H2CO3 are called diprotic because they can donate two protons. Phosphoric acid, H3PO4, is triprotic—it can donate three protons. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.10%3A_Weak_Acids.txt |
By analogy with weak acids, weak bases are not strong enough proton acceptors to react completely with water. A typical example is ammonia, which reacts only to a limited extent:
$\text{NH}_{3} + \text{H}_{2}\text{O} \rightleftharpoons \text{NH}_{4}^{+} + \text{OH}^{-} \nonumber$
As in the case of acetic acid, the current conducted by 0.001 M NH3 (Table 1 from Ions in Solution) is slightly above 10 percent that of 0.001 M NaOH or KOH, indicating that somewhat more than one-tenth of the NH3 molecules have been converted to NH4+ and OH ions.
To verify that NH3 is a weak base, a litmus test can be done. The NH3 is added to litmus paper, which indicates it's pH based on the color of the litmus paper. In the image below you can see the litmus paper turns a greenish blue when exposed to NH3 (pH 9-10). For comparison, a strong base of the same molarity like NaOH or KOH would have a darker blue coloring.
Weak bases fall into two main categories.
Ammonia and amines
Amines may be derived from ammonia by replacing one or more hydrogens with carbon chains. They react with water in the same way as ammonia. Trimethyl amine behaves as follows:
$(\text{CH}_{3})_{3}\text{N} + \text{H}_{2}\text{O} \rightleftharpoons (\text{CH}_{3})_{3}\text{NH}^{+} + \text{OH}^{-} \nonumber$ Anions of weak acids
The fact that the molecules of a weak acid are not entirely converted into hydronium ions and anions implies that those anions must have considerable affinity for protons. For example, the anion of acetic acid, acetate ion, can accept protons as follows:
$\text{C}_{2}\text{H}_{3}\text{O}_{2}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HC}_{2}\text{H}_{3}\text{O}_{2} + \text{OH}^{-} \nonumber$
Thus when a compound like sodium acetate,dissolves, some hydroxide ions are produced and the solution becomes slightly basic. (Sodium ions do not react with water at all, and so they have no effect on acidity or basicity.) Other examples of weak bases which are anions of weak acids are CN, CO32, PO43.
11.12: Amphiprotic Species
Molecules or ions which can either donate or accept a proton, depending on their circumstances, are called amphiprotic species. The most important amphiprotic species is water itself. When an acid donates a proton to water, the water molecule is a proton acceptor, and hence a base. Conversely, when a base reacts with water, a water molecule donates a proton, and hence acts as an acid.
Another important group of amphiprotic species is the amino acids. Each amino acid molecule contains an acidic carboxyl group and a basic amino group. In fact the amino acids usually exist in zwitterion (German for “double ion”) form, where the proton has transferred from the carboxyl to the amino group. In the case of glycine, for example, the zwitterion is
The zwitterion can donate one of the protons from the N, just as an NH4+ ion can donate a proton. On the other hand, its COO end can accept a proton, just as a CH3COO ion can. Other common amphiprotic species are HCO3, H2PO4, HPO42, and other anions derived from diprotic or triprotic acids.
Example $1$ : Equations
Write equations to show the amphiprotic behavior of (a) H2PO4 and (b) H2O.
Solution
To make an amphiprotic species behave as an acid requires a fairly good proton acceptor. Conversely, to make it behave as a base requires a proton donor.
a)
Acid: $\text{H}_2 \text{PO}_4^- + \text{OH}^- \rightarrow \text{HPO}_4^{2-} + \text{H}_2 \text{O}$
Base: $\text{H}_2 \text{PO}_4^- + \text{H}_3 \text{O}^+ \rightarrow \text{H}_3 \text{PO}_4 + \text{H}_2 \text{O}$
b)
Acid: $\text{H}_2 \text{O} + \text{S}^{2-} \rightarrow \text{OH}^- + \text{HS}^-$
Base: $\text{H}_2\text{O} + \text{H}_2\text{SO}_4 \rightarrow \text{H}_3\text{O}^+ + \text{HSO}_4^-$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.11%3A_Weak_Bases.txt |
Through examples found in the sections on acids and bases proton-transfer processes are broken into two hypothetical steps: (1) donation of a proton by an acid, and (2) acceptance of a proton by a base. (Water served as the base in the acid example and as the acid in the base example [amphiprotic]). The hypothetical steps are useful because they make it easy to see what species is left after an acid donated a proton and what species is formed when a base accepted a proton. We shall use hypothetical steps or half-equations in this section, but you should bear in mind that free protons never actually exist in aqueous solution.
Suppose we first consider a weak acid, the ammonium ion. When it donates a proton to any other species, we can write the half-equation:
$\text{NH}_4^+ \rightarrow \text{H}^+ +\text{NH}_3 \nonumber$
The submicroscopic representations below show the donation of the proton of ammonium. The removal of this proton results in NH3, which is easily seen at the submicroscopic level.
But NH3 is one of the compounds we know as a weak base. In other words, when it donates a proton, the weak acid NH4+ is transformed into a weak base NH3. Another example, this time starting with a weak base, is provided by fluoride ion:
$\text{F}^{-} + \text{H}^{+} \rightarrow \text{HF} \nonumber$
The submicroscopic representation above shows how the addition of a proton to fluoride converts a weak base (F- in green) into a weak acid (HF).
The situation just described for NH4+ and NH3 or for F and HF applies to all acids and bases. Whenever an acid donates a proton, the acid changes into a base, and whenever a base accepts a proton, an acid is formed. An acid and a base which differ only by the presence or absence of a proton are called a conjugate acid-base pair. Thus NH3 is called the conjugate base of NH4+, and NH4+ is the conjugate acid of NH3. Similarly, HF is the conjugate acid of F, and F the conjugate base of HF.
Example $1$ : Conjugate Pairs
What is the conjugate acid or the conjugate base of (a) HCl; (b) CH3NH2; (c) OH; (d) HCO3.
Solution:
1. HCl is a strong acid. When it donates a proton, a Cl ion is produced, and so Cl is the conjugate base.
2. CH3NH2 is an amine and therefore a weak base. Adding a proton gives CH3NH3+, its conjugate acid.
3. Adding a proton to the strong base OH gives H2O its conjugate acid.
4. Hydrogen carbonate ion, HCO3, is derived from a diprotic acid and is amphiprotic. Its conjugate acid is H2CO3, and its conjugate base is CO32.
The use of conjugate acid-base pairs allows us to make a very simple statement about relative strengths of acids and bases. The stronger an acid, the weaker its conjugate base, and, conversely, the stronger a base, the weaker its conjugate acid.
Table $1$ gives a list of some of the more important conjugate acid-base pairs in order of increasing strength of the base. This table enables us to see how readily a given acid will react with a given base. The reactions with most tendency to occur are between the strong acids in the top left-hand comer of the table and the strong bases in the bottom right-hand comer.
If a line is drawn from acid to base for such a reaction, it will have a downhill slope. By contrast, reactions with little or no tendency to occur (between the weak acids at the bottom left and the weak bases at the top right) correspond to a line from acid to base with an uphill slope. When the slope of the line is not far from horizontal, the conjugate pairs are not very different in strength, and the reaction goes only part way to completion. Thus, for example, if the acid HF is compared with the base CH3COO, we expect the reaction to go part way to completion since the line is barely downhill.
A strong acid like HCl donates its proton so readily that there is essentially no tendency for the conjugate base Cl to reaccept a proton. Consequently, Cl is a very weak base. A strong base like the H ion accepts a proton and holds it so firmly that there is no tendency for the conjugate acid H2 to donate a proton. Hence, H2 is a very weak acid.
Example $2$ : Balanced Equation
Write a balanced equation to describe the reaction which occurs when a solution of potassium hydrogen sulfate, KHSO4, is mixed with a solution of sodium bicarbonate, NaHCO3.
Solution
The Na+ ions and K+ ions have no acid-base properties and function purely as spectator ions. Therefore any reaction which occurs must be between the hydrogen sulfate ion, HSO4 and the hydrogen carbonate ion, HCO3. Both HSO4 and HCO3 are amphiprotic, and either could act as an acid or as a base. The reaction between them is thus either
$\text{HCO}_3^- + \text{HSO}_4^- \rightarrow \text{CO}_3^{2-} + \text{H}_2\text{SO}_4 \nonumber$
or
$\text{HSO}_4^- + \text{HCO}_3^- \rightarrow \text{SO}_4^{2-} + \text{H}_2\text{CO}_3 \nonumber$
Table $1$ tells us immediately that the second reaction is the correct one. A line drawn from HSO4 as an acid to HCO3 as a base is downhill. The first reaction cannot possibly occur to any extent since HCO3 is a very weak acid and HSO4 is a base whose strength is negligible
Example $3$ : Reaction Prediction
What reactions will occur when an excess of acetic acid is added to a solution of potassium phosphate, K3PO4?
Solution
The line joining CH3COOH to PO43 in Table $1$ is downhill, and so the reaction
$\text{CH}_3\text{COOH} + \text{PO}_4^{3-} \rightarrow \text{CH}_3\text{COO}^- \text{HPO}_4^{2-} \nonumber$
should occur. There is a further possibility because HPO42 is itself a base and might accept a second proton. The line from CH3COOH to HPO42 is also downhill, but just barely, and so the reaction
$\text{CH}_3\text{COOH} + \text{HPO}_4^{2-} \rightarrow \text{CH}_3\text{COO}^- + \text{H}_2\text{PO}_4^- \nonumber$
can occur, but it does not go to completion. Hence double arrows are used. Although H2PO4 is a base and might be protonated to yield phosphoric acid, H3PO4, a line drawn from CH3COOH to H2PO4 is uphill, and so this does not happen. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.13%3A_Conjugate_Acid-Base_Pairs.txt |
Many oxyacids are rather unstable and cannot be isolated in pure form. An example is carbonic acid, H2CO3, which decomposes to water and carbon dioxide:
$\text{H}_2 \text{CO}_3(aq) \rightleftharpoons \text{H}_2\text{O}(l) + \text{CO}_2(g) \nonumber$
This decomposition process is familiar to us, as it is responsible for the fizzy nature of soda. When the CO2 is dissolved in the soda, it becomes H2CO3, but when the soda is released from high pressure, the decomposition process occurs rapidly, forming bubbles of CO2 and water.
Since it can be made by removing H2O from H2CO3, CO2 is called the acid anhydride of H2CO3. (The term anhydride is derived from anhydrous, meaning “not containing water.”) Acid anhydrides are usually oxides of nonmetallic elements. Some common examples and their corresponding oxyacids are SO2—H2SO3; SO3—H2SO4; P4O10—H3PO4; N2O5—HNO3. Any of these anhydrides increases the hydronium-ion concentration when dissolved in water; for example,
$\text{P}_4\text{O}_{10}(s) + 6\text{H}_2\text{O}(l) \rightarrow 4 \text{H}_3\text{PO}_4(aq) \nonumber$
$\text{H}_3\text{PO}_4(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{H}_3\text{O}^+(aq) + \text{H}_2\text{PO}_4^-(aq) \nonumber$
In the Arrhenius sense, then, acid anhydrides are acids, but according to the Brönsted-Lowry definition, they are not acids because they contain no hydrogen.
In 1923, at the same time that the Brönsted-Lowry definition was proposed, G. N. Lewis suggested another definition which includes the acid anhydrides and a number of other substances as acids. According to the Lewis definition, an acid is any species which can accept a lone pair of electrons, and a base is any species which can donate a lone pair of electrons. An acid-base reaction in the Lewis sense involves formation of a coordinate covalent bond (where one atom provides both shared electrons).
The Lewis definition has little effect on the types of molecules we expect to be basic. All the Brönsted-Lowry bases, for example, NH3, O2–, H, contain at least one lone pair. Lewis’ idea does expand the number of acids, though. The proton is not the only species which can form a coordinate covalent bond with a lone pair. Cations of the transition metals, which are strongly hydrated, do the same thing:
\begin{align} & \underset{\text{Lewis acid}}{\text{Cr}^{\text{3+}}}\text{ + }\underset{\text{Lewis base}}{\text{6H}_{\text{2}}\text{O}}\text{ }\to \text{ Cr(H}_{\text{2}}\text{O)}_{\text{6}}^{\text{3+}} \ & \underset{\text{Lewis acid}}{\text{Cu}^{\text{2+}}}\text{ + }\underset{\text{Lewis base}}{\text{4NH}_{\text{3}}}\text{ }\to \text{ Cu(NH}_{\text{3}}\text{)}_{\text{4}}^{\text{2+}} \ \end{align} \nonumber
So can electron deficient compounds such as boron trifluoride:
Many Lewis acid-base reactions occur in media other than aqueous solution. The Brönsted-Lowry theory accounts for almost all aqueous acid-base chemistry. Therefore the Brönsted-Lowry concept is most often intended when the words acid or base are used. The Lewis definition is useful when discussing transition-metal ions, however, and is discussed again in the sections on Metals.
Example $1$ : Lewis Acids and Bases
Identify the Lewis acids and bases in the following list. Write an equation for the combination of each acid with the Lewis base H2O.(a) BeCl2(g); (b) CH3OH; (c) SO2; (d) CF4.
Solution
a) The Lewis diagram
shows that Be is electron deficient. Therefore BeCl2(g) is a Lewis acid. Because of the lone pairs on the Cl atoms, BeCl2 can also act as a Lewis base, but Cl is rather electronegative and reluctant to donate electrons, so the Lewis base strength of BeCl2 is less than the Lewis acid strength.
b) There are lone pairs on O in CH3OH, and so it can serve as a Lewis base.
c) The S atom in SO2 can accept an extra pair of electrons, and so SO2 is a Lewis acid. The O atoms have lone pairs but are only weakly basic for the same reason as the Cl atoms in part (a).
d) Although there are lone pairs on the F atoms, the high electronegativity of F prevents them from being donated to form coordinate covalent bonds. Consequently CF4 has essentially no Lewis-base character. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.14%3A_Lewis_Acids_and_Bases.txt |
In addition to precipitation and acid-base reactions, a third important class called oxidation-reduction reactions is often encountered in aqueous solutions. The terms reduction and oxidation are usually abbreviated to redox. Such a reaction corresponds to the transfer of electrons from one species to another.
A simple redox reaction occurs when copper metal is immersed in a solution of silver nitrate. The solution gradually acquires the blue color characteristic of the hydrated Cu2+ ion, while the copper becomes coated with glittering silver crystals. The reaction may be described by the net ionic Equation
$\ce{Cu(s) + 2Ag^+(aq) -> Cu^{2+}(aq) + Ag(s)}\label{1}$
We can regard this Equation as being made up from two hypothetical half-equations. In one, each copper atom loses 2 electrons:
$\ce{Cu -> Cu^{2+} + 2e^{-}}\label{2}$
while in the other, 2 electrons are acquired by 2 silver ions:
$\ce{2e^{-} + 2Ag^+ -> 2Ag}\label{3}$
If these two half-equations are added, the net result is Equation $\ref{1}$. In other words, the reaction of copper with silver ions, described by Equation $\ref{1}$, corresponds to the loss of electrons by the copper metal, as described by half-equation $\ref{2}$, and the gain of electrons by silver ions, as described by Equation $\ref{3}$.
A species like copper which donates electrons in a redox reaction is called a reducing agent, or reductant. (A mnemonic for remembering this is remember, electron donor = reducing agent.) When a reducing agent donates electrons to another species, it is said to reduce the species to which the electrons are donated. In Equation $\ref{1}$, for example, copper reduces the silver ion to silver. Consequently the half-equation
$\ce{2Ag^+ + 2e^{-} -> 2Ag} \nonumber$
is said to describe the reduction of silver ions to silver.
Species which accept electrons in a redox reaction are called oxidizing agents, or oxidants. In Equation $\ref{1}$ the silver ion, Ag+, is the oxidizing agent. When an oxidizing agent accepts electrons from another species, it is said to oxidize that species, and the process of electron removal is called oxidation. The half-equation
$\ce{Cu -> Cu^{2+} + 2e^{-}} \nonumber$
thus describes the oxidation of copper to Cu2+ ion.
In summary, then, when a redox reaction occurs and electrons are transferred, there is always a reducing agent donating electrons and an oxidizing agent to receive them. The reducing agent, because it loses electrons, is said to be oxidized. The oxidizing agent, because it gains electrons, is said to be reduced.
Copper is also oxidized by the oxygen present in air. The following video shows an example of this oxidation occurring.
Example $1$ : half-equations
Write the following reaction in the form of half-equations. Identify each half-equation as an oxidation or a reduction. Also identify the oxidizing agent and the reducing agent in the overall reaction
$\ce{Zn + 2Fe^{3+} -> Zn^{2+} +2Fe^{2+}} \nonumber$
Solution
The half-equations are
$\ce{Zn -> Zn^{2+} + 2e^{-}}$ oxidation—loss of electrons
$\ce{2e^{-} + 2Fe^{3+} -> 2Fe^{2+}}$ reduction—gain of electrons
Since zinc metal (Zn) has donated electrons, we can identify it as the reducing agent. Conversely, since iron(III) ion (Fe3+) has accepted electrons, we identify it as the oxidizing agent. An alternative method of identification is to note that since zinc has been oxidized, the oxidizing agent must have been the other reactant, namely, iron(III). Also, since the iron(III) ion has been reduced, the zinc must be the reducing agent. Observe also that both the oxidizing and reducing agents are the reactants and therefore appear on the left-hand side of an Equation.
A more complex redox reaction occurs when copper dissolves in nitric acid. The acid attacks the metal vigorously, and large quantities of the red-brown gas, nitrogen dioxide (NO2) are evolved. (NO2 is poisonous, and so this reaction should be done in a hood.) The solution acquires the blue color characteristic of the hydrated Cu2+ ion. The reaction which occurs is
$\ce{Cu(s) + 2NO3^{-}(aq) + 4H3O^+(aq) -> Cu^{2+}(aq) + 2 NO2(g) + 6H2O(l)}\label{7}$
Merely by inspecting this net ionic Equation, it is difficult to see that a transfer of electrons has occurred. Clearly the copper metal has lost electrons and been oxidized to Cu2+, but where have the donated electrons gone? The matter becomes somewhat clearer if we break up Equation $\ref{7}$ into half-equations. One must be
$\ce{Cu(s) -> Cu^{2+}(aq) +2e^{-}} \nonumber$
and the other is
$\ce{2e^{-} + 4H3O^+(aq) + 2NO3^{-}(aq) -> 2NO2(g) + 6H2O(l)}\label{9}$
You can verify that these are correct by summing them to obtain Equation $\ref{7}$.
The second half-equation shows that each NO3 ion has not only accepted an electron, but it has also accepted two protons. To further complicate matters, a nitrogen-oxygen bond has also been broken, producing a water molecule. With all this reshuffling of nuclei and electrons, it is difficult to say whether the two electrons donated by the copper ended up on an NO2 molecule or on an H2O molecule. Nevertheless, it is still meaningful to call this a redox reaction. Clearly, copper atoms have lost electrons, while a combination of hydronium ions and nitrate ions have accepted them. Accordingly, we can refer to the nitrate ion (or nitric acid, HNO3) as the oxidizing agent in the overall reaction. half-equation $\ref{9}$ is a reduction because electrons are accepted. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.15%3A_Redox_Reactions.txt |
Redox reactions may involve proton transfers and other bond-breaking and bond-making processes, as well as electron transfers, and therefore the equations involved are much more difficult to deal with than those describing acid-base reactions. In order to be able to recognize redox reactions, we need a method for keeping a careful account of all the electrons. This is done by assigning oxidation numbers to each atom before and after the reaction.
For example, in NO3 the nitrogen is assigned an oxidation number of +5 and each oxygen an oxidation number of –2. This arbitrary assignment corresponds to the nitrogen’s having lost its original five valence electrons to the electronegative oxygens. In NO2, on the other hand, the nitrogen has an oxidation number of + 4 and may be thought of as having one valence electron for itself, that is, one more electron than it had in NO3.
This arbitrarily assigned gain of one electron corresponds to reduction of the nitrogen atom on going from NO3 to NO2. As a general rule, reduction corresponds to a lowering of the oxidation number of some atom. Oxidation corresponds to increasing the oxidation number of some atom. Applying the oxidation number rules to the following equation, we have
Since the oxidation number of copper increased from 0 to +2, we say that copper was oxidized and lost two negatively charged electrons. The oxidation number of nitrogen went down from 5 to 4, and so the nitrogen (or nitrate ion) was reduced. Each nitrogen gained one electron, so 2e were needed for the 2 NO3. The nitrogen was reduced by electrons donated by copper, and so copper was the reducing agent. Copper was oxidized because its electrons were accepted by an oxidizing agent, nitrogen (or nitrate ion).
Although they are useful and necessary for recognizing redox reactions, oxidation numbers are a highly artificial device. The nitrogen atom in NO3 does not really have a +5 charge which can be reduced to +4 in NO2. Instead, there are covalent bonds and electron-pair sharing between nitrogen and oxygen in both species, and nitrogen has certainly not lost its valence electrons entirely to oxygen. Even though this may (and indeed should) make you suspicious of the validity of oxidation numbers, they are undoubtedly a useful tool for spotting electron-transfer processes. So long as they are used for that purpose only, and not taken to mean that atoms in covalent species actually have the large charges oxidation numbers often imply, their use is quite valid.
The general rules for oxidation numbers are seen below, taken from the following page in the Analytical Chemistry Core Textbook: Oxidation States
Determining Oxidation States
Counting the number of electrons transferred is an inefficient and time-consuming way of determining oxidation states.These rules provide a simpler method:
• The oxidation state of an uncombined element is zero. This applies regardless of the structure of the element: Xe, Cl2, S8, and large structures of carbon or silicon each have an oxidation state of zero.
• The sum of the oxidation states of all the atoms or ions in a neutral compound is zero.
• The sum of the oxidation states of all the atoms in an ion is equal to the charge on the ion.
• The more electronegative element in a substance is assigned a negative oxidation state. The less electronegative element is assigned a positive oxidation state. Remember that electronegativity is greatest at the top-right of the periodic table and decreases toward the bottom-left.
• Some elements almost always have the same oxidation states in their compounds:
Element Usual oxidation state Exceptions
Group 1 metals Always +1
Group 2 metals Always +2
Oxygen Usually -2 Peroxides and F2O (see below)
Hydrogen Usually +1 Metal hydrides (-1) (see below)
Fluorine Always -1
Chlorine usually -1 Compounds with O or F (see below)
Exceptions:
Hydrogen in the metal hydrides: Metal hydrides include compounds like sodium hydride, NaH. Here the hydrogen exists as a hydride ion, H-. The oxidation state of a simple ion like hydride is equal to the charge on the ion—in this case, -1.
Alternatively, the sum of the oxidation states in a neutral compound is zero. Because Group 1 metals always have an oxidation state of +1 in their compounds, it follows that the hydrogen must have an oxidation state of -1 (+1 -1 = 0).
Oxygen in peroxides: Peroxides include hydrogen peroxide, H2O2. This is an electrically neutral compound, so the sum of the oxidation states of the hydrogen and oxygen must be zero.
Because each hydrogen has an oxidation state of +1, each oxygen must have an oxidation state of -1 to balance it.
Oxygen in F2O: The deviation here stems from the fact that oxygen is less electronegative than fluorine; the fluorine takes priority with an oxidation state of -1. Because the compound is neutral, the oxygen has an oxidation state of +2.
Chlorine in compounds with fluorine or oxygen: Because chlorine adopts such a wide variety of oxidation states in these compounds, it is safer to simply remember that its oxidation state is not -1, and work the correct state out using fluorine or oxygen as a reference.
Example \(1\) : Redox Reactions
Identify the redox reactions and the reducing and oxidizing agents from the following:
1. \(\ce{2MnO4^{–} + 5SO2 + 6H2O -> 5SO4^{2–} + 2Mn^{2+} + 4H3O^{+}}\)
2. \(\ce{NH4^+ + PO4^{3–} -> NH3 + PO4^{2–}}\)
3. \(\ce{HClO + H2S -> H3O^+ + Cl^{–} + S}\)
Solution:
a) The appropriate oxidation numbers are
The only atoms which change are Mn, from +7 to +2, a reduction, and S, from +4 to +6, an oxidation. The reaction is a redox process. SO2 has been oxidized by MnO4, and so MnO4is the oxidizing agent. MnO4 has been reduced by SO2, and so SO2 is the reducing agent.
b) The oxidation numbers
show that no redox has occurred. This is an acid-base reaction because a proton, but no electrons, has been transferred.
c)
H2S has been oxidized, losing two electrons to form elemental S. Since H2S donates electrons, it is the reducing agent. HClO accepts these electrons and is reduced to Cl. Since it accepts electrons, HClO is the oxidizing agent. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.16%3A_Oxidation_Numbers_and_Redox_Reactions.txt |
Some redox equations may be balanced using the methods developed in Balancing Chemical Equations, but most are rather difficult to handle. Therefore it is useful to have some rules, albeit somewhat arbitrary ones, to help find appropriate coefficients. These rules depend on whether the reaction occurs in acidic or basic solution. In either situation we must make sure that the number of electrons accepted by the oxidizing agent exactly equals the number of electrons donated by the reducing agent.
In acid solution
We shall apply the rules to the equation
$\overset{\text{+5 }-\text{2}}{\mathop{\text{IO}_{\text{3}}^{-}}}\,\text{ + }\overset{\text{+4 }-\text{2}}{\mathop{\text{SO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{O}}}\,\text{ }\to \text{ }\overset{\text{0}}{\mathop{\text{I}_{\text{2}}}}\,\text{ + }\overset{\text{+6 }-\text{2}}{\mathop{\text{SO}_{\text{4}}^{\text{2}-}}}\,\text{ + }\overset{\text{+1 }-\text{2 }}{\mathop{\text{H}_{\text{3}}\text{O}^{\text{+}}}}\,\text{ (1)}$ The changes in oxidation numbers verify that it is a redox equation, and the presence of H3O+ indicates that it occurs in acidic solution. The rules are 1 Write unbalanced half-equations for the oxidation of the reducing agent and for the reduction of the oxidizing agent. Oxidation: $\ce{SO2 -> SO4^{2–}}$
Reduction: $\ce{IO^{3–} -> I2}$
2 Balance the element reduced or oxidized in each half-equation. Oxidation: $\ce{SO2 -> SO4^{2–}}$...S already balanced
Reduction: $\ce{2IO3^{–} -> I2}$
3 Balance oxygen atoms by adding water (solvent) molecules. Oxidation: $\ce{SO2 + 2H2O -> SO4^{2–}}$
Reduction: $\ce{2IO3^{–} -> I2 + 6H2O}$
4 Balance hydrogen atoms by adding hydrogen ions (available from the acidic solution). Oxidation: $\ce{SO2 + 2H2O -> SO4^{2–} + 2H^+}$
Reduction: $\ce{12H^+ + 2IO3^{–} -> I2 + 6H2O}$
5 Balance electrical charges by adding electrons. Oxidation: $\ce{SO2 + 2H2O -> SO4^{2–} + 2H^+ + 2e^{–}}$
(The total charge on the left side was 0, but on the right it was – 2 + 4 = +2. Therefore 2e were needed on the right.)
Reduction: $\ce{10e^{–} + 12H^+ + 2IO3^{–} -> I2 + 6H2O}$
(The total charge on the left was 12 - 2 = +10, but on the right it was 0. Therefore 10e were needed on the left.)
6 Use oxidation numbers to check that the number of electrons is correct.
Oxidation: The oxidation number of electrons increases from +4 to +6, corresponding to a loss of 2e.
Reduction: The oxidation number of I falls from +5 to 0, corresponding to a gain of 5e for each I. Since there are 2 I atoms, 10e must be added.
7 Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent. Since only 2 electrons are donated in the oxidation half-equation, while 10 are required by the reduction, the oxidation must occur 5 times for each reduction. That is, both sides of the oxidation half-equation must be multiplied by 5: Oxidation: $\ce{5SO2 + 10H2O -> 5SO4^{2–} + 20H^+ + \cancel{10e^–}}$
Reduction: $\cancel{10e^–} + \ce{12H^+ + 2IO3^{–} -> I2 + 6H2O}$
8 Sum the half-equations. The net equation which result is $\ce{5SO2 + 4H2O + 2IO3^{–} -> 5SO4^{2–} + 8H^+ + I2}$
Note that when the half-equations were summed, the number of electrons was the same on both sides, and so no free electrons (which could not exist in aqueous solution) appear in the final result. It also would be more accurate to write H3O+ instead of H+ for the hydronium ion. This can be done by adding 8H2O to both sides of the equation:
$\ce{5SO2 + 12H2O + 2IO3^{–} -> 5SO4^{2–} + 8 H3O^+ + I2}$
(On the right, the 8H2O molecules are protonated to 8H3O+. It is also a good idea at this point to check that all atoms, as well as the electrical charges, balance.
In basic solution
Potassium permanganate KMnO4, can be used to oxidize alcohols to carboxylic acids. An example is
$\overset{\text{+7 }-\text{2}}{\mathop{\text{2MnO}_{\text{4}}^{-}}}\,\text{ + }\overset{-\text{2 +1 }-\text{2 +1}}{\mathop{\text{CH}_{\text{3}}\text{OH}}}\,\text{ + }\to \text{ }\overset{\text{+4 }-\text{2}}{\mathop{\text{MnO}_{\text{2}}}}\,\text{ + }\overset{\text{+1 }-\text{2}}{\mathop{\text{H}_{\text{2}}\text{O}}}\,\text{ + }\overset{\text{+1 +2 }-\text{2 }-\text{2 }}{\mathop{\text{HCOO}^{-}}}\,\text{ }\overset{\text{+2}}{\mathop{\text{2Mn}^{\text{2+}}}}\,\text{ + }\overset{-\text{2 +1}}{\mathop{\text{OH}^{-}}}\,\text{ (2)}$
Since OH is produced, the reaction occurs in basic solution. It clearly involves redox. 1 Write unbalanced equations for the oxidation of the reducing agent and the reduction of the oxidizing agent (same as for acid solution). Oxidation: $\ce{CH3OH -> HCOO^{–}}$
Reduction: $\ce{MnO4^{–} -> MnO2}$
2 Balance the element reduced or oxidized in each half-equation (same as for acid solution). (Both C and Mn are already balanced.) 3 Balance oxygen atoms by adding hydroxide ions (available from the basic solution). Oxidation: $\ce{CH3OH + OH^{–} -> HCOO^{–}}$
Reduction: $\ce{MnO4^{–} -> MnO2 + 2OH^–}$
4 To the side of each half-equation which lacks hydrogen, add one water molecule for each hydrogen needed. Add an equal number of hydroxide ions to the opposite side.
Oxidation: $\ce{CH3OH + 5OH^{–} -> HCOO^{–} + 4H2O}$
(Four hydrogens were needed on the right, and so 4 water molecules were added on the right and 4 hydroxide ions on the left.)
Reduction: $\ce{MnO4^{–} + 2H2O -> MnO2 + 4OH^{–}}$
(Two hydrogens were needed on the left, and so 2 water molecules were added on the left and 2 hydroxide ions were added on the right. Note that the added hydroxide ions are to maintain the balance of oxygen atoms.) 5 Balance electrical charges by adding electrons (same as for acid solution).
Oxidation: $\ce{CH3OH + 5OH^{–} -> HCOO^{–} + 4H2O + 4e^{–}}$
(The total charge on the left was -5, but on the right it was -1, and so 4e were added on the right.)
Reduction: $\ce{MnO4^{–} + 2H2O + 3e^{–} -> MnO2 + 4OH^{–}}$
(The total charge on the left was -1, but on the right it was -4, and so 3e were added on the left.) 6 Use oxidation numbers to check that the number of electrons is correct (same as for acid solution).
Oxidation: C goes from -2 to +2, corresponding to a loss of 4e.
Reduction: Mn goes from +7 to +4, corresponding to a gain of 3e.
7 Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent (same as for acid solution). Multiplying the oxidation half-equation by 3 and the reduction half-equation by 4 adjusts each so it involves 12e. Oxidation: $\ce{3CH3OH + 15OH^{–} -> 3HCOO^{–} + 12H2O + 12e^{–}}$
Reduction: $\ce{4MnO4^{–} + 8H2O + 12e^{–} -> 4MnO2 + 16OH^{–}}$
8 Sum the half-equations (same as for acid solution). The net equation which results is $\ce{3CH3OH +4MnO4^{–} -> 3HCOO^{–} + 4MnO2 + OH^{–}}$ Again, it is worthwhile to check that all atoms and charges balance. The rules for balancing redox equations involve adding H+, H2O, and OH to one side or the other of the half-equations. Since these species are present in the solution, they may participate as reactants or products, but usually there is no experiment which can tell whether they do participate. However, the balanced equation derived from our rules does indicate just what role H+, H2O, or OH play in a given redox process.
Reference
The steps for balancing a redox reaction in an acidic or basic solution are summarized below for reference.
In Acidic Solution
1. Write unbalanced half-equations for the oxidation of the reducing agent and for the reduction of the oxidizing agent.
2. Balance the element reduced or oxidized in each half-equation.
3. Balance oxygen atoms by adding water (solvent) molecules.
4. Balance hydrogen atoms by adding hydrogen ions (available from the acidic solution).
5. Balance electrical charges by adding electrons.
6. Use oxidation numbers to check that the number of electrons is correct.
7. Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent.
8. Sum the half-equations.
In Basic Solution
1. Write unbalanced equations for the oxidation of the reducing agent and the reduction of the oxidizing agent (same as for acid solution).
2. Balance the element reduced or oxidized in each half-equation (same as for acid solution).
3. Balance oxygen atoms by adding hydroxide ions (available from the basic solution).
4. To the side of each half-equation which lacks hydrogen, add one water molecule for each hydrogen needed. Add an equal number of hydroxide ions to the opposite side.
5. Balance electrical charges by adding electrons (same as for acid solution).
6. Use oxidation numbers to check that the number of electrons is correct (same as for acid solution).
7. Adjust both half-equations so that the number of electrons donated by the reducing agent equals the number of electrons accepted by the oxidizing agent (same as for acid solution).
8. Sum the half-equations (same as for acid solution). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.17%3A_Balancing_Redox_Equations.txt |
Oxidizing agents must be able to accept electrons readily. Elements with high electronegativity readily accept electrons, as can molecules or ions which contain relatively electronegative elements and even some metals which have high oxidation numbers. Bear these general rules in mind as we examine examples of common oxidizing agents in the following paragraphs.
Oxidizing Agents
Halogens (group VllA elements)
All four elemental halogens, F2, Cl2, Br2, and I2, are able to accept electrons according to the half-equation
$\ce{X2 + 2e^{–} -> 2X^{–}} \nonumber$
with $X = F, Cl, Br, I$
As we might expect from the periodic variation of electronegativity, the oxidizing power of the halogens decreases in the order F2 > Cl2 > Br2 > I2. Fluorine is such a strong oxidizing agent that it can react with water (water is very difficult to oxidize):
$\ce{2F2 + 6H2O -> 4H3O^+ + 4F^{–} + O2} \nonumber$
Chlorine also reacts with water, but only in the presence of sunlight. Bromine is weaker, and iodine has only mild oxidizing power.
Oxygen
Oxygen gas, which constitutes about 20 percent of the earth’s atmosphere, is another electronegative element which is a good oxidizing agent. It is very slightly weaker than chlorine, but considerably stronger than bromine. Because the atmosphere contains such a strong oxidant, few substances occur in reduced form at the earth’s surface. An oxidized form of silicon, SiO2, is one of the most plentiful constituents of the crust of the earth. Most metals, too, occur as oxides and must be reduced before they can be obtained in elemental form. When iron rusts, it forms the red-brown oxide Fe2O3xH2O, seen below, which always contains an indeterminate amount of water.
Iron Oxide (Fe2O3•xH2O)
Oxyanions and oxyacids
In aqueous solution NO3, IO3, MnO4, Cr2O72, and a number of other oxyanions serve as convenient, strong oxidizing agents. The structure of the last oxyanion mentioned above is shown in Figure 1. The most strongly oxidizing oxyanions often contain an element in its highest possible oxidation state, that is, with an oxidation number equal to the periodic group number. For example, NO3 contains nitrogen in a +5 oxidation state, Cr2O72 (seen below) contains chromium +6, and has manganese +7. The oxidizing power of the dichromate ion is employed in laboratory cleaning solution, a solution of Na2Cr2O7 in concentrated H2SO4. This readily oxidizes the organic compounds in grease to carbon dioxide. It is also highly corrosive, eats holes in clothing, and must be handled with care. Dark purple permanganate ion is another very common oxidizing agent (seen below). In basic solution it is reduced to solid dark brown MnO2. In acidic solution, however, it forms almost colorless Mn2+(aq).
11.19: Common Reducing Agents
A good reducing agent must be able to donate electrons readily, meaning it must not have a high electronegativity. Among the elements, low electronegativity is characteristic of good reducing agents. Molecules and ions which contain relatively electropositive elements which have low oxidation numbers are also good reducing agents. Bear these general rules in mind as we examine examples of common reducing agents in the following paragraphs.
Reducing Agents
Metals
All metals have low ionization energies and are relatively electropositive, and so they lose electrons fairly easily. Therefore, most metals are good reducing agents. Metals on the left of the periodic table exhibit this property to the greatest extent, and some of them, such as Li or Na, can even reduce H2O:
$\ce{2Li(s) + 2H2O(l) -> 2Li^+(aq) + 2OH^{–}(aq) + H2(g)} \nonumber$
Skip to the 1 minute mark to get directly to the reaction.
Other metals, such as Fe or Zn, cannot reduce H2O but can reduce hydronium ions, and so they dissolve in acid solution:
$\ce{Zn(s) + 2H3O^+(aq) -> Zn^{2+}(aq) + 2H2O(l) + H2(g)} \nonumber$
This is one of the characteristic reactions of acids. There are a few metals that will not dissolve in just any acid but instead require an acid like HNO3 whose anion is a good oxidizing agent. Cu and Hg are examples: $\ce{3Hg(s) + 8 H3O^+(aq) + 2NO3^{–}(aq) -> 3Hg^{2+}(aq) + NO(g) + 12H2O} \nonumber$ Finally, a few metals, such as Au and Pt, are such poor reducing agents that even an oxidizing acid like HNO3 will not dissolve them. This is the origin of the phrase “the acid test.” If a sample of an unknown yellow metal can be dissolved in acid, then the metal is not gold. Kings who collected tax payments in gold kept a supply of HNO3 available to make sure they were not being cheated. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.18%3A_Common_Oxidizing_Agents.txt |
In the section on acids and bases, we saw that some substances can act as both an acid and a base (amphiprotic). In the world of redox chemistry there exist substances that can act as both a reducing agent and oxidizing and a couple of examples are given below.
Water
We have seen that some oxidizing agents, such as fluorine, can oxidize water to oxygen. There are also some reducing agents, such as lithium, which can reduce water to hydrogen. In terms of redox, water behaves much as it did in acid-base reactions, where we found it to be amphiprotic. In the presence of a strong electron donor (strong reducing agent), water serves as an oxidizing agent. In the presence of a strong electron acceptor (strong oxidizing agent), water serves as a reducing agent. Water is rather weak as an oxidizing or as a reducing agent, however; so there are not many substances which reduce or oxidize it. Thus it makes a good solvent for redox reactions. This also parallels water’s acid-base behavior, since it is also a very weak acid and a very weak base.
Hydrogen peroxide ($H_2O_2$)
In this molecule the oxidation number for oxygen is –1. This is halfway between O2(0) and H2O(–2), and so hydrogen peroxide can either be reduced or oxidized. When it is reduced, it acts as an oxidizing agent:
$\ce{H2O2 + 2H^+ + 2e^{–} -> 2H2O} \nonumber$
When it is oxidized, it serves as a reducing agent:
$\ce{H2O2 -> O2 + 2H^+ + 2e^{–}} \nonumber$
Hydrogen peroxide is considerably stronger as an oxidizing agent than as a reducing agent, especially in acidic solutions.
11.21: Redox Couples
When a reducing agent donates one or more electrons, its oxidation number goes up, and the resulting species is capable of reaccepting the electrons. That is, the oxidized species is an oxidizing agent. For example, when copper metal dissolves, the copper(II) ion formed can serve as an oxidizing agent:
Similarly, when an oxidizing agent such as silver ion is reduced, the silver metal can donate an electron, serving as a reducing agent:
This is analogous to what we observe in the case of conjugate acids and bases. For every oxidizing agent, there corresponds some reducing agent, and for every reducing agent, there corresponds an oxidizing agent.
TABLE $1$: Selected Redox Couples Arranged in Order of Decreasing Strength of Oxidizing Agent.
An oxidizing and reducing agent which appear on opposite sides of a half-equation constitute a redox couple. Redox couples are analogous to conjugate acid-base pairs and behave in much the same way. The stronger an oxidizing agent, the weaker the corresponding reducing agent, and the stronger a reducing agent, the weaker the corresponding oxidizing agent. Thus the strong oxidizing agent F2 produces the weak reducing agent F. Conversely, the strong reducing agent Li corresponds with the weak oxidizing agent Li+.
This type of relationship is demonstrated in Table $1$, where selected redox couples have been arranged in order of increasing strength of the reducing agent. As in the case of acids and bases, this table of half-equations can be used to predict which way a redox reaction will proceed. The reactions with the greatest tendency to occur are between strong oxidizing agents from the upper left and strong reducing agents from the lower right. If a line is drawn from oxidizing agent to reducing agent for such a reaction, it will have a downhill slope. Reactions with little tendency to occur involve the weak oxidizing agents at the lower left and weak reducing agents at the upper right. In such cases a line from oxidizing to reducing agent has an uphill slope. When the slope of the line is not far from horizontal, the oxidizing and reducing agents are similar in strength and their reaction will go only part way to completion.
Example $1$: Oxidation
Predict whether iron(III) ion, Fe3+, will oxidize copper metal. If
so, write a balanced equation for the reaction.
Solution
A line from the oxidizing agent Fe3+ to the reducing agent Cu is downhill, and so the reaction will occur. The balanced equation is the sum of the two half-equations adjusted to equalize the number of electrons transferred.
$\ce{2Fe^{3+} + 2e^{–} -> 2Fe^{2+}}$
$\underline{\text{Cu} \rightarrow \text{Cu}^{2+} + \cancel{2e^–}}$
$\ce{2Fe^{3+} + Cu -> 2Fe^{2+} + Cu^{2+}}$
Note that the half-equation for Cu is reversed from that in Table 1 because the reactant Cu is a reducing agent. All half-equations in Table $1$ have the oxidizing agent on the left.
Example $2$: Oxidizing Agent
Use Table 11.15 to find a reagent that will oxidize H2O (other than F2 or Cl2, which were mentioned earlier).
Solution: We must find an oxidizing agent from which a line to the reducing agent H2O has a downhill slope. The only possibilities are MnO4 or CrO72. The latter reacts extremely slowly, but aqueous permanganate solutions decompose over a period of weeks. The concentration of MnO4 decreases because of the reaction
$\ce{4MnO4^{–} + 12H3O^+ -> 4Mn^{2+} + 5O2 + 18H2O}$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/11%3A_Reactions_in_Aqueous_Solutions/11.20%3A_Substances_Which_Are_Both_Oxidizing_and_Reducing_Agents.txt |
We will describe some of the chemistry of the representative elements, showing as we do so how their properties may be rationalized on the basis of concepts and principles. Explanation of these properties is organized according to the concept of periodicity, with each subsequent section corresponding to one of the eight groups of representative elements. As you read, you should try to see why a certain reaction occurs, as well as what actually happens, and instead of memorizing each specific equation, you should try to organize the chemistry of these elements according to the generalizations you have already learned. At this point it is useful to look and consolidate some of the general trends observed for the representative elements.
12: Chemistry of the Representative Elements
So far we have been devoted to explaining fundamental concepts and principles such as the atomic theory, electronic structure and chemical bonding, intermolecular forces and their effects on solids, liquids, or gases, and classes of reactions such as redox or acid-base. It is well to remember, though, that all these concepts and principles have been developed and used by chemists in order to better understand, recall, and systematize macroscopic laboratory observations. In other words, although the concepts we have described so far have their own inherent beauty as great ideas, they are primarily important because they reduce the quantity of memorization which is necessary to master descriptive chemistry, allowing people to recall facts that otherwise might be forgotten.
Such a reduction in memory work is only possible if you know how to apply principles to specific elements, their compounds, and the reactions they undergo. This is not as easy as it may seem, but neither is it impossibly difficult. We will describe some of the chemistry of the representative elements, showing as we do so how their properties may be rationalized on the basis of concepts and principles. Explanation of these properties is organized according to the concept of periodicity, with each subsequent section corresponding to one of the eight groups of representative elements. As you read, you should try to see why a certain reaction occurs, as well as what actually happens, and instead of memorizing each specific equation, you should try to organize the chemistry of these elements according to the generalizations you have already learned.
At this point it is useful to look and consolidate some of the general trends observed for the representative elements. First, metals on the far left of the periodic table are good reducing agents, while nonmetals on the far right (excluding noble gases) are strong oxidizing agents. Thus these elements are quite reactive, especially when one from the left combines with one from the right. Hydrogen compounds (hydrides) of the alkali and alkaline-earth metals contain strongly basic H ions and produce basic solutions. Toward the middle of the periodic table acid-base properties of hydrogen compounds are harder to predict. Some, like CH4 in Group IVA, are neither acids nor bases, but others, like NH3, have lone pairs of electrons and can accept protons. Protons can be easily donated and are acidic only when they are bound to halogens or oxygen.
The acidic behavior of oxides also increases from left to right across the periodic table and decreases from top to bottom. The situation is complicated by the fact that the higher the oxidation state of an atom, the more covalent its oxide and the more acidic it will be. Thus SO3 dissolves in water to give a strong acid, while SO2 gives a weak one. Taking account of both of these trends, one can fairly well predict which oxides are likely to be basic, which amphoteric, and which acidic.
General rules can also be used to predict which oxidation states will be most common. On the left of the periodic table the group number gives the most common oxidation state. From group IIIA on, the group number minus 2 (for the ns2 electrons) is also common, especially for elements near the bottom of the table. .The group number is a good choice when an element combines with a highly electronegative element, but the group number minus 2 is more common when one element is bonded to another element of intermediate electron-withdrawing power. For example from the Chalcogens, SF4 and SF6 are both stable, but SF4 is the most stable sulfur chloride.
From group VA on to the right of the table, the group number minus 8 is an important oxidation state, especially for the first member of a group. Oxidation numbers other than those already mentioned usually differ by increments of 2. For example, chlorine exhibits –1, +1, +3, +5 and +7 oxidation states in stable compounds, and sulfur is found in –2, +2, +4, and +6 states.
In conclusion, do not overlook the forest by concentrating too much on individual trees. Look for and try to understand and use the generalizations and correlations that have been developed in this chapter. If you do this, you will retain the facts presented here much more efficiently. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.01%3A_Prelude_to_Descriptive_Chemistry.txt |
Li, Na, K, Rb, and Cs are all group IA elements, also known as the alkali metals. The seventh member of the group, francium (Fr) is radioactive and so rare that only 20 atoms of Fr may exist on Earth at any given moment[1]. The term alkali is derived from an Arabic word meaning “ashes.” Compounds of potassium as well as other alkali metals were obtained from wood ashes by early chemists. All the alkali metals are soft and, except for Cs which is yellow, are silvery-gray in color.
Lithium, sodium, potassium, rubidium, and cesium have a great many other properties in common. All are solids at 0°C and melt below 200°C. Each has metallic properties such as good conduction of heat and electricity, malleability (the ability to be hammered into sheets), and ductility (the ability to be drawn into wires). The high thermal (heat) conductivity and the relatively low melting point (for a metal) of sodium make it an ideal heat-transfer fluid. It is used to cool certain types of nuclear reactors (liquid-metal fast breeder reactors, LMFBRs) and to cool the valves of high-powered automobile engines for this reason.
Some general properties of the alkali metals are summarized in the table below. All these metal atoms contain a singles electron outside a noble-gas configuration, and so the valence electron is-well shielded from nuclear charge and the atomic radii are relatively large. The large volume of each atom results in a low density—small enough that Li, Na, and K float on water as they react with it.
Table $1$ Properties of the Group IA Alkali Metals
Element Symbol Electron Configuration Usual Oxidation State Atomic Radius/pm Ionic (M+) Radius/pm
Lithium Li [He]2s1 +1 122 60
Sodium Na [Ne]3s1 +1 157 95
Potassium K [Ar]4s1 +1 202 133
Rubidium Rb [Kr]5s1 +1 216 148
Cesium Cs [Xe]6s1 +1 235 169
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electronegativity Melting Point (in °C)
First Second
Li 0.526 7.305 0.534 1.0 179
Na 0.502 4.569 0.97 0.9 98
K 0.425 3.058 0.86 0.8 64
Rb 0.409 2.638 1.52 0.8 39
Cs 0.382 2.430 1.87 0.7 28
The atoms do not have a strong attraction for the single valence electron, and so it is easily lost (small first ionization energy) to from a +1 ion. Because they readily donate electrons in this way, all the alkali metals are strong reducing agents. They are quite reactive, even reducing water.
Weak attraction for the valence electron also results in weak metallic bonding, because it is attraction among nuclei and numerous valence electrons that holds metal atoms together. Weak metallic bonding results in low melting points, especially for the larger atoms toward the bottom of the group. Cs, for example, melts just above room temperature. Weak metallic bonding also accounts for the fact that all these metals are rather soft.
That the chemistry of alkali metals is confined to the +1 oxidation state is confirmed by the large second-ionization energies. Removing the first electron from the large, diffuses orbital is easy, but removing a second electron from an octet in an M+ ion is much too difficult for any oxidizing agent to do.
Two other elements are found in group IA. Hydrogen, although many of its compounds have formulas similar to the alkali metals, is a nonmetal and is almost unique in its chemical behavior. Therefore it is not usually included in this group. Francium (Fr) is quite radioactive, and only small quantities are available for study; so it too is usually omitted. Its properties, however, appear to be similar to those of Cs and the other alkali metals.
Chemical Reactions and Compounds
The element lithium combines violently and spectacularly with water. Hydrogen gas is given off, which propels the the lithium metal across the water as it reacts. If the excess water is evaporated, the compound lithium hydroxide (LiOH) remains behind. LiOH is visualized by phenolphthalein indicator, which turns pink as LiOH, a base, is produced. Thus the equation for this reaction is
$\text{2Li}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2LiOH}(aq) + \text{H}_\text{2}(g)\nonumber$
The elements sodium, potassium, rubidium, and cesium also combine violently with water to form hydroxides. The equations for their reactions are
$\text{2Na}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2NaOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2K}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2KOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2Rb}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2RbOH}(aq) + \text{H}_\text{2} (g)\nonumber$
$\text{2Cs}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2CsOH}(aq) + \text{H}_\text{2} (g)\nonumber$
Since the alkali metals all react with water in the same way, a general equation may be written:
$\text{2M}(s) + \text{2H}_\text{2}\text{O}(l) \rightarrow \text{2MOH}(aq) + \text{H}_\text{2} (g)\nonumber$
with M = K, Li, Na, Rb, or Cs.
The symbol M represents any one of the five elements.
In addition to their behavior when added to water, the alkali metals react directly with many elements. All combine swiftly with oxygen in air to form white oxide:
$\text{4M}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
(Li2O is lithium oxide, Na2O is sodium oxide, etc.)
All except lithium react further to form yellow peroxides, M2O2:
$\text{2M}_2 \text{O}(s) + \text{O}_2(g) \rightarrow \text{2M}_2 \text{O}_2(s)\nonumber$ M = Na, K, Rb, or Cs
(Na2O2 is sodium peroxide, etc.)
Potassium, rubidium, and cesium are sufficiently reactive that yellow superoxides (whose general formula is MO2) can be formed:
$\text{2M}_2 \text{O}_2(s) + \text{O}_2(g) \rightarrow \text{2MO}_2(s)\nonumber$
with M = K, Rb, or Cs
Unless the surface of a sample of an alkali metal is scraped clean, it will appear white or gray instead of having a silvery metallic luster. This is due to the oxide, peroxide, or superoxide coating that forms after a few seconds of exposure to air. The following movie shows how a freshly cut piece of lithium is shiny, but dulls to gray when exposed to oxygen in the air. The video also focuses on another important property of alkali metals: they are soft, and easy to cut, compared to other metals.
A dull gray oxidized cylinder of lithium metal is cut, revealing a shiny silvery surface. After 1 minute, the surface has dulled, and after 10 minutes, the cut surface has returned to the dull gray of the rest of the lithium metal. Since the alkali metal is lithium, the only reaction with oxygen that occurs is:
$\text{4Li}(s) + \text{O}_2(g) \rightarrow \text{2Li}_2\text{O}(s)\nonumber$
The alkali also combine directly with hydrogen gas to form compounds known as hydrides, MH:
$\text{2M}(s) + \text{H}_2(g) \rightarrow \text{2MH}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
They react with sulfur to form sulfides, M2S:
$\text{2M}(s) + \text{S}(g) \rightarrow \text{M}_2\text{S}(s)\nonumber$
with M = Li, Na, K, Rb, or Cs
These oxides, hydrides, hydroxides, and sulfides all dissolve in water to give basic solutions, and these compounds are among the strong bases.
The peroxides and superoxides formed when the heavier alkali metals react with O2 also dissolve to give basic solutions:
$\text{2NaO}_2(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4Na}^{+}(aq) + \text{4OH}^{-}(aq) + \text{O}_2(g)\nonumber$
$\text{4K}_2\text{O}(s) + \text{2H}_2\text{O}(l) \rightarrow \text{4K}^{+}+ \text{4OH}^{-} + \text{3O}_2(g)\nonumber$
Both of the latter equations describe redox as well as acid-base processes, as you can confirm by assigning oxidation numbers. The peroxide and superoxide ions contain O atoms in the unusual (for O) –1 and –½ oxidation states:
Therefore disproportionation (simultaneous oxidation and reduction) of O22 or O2 to the more common oxidation states of 0 (in O2) and –2 (in OH) is possible.
The alkali metals also react directly with the halogens, for instance with chlorine, forming chlorides,
$\text{2M}(s) + \text{Cl}_2(g) \rightarrow \text{2MCl}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
Below is an example of the reaction of Na with Cl2
A piece of sodium metal is added to a flask containing chlorine gas. Initially no reaction takes place, but when a drop of water is added, sodium and chlorine react, violently flaring up and producing so much heat that sand is needed in the bottom of the flask to absorb the heat and prevent the glass from cracking. This equation for this reaction is:
$\text{2Na}(s) + \text{Cl}_2(g) \rightarrow \text{2NaCl}(s)\nonumber$
with fluorine to form fluorides, MF:
$\text{2M}(s) + \text{F}_2(g) \rightarrow \text{2MF}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
and with bromine to form bromides, MBr:
$\text{2M}(s) + \text{Br}_2(g) \rightarrow \text{2MBr}(s)\nonumber$ M = Li, Na, K, Rb, or Cs
Below is an example of K reacting with Br2
In this video, potassium, which is stored in inert mineral oil due to its high reactivity, is placed in a beaker of liquid bromine after the protective layer of mineral oil has been removed. The potassium reacts explosively with the bromine. The container is covered during the whole process to prevent reactants and products from entering the environment. The chemical equation for this reaction is:
$\text{2K}(s) + \text{Br}_2(g) \rightarrow \text{2KBr}(s)\nonumber$
Sodium and potassium are quite abundant, ranking sixth and seventh among all elements in the earth’s crust, but the other alkali metals are rare. Sodium and potassium ions are components of numerous silicate crystal lattices seen in the Earth's crust, but since most of their compounds are water soluble, they are also important constituents of seawater and underground deposits of brine. Sodium chloride obtained from such brines is the chief commercial source of sodium, while potassium can be obtained from the ores sylvite (KCl) or carnallite (KCl•MgCl2•6H2O).
Both sodium (Na+) and potassium (K+) ions are essential to living systems. Na+ is the main cation in fluids surrounding the cells, while K+is most important inside the cells. Na+ plays a role in muscle contraction, and both K+ and Na+ play a role in transmitting nerve impulses. K is more important than Na in plants, and it is one of three elements (K, P, N) which must be supplied in fertilizer to maintain high crop yields. K is especially abundant in trees—wood ashes from kitchen fires (potash) were the major source of this element as recently as a century ago, and they still make good fertilizer for your garden. Wood ashes contain a mixture of potassium oxide and potassium carbonate, the latter formed by combination of K2O with CO2 produced when C in the wood combines with O2:
$\text{K}_2\text{O} + \text{CO}_2 \rightarrow \text{K}_2\text{CO}_3\nonumber$
Na compounds are obtained commercially from brine or from seawater. When an electrical current is passed through an NaCl solution (a process called electrolysis), Cl2(g), H2(g), and a concentrated solution of NaOH (caustic soda or lye) are obtained:
$\text{Na}^{+}(aq) + \text{2Cl}^{-}(aq) + \text{2H}_2\text{O}(l) \xrightarrow{\text{electrolysis}} \text{Cl}_2(g) + \text{H}_2(g) + \text{Na}^{+}(aq) + \text{2OH}^{-}(aq)\nonumber$
This process is described in more detail in the section on electrochemical cells, but you can see from the equation that the electrical current oxidizes Cl to Cl2 and reduces H2O to H2. NaOH(aq) is used as a strong base in numerous industrial processes to make soap, rayon, cellophane, paper, dyes, and many other products. Lye is also used in home drain cleaners. It must be handled with care because it is strongly basic, highly caustic, and can severely burn the skin. A second important industrial use of brine is the Solvay process:
$\text{CO}_2(g) + \text{NH}_3(aq) + \text{Na}^{+}(aq) + \text{Cl}^{-}(aq) + \text{H}_2\text{O}(l) \rightarrow \text{NaHCO}_3(s) + \text{NH}_4^{+}(aq) + \text{Cl}^{-}(aq)\nonumber$
The Solvay process is an acid-base reaction combined with a precipitation. The acid anhydride, CO2, reacts with H2O to produce H2CO3. This weak acid donates a proton to NH3, yielding NH4+ and HCO3, and the latter ion precipitates with Na+. The weakly basic sodium hydrogen carbonate produced by the Solvay process can be purified for use as an antacid (bicarbonate of soda), but most of it is converted to sodium carbonate (soda ash) by heating:
$\text{2NaHCO}_3(s) \xrightarrow{\Delta } \text{Na}_2\text{CO}_3(s) + \text{H}_2\text{O}(g) + \text{CO}_2(g)\nonumber$
(The Δ in this equation indicates heating of the reactant.) Sodium carbonate (Na2CO3) is used in manufacturing glass and paper, and in some detergents. The carbonate ion is a rather strong base, however, and detergents containing Na2CO3 (washing soda) have resulted in severe chemical burns to some small children who, out of curiosity, have eaten them. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.02%3A_Group_IA_-_Alkali_Metals.txt |
Group IIA also known as the alkaline-earth metals, include beryllium, magnesium, calcium, strontium, barium, and radium. The last member of the group, Ra, is radioactive and will not be considered here. All alkaline earths are silvery-gray metals which are ductile and relatively soft. However, the following table shows that they are much denser than the group IA metals, and their melting points are significantly higher. They are also harder than the alkali metals. This may be attributed to the general valence electron configuration ns2 for the alkaline earths, which involves two electrons per metal atom in metallic bonding (instead of just one as in an alkali metal).
Table $1$: Properties of the Group IIA Alkaline-Earth Metals
Element Symbol Electron Configuration Usual Oxidation State< Radius/pm
Atomic Ionic (M2+)
Barium Ba [Xe]6s2 +2 198 135
Beryllium Be [He]2s2 +2 89 31
Calcium Ca [Ar]4s2 +2 174 99
Magnesium Mg [Ne]3s2 +2 136 65
Strontium Sr [Kr]5s2 +2 191 113
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electro negativity Melting Point (in °C)
First Second Third
Be 0.906 1.763 14.86 1.86 1.5 1278
Mg 0.744 1.467 7.739 1.74 1.2 651
Ca 0.596 1.152 4.918 1.54 1.0 839
Sr 0.556 1.071 4.21 2.60 1.0 769
Ba 0.509 0.972 3.43 3.51 0.9 725
First and second ionization energies for the alkaline earths (corresponding to removal of the first and second valence electrons) are relatively small, but the disruption of an octet by removal of a third electron is far more difficult. Like the alkali metals, the alkaline-earth atoms lose electrons easily, and so they are good reducing agents. Other trends among the data in the table are what we would expect. Ionization energies and electronegativities decrease from top to bottom of the group, and atomic and ionic radii increase. The radii of +2 alkaline-earth ions are much smaller than the +1 alkali-metal ions of the same period, because the greater nuclear charge holds the inner shells more tightly. This effect is sufficiently large that an alkaline earth below and to the right of a given alkali metal in the periodic table often has nearly the same ionic radius. Thus Na+ (95 pm), can fit into exactly the same type of crystal lattice as Ca2+ (99 pm), and these two elements are often found in the same minerals. The same is true of K+ and Ba2+. Below is the table for alkali metals, to compare with the table of alkaline earth metals.
Table $2$: Properties of the Group IA Alkali Metals
Element Symbol Electron Configuration Usual Oxidation State Radius/pm
Atomic Ionic (M+)
Lithium Li [He]2s1 +1 122 60
Sodium Na [Ne]3s1 +1 157 95
Potassium K [Ar]4s1 +1 202 133
Rubidium Rb [Kr]5s1 +1 216 148
Cesium Cs [Xe]6s1 +1 235 169
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electronegativity Melting Point (in °C)
First Second
Li 0.526 7.305 0.534 1.0 179
Na 0.502 4.569 0.97 0.9 98
K 0.425 3.058 0.86 0.8 64
Rb 0.409 2.638 1.52 0.8 39
Cs 0.382 2.430 1.87 0.7 28
Similarity of ionic radii also leads to related properties for Li and Mg. Since these two elements are adjacent along a diagonal line from the upper left to the lower right in the periodic table, their similarity is called a diagonal relationship. Diagonal relationships are mainly evident in the second and third periods: Be is similar to Al, and B is like Si in many ways.
Farther toward the right-hand side of the table such relationships are less pronounced. The most striking similarity between Li and Mg is their ability to form covalent bonds with elements of average electronegativity, such as C, while forming fairly ionic compounds with more electronegative elements, such as O or F. Two examples of covalent compounds are ethyllithium, CH3CH2Li, and diethylmagnesium, (CH3CH2)2Mg. Such compounds are likely in the case of Li and Mg but not the alkali or alkaline earths below them, because Li+ and Mg2+ are small enough to be strongly polarizing and thus form bonds with considerable covalent character.
Chemical Reactions and Compounds
The alkaline earth metals react directly with most nonmetallic elements. forming Except for beryllium, the alkaline earths react directly with hydrogen gas to form hydrides, MH2; M = Mg, Ca, Sr, Ba, or Ra. Beryllium hydride, BeH2 can also be prepared, but not directly from the elements. Alkaline-earth metals combine readily with oxygen from the air to form oxides, MO. This follows the general reaction:
$\text{2M}(s) + \text{O}_2(g) \rightarrow \text{2MO}_2(s) \nonumber$ M = Be, Mg, Ca, Sr, Ba, or Ra
The following video shows the reaction of magnesium with oxygen:
In the video, magnesium is burned in air, and emits a bright white flame. A white powder of MgO remains after the reaction described by the equation:
$\text{2Mg}(s) + \text{O}_2(g) \rightarrow \text{2MgO}_2(s) \nonumber$
It should also be noted that while MgO is the main product, nitrogen is also present in the air, and so some magnesium nitride is also produced according to the chemical equation:
$\text{Mg}(s) + \text{N}_2(g) \rightarrow \text{Mg}_3\text{N}_2(s) \nonumber$
These oxides will coat the surface of the metal and prevent other substances from contacting and reacting with it. A good example of the effect of such an oxide coating is the reaction of alkaline-earth metals with water. Beryllium and magnesium react much more slowly than the others because their oxides are insoluble and prevent water from contacting the metal.
Alkaline-earth metals react directly with halogens to form salts:
$\text{M}(s) + \text{Cl}_2(g) \rightarrow \text{MCl}_2(s) \nonumber$ M = Be, Mg, Ca, Sr, Ba, or Ra
Salt obtained by evaporating seawater (sea salt) contains a good deal of magnesium chloride and calcium chloride as well as sodium chloride. It also has small traces of iodide salts, accounting for the absence of simple goiter in communities which obtain their salt from the oceans. Simple goiter is an enlargement of the thyroid gland caused by iodine deficiency.
Alkaline earths also form sulfides: MS. In all these compounds the alkaline-earth elements occur as dipositive ions, Mg2+, Ca2+, Sr2+, or Ba2+.
Similar compounds of Be can be formed by roundabout means, but not by direct combination of the elements. Moreover, the Be compounds are more covalent than ionic. The Be2+ ion has a very small radius (31 pm) and is therefore capable of distorting (polarizing) the electron cloud of an anion in its vicinity. Therefore all bonds involving Be have considerable covalent character, and the chemistry of Be is significantly different from that of the other members of group IIA.
As in the case of the alkali metals, the most important and abundant alkaline earths, Mg and Ca, are in the third and fourth periods. Be is rare, although its strength and low density make it useful in certain special alloys. Sr and Ba occur naturally as the relatively insoluble sulfates SrSO4 (strontianite) and BaSO4 (barite), but these two elements are of minor commercial importance.
The most common ores of Mg and Ca are dolomite, MgCO3•CaCO3, after which an entire mountain range in Italy is named, and limestone, CaCO3, an important building material. Mg is also recovered from seawater on a wide scale. The oxides of the alkaline earths are commonly obtained by heating the carbonates. For example, lime, CaO, is obtained from limestone as follows:
$\text{CaCO}_3(s) \xrightarrow{\Delta } \text{CaO}(s) + \text{CO}_2(g) \nonumber$
Except for BeO, which is covalently bonded, alkaline-earth oxides contain O2– ions and are strongly basic. When treated with water (a process known as slaking), they are converted to hydroxides:
$\text{CaO}(s) + \text{H}_2\text{O}(l) \rightarrow \text{Ca(OH)}_2(s) \nonumber$
Ca(OH)2 (slaked lime) is an important strong base for industrial applications, because it is cheaper than NaOH.
MgO has an extremely high melting point (2800°C) because of the close approach and large charges of its constituent Mg2+ and O2– ions in the crystal lattice. As a solid it is a good electrical insulator, and so it is used to surround metal-resistance heating wires in electric ranges. MgO is also used to line high-temperature furnaces. When converted to the hydroxide, Mg finds a different use. Mg(OH)2 is quite insoluble in water, and so it does not produce a high enough concentration of hydroxide ions to be caustic. It is basic, however, and gram for gram can neutralize nearly twice the quantity of acid that NaOH can. Consequently a suspension of Mg(OH)2 in water (milk of magnesia) makes an excellent antacid, for those who can stand its taste.
Because the carbonate ion behaves as a Brönstedt-Lowry base, carbonate salts dissolve in acidic solutions. In nature, water often becomes acidic because the acidic oxide CO2 is present in the atmosphere. When CO2 from the air dissolves in water, it can help dissolve limestone:
$\text{CO}_2(g) + \text{H}_2\text{O}(l) + \text{CaCO}_3(s) \rightleftharpoons \text{Ca}^{2+}(aq) + \text{HCO}_3^{-}(aq) \nonumber$
This reaction often occurs underground as rainwater saturated with CO2 seeps through a layer of limestone. Caves from which the limestone has been dissolved are often prevalent in areas where there are large deposits of CaCO3. In addition, the groundwater and well water in such areas becomes hard. Hard water contains appreciable concentrations of Ca2+, Mg2+ , and certain other metal ions. These form insoluble compounds with soap, causing curdy, scummy precipitates. Hard water can be softened by adding Na2CO3, washing soda, which precipitates CaCO3, or by ion exchange, a process in which the undesirable Ca2+ and Mg2+ ions are replaced in solution by Na+ ions, which do not precipitate soap. Most home water softeners work on the latter principle. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.03%3A_Group_II-_Alkaline_Earths.txt |
The elements of group IIIA show considerable variability in properties from top to bottom of the periodic table. B is a semimetal or metalloid, and the element has a covalent network structure in which icosahedrons of boron atoms (Figure $1$) are linked together. None of its compounds contain B3+ ions—covalent bonding is the rule there as well. Many B compounds, especially those containing O, are similar to those of Si (another diagonal relationship) instead of Al. In elemental Al the atoms are closest packed. Al is definitely metallic, but like Be, to which it is related diagonally, it forms many compounds with extensive covalent character.
The other three elements in the group, Ga, In, and Tl, are also metals, but their chemistry is affected to some extent by the fact that they follow the transition elements in the periodic system (Table $1$). The presence of a filled d subshell, as opposed to only s and p subshells in B and Al, introduces some new atomic properties.
Table $1$: Properties of the Group IIIA Elements
Element Symbol Electron Configuration Usual Oxidation State Radius/pm
Atomic Ionic (M3+)
Boron B [He]2s22p1 +3 80 -
Aluminum Al [Ne]3s23p1 +3 125 50
Gallium Ga [Ar]4s23d104p1 +3 124 62
Indium In [Kr]5s24d105p1 +3, +1 142 81
Thallium Tl [Xe]6s24f145d106p1 +1, +3 144 95
Some properties of the group IIIA elements are shown in Table $2$. In this case the trends are not quite what might have been expected on the basis of previous experience with the periodic table. Both ionization energy and electronegativity decrease significantly from B to Al, but as one proceeds on down the group, these atomic properties change very little. Indeed electronegativity increases from Ga to In to Tl. There is also a break in the expected steady increase in atomic radius down the group. Al and Ga have nearly identical radii, and the ionic radii of A13+ and Ga3+ differ by only 8 pm. In the case of Tl the most common oxidation state is +1,corresponding to loss of only the 6p valence electron, rather than an oxidation state of +3, which would entail loss of the 6s2 pair as well.
Table $2$: More Properties of the Group IIIA Elements
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electro- negativity Melting Point (in °C)
First Second Third
B 0.807 2.433 3.666 2.34 2.0 2300
Al 0.584 1.823 2.751 2.70 1.5 660
Ga 0.585 1.985 2.969 5.90 1.6 30
In 0.565 1.827 2.711 7.31 1.7 157
Tl 0.596 1.977 2.884 11.83 1.8 304
Many of these anomalies can be understood if we look more closely at the electron configurations of Ga, In, and Tl. In each case a d subshell has been filled between the ns2 pair and the final np1 valence electron. The 4f subshell also intervenes in the building-up process for Tl. Since d and f electrons are not as efficient as s or p electrons at screening nuclear charge, the valence electrons of Ga, In, and Tl are more strongly attracted, the atoms are smaller than would be expected, and ionization energies are unusually large.
Another factor affecting the chemistry of the group IIIA elements is the relative sizes of the first, second, and third ionization energies. These are roughly in the ratio 1:3:4.5 for all elements, and the large increase from first to second ionization energy becomes more pronounced toward the bottom of the group. In the case of Tl, whose large radius prevents close approach and strong bonding to other atoms, the energy required to unpair or ionize the 6s2 pair of electrons and use them to form bonds is often too large to be compensated by the bond energies of the two additional bonds. Consequently Tl+ is the more common oxidation state, rather than Tl3+. This reluctance of the 6s2 pair of electrons to be used in bonding is called the inert-pair effect. It also affects the chemistry of Hg, Pb, and Bi, elements which are adjacent to Tl in the sixth period.
Chemical Reactions and Compounds
Unlike groups IA and IIA, none of the group IIIA elements react directly with hydrogen to form hydrides. The halides of B, Al, and Ga will react with sodium hydride, however, to form tetrahydro anions:
$\text{4NaH} + \text{BF}_3 \rightarrow \text{NaBH}_4 + \text{3NaF} \nonumber$
The compounds NaBH4, NaAlH4, and NaGaH4 do not contain H ions. Instead the hydrogens are covalently bonded to the group IIIA atom:
All these anions are tetrahedral, as would be predicted by VSEPR theory.
The tetrahydroaluminate and tetrahydrogallate ions react readily with H2O, splitting the H2O molecule so that the H ends up with another H to form H2, while the OH ends up with the group IIIA atom:
$\text{AlH}_4^{-}(aq) + \text{4HOH}(l) \rightarrow \text{Al(OH)}_3(s) + \text{OH}^{-}(aq) + \text{4H}_2(g) \nonumber$
Splitting of an H2O molecule (hydrolysis) by these compounds is similar to the hydrolysis of esters. The hydrolysis of AlH4 is violently explosive, making it necessary to handle NaAlH4 in a dry environment. The anion in NaBH4 by contrast, involves bonds with greater covalent character and is much less readily hydrolyzed.
Only B forms a wide variety of hydride compounds. The simplest of these is diborane, B2H6, a volatile, readily hydrolyzed compound which may decompose explosively. Diborane is made as follows:
$\ce{3NaBH_{4} + 4BF_{3} \rightarrow 3NaBF_{2} + 2B_{2}H_{6}} \nonumber$
Diborane is electron deficient—if you try to draw a Lewis diagram, you will soon find it to be impossible—and for a number of years theoretical chemists were mystified about what held the atoms together.
The current picture of bonding in diborane is shown in the Figure $2$ and the accompanying Jmol Applet. Each B is assumed to be surrounded by four sp3 hybrid orbitals. Four of the eight sp3 hybrids from the two B’s overlap the 1s orbitals of four H’s, forming four B—H bonds, all of which lie in a plane. The other four sp3 hybrids overlap between the B’s, forming two banana bonds, a form we described when discussing the double bonds of ethene. Each of the remaining two H nuclei is embedded in the middle of one of the banana bonds. In this way the two electrons in each banana bond hold three nuclei (two B’s and one H) together, and the bond is called a three-center two-electron bond.
Example$1$: Orbitals
Below is a 3D Jmol model of diborane. We've added a molecular electrostatic potential surface function, as well as molecular orbitals. Let us look at the MEP options first.
1. Before looking at any of the surface options, set the minimum cutoff, at the bottom of the MEP options to -0.0100 and the maximum cutoff to 0.0100. Then click "set this cutoff on surface' to apply it. Look at the "MEP on isopotential surface." This will give a good sense of the shape of the potential distribution. You may also want to look at "MEP on Van der Waals Surface". There are two informative "MEP on a plane" surfaces to look at. Click "XY" and "Set Plane Equation" to see the electrostatic potential in the plane of the hydrogen atoms not in the banana bonds. Click "XZ" and "Set Plane Equation" to see the electrostatic potential in the plane containing the banana bond. What do these electrostatic visualizations tell you about diborane?
2. While each of the molecular orbitals is informative, for our purposes, look at N5. What does the shape of this orbital mean for the nature of the banana bond?
Solution
a) All three show that the boron atoms are surrounded by negative charge, the free hydrogen atoms are neutral, and the two hydrogen atoms in the banana bonds are electron deficient. This picture makes sense in terms of our discussion on diborane structure. First, the two banana bond hydrogen atoms are sharing an electron pair not just with one boron atom, but with two. As you can see in the dot density diagram, the electron density for this bond is pulled away from the hydrogen atoms. The negative charge on boron arises from their being four, instead of three, bonds to the atom.
b)This orbital spans the region between the hydrogen atom and both borons. Since orbitals contain only 2 electrons, this means that there is a strong bonding presence of only two electrons shared between 3 atoms (rather than 2 individual bonds of 2 electrons each).
All the group IIIA elements form trihalides, although Tl does so reluctantly, preferring to remain in the +1 oxidation state. The general reaction is
$\text{2M}(s) + \text{3X}_2 \rightarrow \text{2MX}_3 \nonumber$ M = B, Al, Ga, In, Tl X = F, Cl, Br, I
Below is a video example of one such reaction, that of Al and Br2:
In this video, aluminum foil is cut into smaller squares, and placed in a watch glass with liquid bromine, also giving off gaseous bromine vapor fumes. Initially, no reaction occurs, and the aluminum foil has a protective oxide coating. Soon, white fumes start to come off, signaling the start of the reaction. Since the reaction is exothermic, more bromine fumes also emerge. The aluminum and bromine react violently, causing flashes of light. After the reaction has occurred, a fan is used to remove the fumes, so that the AlBr3 product may be seen. The overall reaction is:
$\text{2Al}(s) + \text{3Br}_2 \rightarrow \text{2AlBr}_3 \nonumber$
All the boron halides consist of discrete BX3 molecules, which are electron deficient. Below is a 3D model of one of these halides,BCl3. As predicted by VSEPR theory the molecule has a trigonal planar structure. On this model, Molecular Electrostatic Potential Surface options have been provided. First, set the minimum cutoff to -0.0100 and the maximum cutoff to 0.0100. Again, any of the three surface potential options are useful, but in this case, "MEP on Isopotential Surface" is most informative. First, all three chlorides have a partial negative charge. The boron halides are strong Lewis acids because each can accept an electron pair to complete an octet on B. Looking at the positive isopotential surface around the boron atom, it is clear how open each side of the molecule is to accepting an electron pair.
The bromides and iodides of Al, Ga, In, and Tl are primarily covalent, and two MX3 molecules usually combine to form a dimer (dimer means two units):
In the dimer structure a lone electron pair from a Br atom in one AlBr3 unit has been donated to the Al of the other AlBr3 unit, and vice versa. This forms two coordinate covalent bonds to hold the dimer together. (These bonds are shown as arrows.)
The fluorides of Al, Ga, In, and Tl are primarily ionic. Only the larger, more easily polarizable Br and I ions can be distorted enough by the small M3+ ions to give mainly covalent bonds. Most of the chlorides are border-line cases. This is reflected in the melting points:
AlF3, 1291°C AlCl3, 190°C Al2Br6, 97.5°C Al2I6, 191°C
Solid AlCl3 consists of A13+ and Cl ions, but in the liquid and gas phases Al2Br6 dimers predominate.
The oxides and oxyanions of B and Al are the main compounds of commercial importance. Borax, Na2B4O7•10H2O, is the principal ore of B, and Al is obtained from bauxite, Al2O3xH2O. Bauxite is an example of a hydrous oxide. It contains an indeterminate amount of water (xH2O) in the crystal lattice. Although Al is the most abundant metal in the earth’s crust, most of it is found in complicated aluminosilicates (feldspars). It is extremely difficult to convert the feldspars to Al metal, and any Si which remains as an impurity greatly degrades the strength and other properties of Al. Consequently the metal is obtained from bauxite, of which there is only a limited supply.
The first step in the recovery of Al is called the Bayer process. This depends on the fact that Al2O3 is amphoteric; that is, it can behave as either an acid or a base. If it behaves as a base, it can react with, and dissolve in, acidic solutions. If it behaves as an acid, it can dissolve in basic solutions. The amphoteric behavior of Al2O3 comes about because of the presence of A13+ ions and O2– ions. Al3+ has a very small radius and a large ionic charge, and so it strongly attracts H2O molecules—so strongly that the Al(H2O)63+ ion can donate protons. (Other cations which do this were described in Sec. 11.3.) Oxide ion, of course, is a good proton acceptor and a strong base.
The Bayer process makes use of the acidic behavior of Al2O3 by dissolving it in strong base:
$\text{Al}_2\text{O}_3(s) + \text{3H}_2\text{O}(l) + \text{2OH}^{-}(aq) \rightarrow \text{2Al(OH)}_4^{-}(aq) \nonumber$
The oxides of most metals are strongly basic and do not dissolve, while the oxides of nonmetals such as Si are acidic and more soluble in base than Al2O3. Thus a fairly pure sample of aluminum hydroxide, Al(OH)3, can be obtained by filtering off the basic oxides and acidifying the solution slightly. CO2, an acidic oxide, is used for this purpose:
$\text{CO}_2(g) + \text{Al(OH)}_4^{-}(aq) \rightarrow \text{Al(OH)}_3(s) + \text{HCO}_3^{-}(aq) \nonumber$
The Al(OH)3 is then heated to drive off H2O:
$\text{2Al(OH)}_3(s) \xrightarrow{\Delta } \text{Al}_2\text{O}_3(s) + \text{3H}_2{O}(g) \nonumber$
And Al is obtained by the Hall-Heroult process:
$\text{Al}_2\text{O}_3(l) \xrightarrow{\text{electrolysis}} \text{2Al}(l) + \text{3H}_2\text{O}(g) \nonumber$
This process is described in greater detail in the section on aluminum production.
The last step in aluminum production requires tremendous quantities of electrical energy. This, as well as the scarcity of its ore, makes aluminum more expensive than iron, the only metal which is more widely used. Nevertheless, aluminum has several advantages over iron. One is its considerably lower density, making possible alloys of comparable strength with considerably lower mass. A second advantage of aluminum is the coating of Al2O3 which forms on its surface when it is exposed to air. This coating sticks to the surface and is insoluble in neutral solutions, and so it prevents further oxidation. Iron, by contrast, forms a hydrous oxide (rust) which flakes off easily, exposing additional metal to the air. Thus under normal environmental conditions aluminum will last much longer than iron. This is a problem when aluminum beverage cans litter roadsides, but in general it makes aluminum a very favorable candidate for recycling. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.04%3A_Group_IIIA.txt |
Near the middle of the periodic table there is greatest variability of properties among elements of the same group. This is certainly true of group IVA, which contains carbon, a nonmetal, silicon and germanium, both semi- metals, and tin and lead, which are definitely metallic. Elemental carbon exists in two allotropic forms, diamond and graphite, whose structures are shown below.
The crystal structure of (a) diamond and (b) graphite.
In diamond there is a three-dimensional network of covalent bonds, while graphite consists of two-dimensional layers covalently bonded. Silicon, germanium, and one allotrope of tin (gray tin) also have the diamond structure—each atom is surrounded by four others arranged tetrahedrally. White tin has an unusual structure in which there are four nearest-neighbor atoms at a distance of 302 pm and two others at 318 pm. Only lead has a typical closest-packed metallic structure in which each atom is surrounded by 12 others.
Properties of the Group IVA Elements
Element Symbol
Electron Configuration
Usual Oxidation State
Radius/pm
Covalent
Ionic (M2+)
Carbon C [He]2s22p2 +4, +2 77
-
Silicon Si [Ne]3s23p2 +4, +2 117
-
Germanium Ge [Ar]4s23d104p2 +4, +2 122
-
Tin Sn [Kr]5s24d105p2 +2, +4 141
122
Lead Pb [Xe]6s24f145d106p2 +2, +4 146 131
Symbol Ionization Energy/MJ mol–1
Density/
g cm–3
Electro-
negativity
Melting
Point (in °C)
First Second Third Fourth
C 1.093 2.359 4.627 6.229 3.51 2.5
3550
Si 0.793 1.583 3.238 4.362 2.33 1.8 1410
Ge 0.768 1.544 3.308 4.407 5.35 1.8 937
Sn 0.715 1.418 2.949 3.937 7.28 1.8 232
Pb 0.722 1.457 3.088 4.089 11.34 1.8 327
Some properties of the group IVA elements are summarized in the table. As in the case of group IIIA, there is a large decrease in ionization energy and electronegativity from carbon to silicon, but little change farther down the group. This occurs for the same reason in both groups, namely, that elements farther down the group have filled d subshells. Note also that ionization energies, especially the third and fourth, are rather large. Formation of true +4 ions is very difficult, and in their +4 oxidation states all group IVA elements form predominantly covalent bonds. The +2 oxidation state, corresponding to use of the np2, but not the ns2, electrons for bonding, occurs for all elements. It is most important in the case of tin and especially lead, the latter having an inert pair like that of thallium. In the +4 oxidation state lead is a rather strong oxidizing agent, gaining two electrons (6s2) and being reduced to the +2 state.
Chemical Reactions and Compounds
Carbon’s ability to form strong bonds with other carbon atoms and the tremendous variety of organic compounds have already been discussed extensively in the section on organic compounds. You may want to review the subsections dealing with hydrocarbons and the other organic compounds. The most important inorganic carbon compounds are carbon monoxide and carbon dioxide. Both are produced by combustion of any fuel containing carbon:
$\text{C} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}$ (1)
$\text{CO} + \frac{1}{2}\text{O}_2 \rightarrow \text{CO}_2$ (2)
The triple bond in is the strongest chemical bond known, and contains two double bonds, and so both molecules are quite stable. Equations (1) and (2) occur stepwise when a fuel is burned, and the strong bond makes Eq. (2) slow unless the temperature is rather high. If there is insufficient O2 or if the products of combustion are cooled rapidly, significant quantities of CO can be produced. This is precisely what happens in an automobile engine, and the exhaust contains between 3 and 4% CO unless pollution controls have been installed.
CO is about 200 times better than O2 at bonding to hemoglobin, the protein which transports O2 through the bloodstream from the lungs to the tissues. Consequently a small concentration of CO in the air you breathe can inhibit transport of O2 to the brain, causing drowsiness, loss of consciousness, and death. (After a few minutes of breathing undiluted auto exhaust, more than half your hemoglobin will be incapable of transporting O2, and you will faint.) CO in automobile exhaust can be used to put animals to sleep. Because CO is colorless and odorless, your senses cannot detect it, and people must constantly be cautioned not to run cars in garages or other enclosed spaces. With the large number of cars and the great number of miles driven, it is important to limit CO emissions from automobiles. In the early 1970s new EPA standards led to the adoption of catalytic converters, which convert the poisonous CO into CO2[1]. Implementation and increasing effectiveness of these converters has caused CO levels to drop since the 1970s, despite the increase in automobiles on the road[2].
Like the organic compounds of carbon, the oxygen compounds of silicon which make up most of the earth’s crust have already been described. These substances illustrate a major contrast between the chemistry of carbon and silicon. The latter element does form a few compounds, called silanes, which are analogous to the alkanes, but the Si—Si bonds in silanes are much weaker than Si—O bonds. Consequently the silanes combine readily with oxygen from air, forming Si—O—Si linkages. Unlike the alkanes, which must be ignited with a spark or a match before they will burn, silanes catch fire of their own accord in air:
$\text{2Si}_4\text{H}_10 + \text{13O}_2 \rightarrow \text{4SiO}_2 + \text{5H}_2\text{O}$
Another important group of silicon compounds is the silicones. These polymeric substances contain Si—O—Si linkages and may be thought of as derived from silicon dioxide, SiO2. To make silicones, one must first reduce silicon dioxide to silicon. This can be done using carbon as the reducing agent in a high-temperature furnace:
$\text{SiO}_2(s) + \text{2C}(s) \xrightarrow{\text{3000}{}^\circ \text{C}} \text{Si}(l) + \text{2CO}(g)$
The silicon is then reacted with chloromethane:
$\text{Si}(s) + \text{2CH}_3\text{Cl}(g) \xrightarrow[\text{Cu catalyst}]{\text{300}{}^\circ \text{C}} (\text{CH}_3)_2\text{SiCl}_2(g)$
The dichlorodimethylsilane obtained in this reaction polymerizes when treated with water:
$\text{n(CH}_3)_2\text{SiCl}_2 + \text{nH}_2\text{O} \rightarrow$ $+ \text{2nHCl}$ The silicone polymer consists of a strongly bonded —Si—O—Si—O—Si—O chain, called a siloxane chain, with two methyl groups (or other organic groups) on each silicon atom. The strong backbone of a silicone polymer makes it stable to heat and difficult to decompose. Silicone oils make good lubricants and heat-transfer fluids, and rubber made from silicone remains flexible at low temperatures.
Besides the metals themselves, some tin and lead compounds are of commercial importance. Tin(II) fluoride (stannous fluoride), SnF2, is added to some toothpastes to inhibit dental caries. Tooth decay involves dissolving of dental enamel [mainly Ca10(PO4)6(OH)2] in acids synthesized by bacteria in the mouth. Fluoride ions from SnF2 inhibit decay by transforming tooth surfaces into Ca10(PO4)6F2, which is less soluble in acid:
$\text{Ca}_{10}(\text{PO}_4)_6\text{(OH)}_2 + \text{SnF}_2 \rightarrow \text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + \text{Sn(OH)}_2$
Since F is a weaker base than OH, the F compound has less tendency to react with acids. Note that when tin or lead are in the +2 oxidation state and are combined with a highly electronegative element like fluorine, the compounds formed are rather ionic.
Lead is found in two main commercial applications. One, the lead-acid storage battery is used to start cars and power golf carts. The other is the lead found in automobile fuel. In the +4 oxidation state lead forms primarily covalent compounds and bonds strongly to carbon. The compound tetraethyllead may be synthesized by reacting with a sodium-lead alloy:
$\text{4NaPb} + \text{4CH}_3\text{CH}_2\text{Cl} \rightarrow (\text{CH}_3\text{CH}_2)\text{Pb} + \text{4Pb} + \text{4NaCl}$
Sodium dissolved in the lead makes the latter more reactive. Tetraethyl-lead prevents gasoline from igniting too soon or burning unevenly in an automobile engine, circumstances which cause the engine to “knock” or “ping.” This is where the term leaded gasoline comes from. A major problem connected to using tetraethyl-lead is the introduction of lead into the atmosphere. Lead is toxic, and thus use of TEL as an antiknock agent has been phased out in favor of other agents less dangerous to public health. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.05%3A_Group_IVA.txt |
Although all the elements in this group form compounds in which their oxidation state of +5 equals the group number, their other properties vary considerably. Nitrogen is clearly nonmetallic and consists of diatomic triply-bonded N2 molecules.
Phosphorus, also a nonmetal, exists as tetrahedral P4 molecules (Figure $\PageIndex{1a}$) in the vapor and the white allotropic form of the solid. On standing, white phosphorus slowly changes to the red allotrope, whose structure is shown in Figure $\PageIndex{1c}$. The most stable form of the element is black phosphorus, which has a layer structure (Figure $\PageIndex{1b}$). Black phosphorus can be made by heating the white form with a mercury catalyst for 8 days at 220 to 370°C.
Arsenic is a semimetal and consists of As4 molecules in the gas phase. When As4(g) is condensed to a solid, three allotropes may form. The most stable of these is metallic arsenic, in which each arsenic atom has three nearest neighbors, with three more arsenic atoms somewhat farther away.
Antimony, also a semimetal, has two allotropes, the more stable one being metallic, like arsenic. In the case of bismuth, only the metallic form occurs. Note that for all the group VA elements the 8 – N rule is followed. The number of bonds or nearest neighbors for each atom is 8 minus the group number in , P4, and even in the metallic forms of As, Sb, and Bi. The table summarizes the atomic properties of the group VA elements. Overall, the trends are what we would expect, based on our experience with previous groups. These elements exhibit a much wider variety of oxidation states, however, especially in the case of nitrogen. This element forms compounds in which it has every possible oxidation number from –3 (the group number minus 8) to +5 (the group number). As in previous groups, the oxidation state in which the ns2 pair of electrons is not used for bonding becomes more prominent toward the bottom of the periodic table. There are a few compounds, Bi(NO3)3, for example, in which discrete Bi3+ ions are present.
Table $1$ : Properties of the Group VA Elements.
Element Symbol
Electron Configuration
Usual Oxidation State
Radius/pm
Covalent
Ionic (Charge)
Nitrogen N [He]2s22p3 +5, +3, -3 70
(3-)171
Phosphorus P [Ne]3s23p3 +5, +3, -3 110
-
Arsenic As [Ar]4s23d104p3 +5, +3 121
-
Antimony Sb [Kr]5s24d105p3 +5, +3 141
-
Bismuth Bi [Xe]6s24f145d106p3 +5, +3 146 (3+)108
Symbol Ionization Energy/MJ mol–1
Density/
g cm–3
Electro-
negativity
Melting
Point (in °C)
First Second Third Fourth Fifth
N 1.407 2.862 4.585 7.482 9.452 1.25×10-3 3.0
-210
P 1.018 1.909 2.918 4.963 6.280 1.82 2.1 44
As 0.953 1.804 2.742 4.843 6.049 5.72 2.0 817
Sb 0.840 1.601 2.450 4.271 5.403 6.69 1.9 631
Bi 0.710 1.616 2.472 4.380 5.417 9.80 1.9 271
The most important compounds of the group VA elements are those of nitrogen and phosphorus. Both elements are essential to all living organisms, and both are progressively removed from soil when plants are cultivated and crops harvested. According to Liebig’s law of the minimum, an insufficient supply of either element can limit plant growth and reduce crop yields, and so these elements are important components of fertilizer. More recently both elements have been implicated in several kinds of pollution problems. As we discuss the properties of nitrogen and phosphorus compounds, their effects on food production and environmental degradation will also be discussed.
Nitrogen
The importance of nitrogen fertilizer was first recognized over a century ago. By the late 1800s the only major ore of nitrogen, Chile saltpeter, NaNO3, was being mined in Chile and shipped to Europe for application to agricultural land, but the supply was obviously limited. Most nitrogen at the earth’s surface is in the form of N2(g), which makes up 78 percent of the atmosphere by volume (or by amount of substance). Therefore chemists began to look for ways of obtaining nitrogen compounds directly from the atmosphere. Any process which does this is called nitrogen fixation.
Nitrogen fixation can occur naturally when an electrical discharge (lightning) heats air to a high temperature. The following reaction occurs:
$\text{N}_2(g) + \text{O}_2(g) \rightarrow \text{2NO}(g) \label{1}$
The nitrogen monoxide (nitric oxide) formed can react further at ordinary temperatures, producing the brown gas, nitrogen dioxide:
$\text{2NO}(g) + \text{O}_2(g) \rightarrow \text{2NO}_2(g)\label{2}$
The Lewis diagrams for these and other important nitrogen compounds are shown in Figure $2$ .
From the figure you can see that both NO and NO2 have an odd number of electrons and violate the octet rule. In such a case it is common for two molecules to combine (dimerize) by pairing their odd electrons. In the case of NO2, dimerization occurs below room temperature, producing colorless dinitrogen tetroxide:
$\text{2NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g) \nonumber$
At room temperature, however, the NO2 and N2O4 are in equilibrium, as evidenced by the brown color of the mixture. NO dimerizes only at very low temperatures in the solid state.
The first industrial nitrogen fixation was done by mimicking nature.
Reaction $\ref{1}$ was carried out in a plant near Niagara Falls, where hydroelectric generation provided inexpensive power to support an electric arc. NO was further oxidized to NO2 which was dissolved in H2O to convert it to nitric acid, HNO3:
$\text{3NO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{2H}^{+}(aq) + \text{2NO}_3^{-}(aq) + \text{NO}(g)\label{4}$
Note that NO2 is not the acid anhydride of HNO3. This reaction involves, dis-proportionation of NO2 (which contains N in the +4 oxidation state) to form HNO3 in the + 5 state) and NO (N in the + 2 state). The NO can be recycled by reoxidizing it to NO2, and so it was not wasted. The HNO3 produced in Equation (12.5) was neutralized with NaOH to make a substitute for Chile saltpeter:
$\text{NaOH}(aq) + \text{HNO}_3(aq) \rightarrow \text{NaNO}_3(aq) + \text{H}_2\text{O}(l) \nonumber$
Fixation of nitrogen by the electric-arc process used a great deal of energy and was rather expensive. Other methods were designed to replace it, and the most successful of these is the Haber process, which is the major one used today. Nitrogen is reacted with hydrogen at a high temperature and extremely high pressure over a catalyst consisting of iron and aluminum oxide:
$\text{N}_2(g) + \text{3H}_2(g) \xrightleftharpoons[\text{1000atm}]{\text{450}{}^\circ \text{C}} \text{2NH}_3(g) \nonumber$
The ammonia produced by the Haber process is used directly as a fertilizer. It can be liquefied under pressure and injected through special nozzles about a foot under the soil surface. This prevents loss of gaseous ammonia which would otherwise irritate the nose, throat, and lungs of anyone near a fertilized field. You are probably familiar with the odor of ammonia since it is the most common weak base encountered in the chemical laboratory.
Prior to the recent development of underground injection techniques, most ammonia was converted to ammonium nitrate for fertilizer use:
$\text{NH}_3(aq) + \text{HNO}_3(aq) \rightarrow \text{NH}_4\text{NO}_3(aq)\label{7}$
Except for ammonia, ammonium nitrate contains a greater mass fraction of nitrogen than any other compound of comparable cost. Ammonium nitrate manufacture requires that half the ammonia produced in the Haber process be converted to nitric acid. The first step is oxidation of ammonia over a catalyst of platinum metal:
$\text{4NH}_3(g) + \text{5O}_2(g) \xrightarrow[\text{800}{}^\circ \text{C}]{\text{Pt}} \text{4NO}(g) + {6H}_2\text{O}(g) \nonumber$
This is called the Ostwald process. It is followed by Eqs. $\ref{2}$ and $\ref{4}$, yielding nitric acid, which can be combined with ammonia (Equation $\ref{7}$).
Nitric acid and nitrates have commercial applications other than fertilizer production. Because NO3 is a strong oxidizing agent, it reacts vigorously with substances whose elements are in low oxidation states. One example of this is black powder, which consists of charcoal (carbon), sulfur, an potassium nitrate, KNO3 (saltpeter or nitre). During the American revolution, for example, both armies had numerous persons whose job was to find caves in which the relatively soluble KNO3 had been deposited as water evaporated. A second example is nitroglycerin which contains carbon and hydrogen in low oxidation states as well as nitrate. Still another example of an explosive nitrate is NH4NO3, which contains nitrogen in its maximum and minimum oxidation state. NH4NO3 decomposes as follows:
$\text{NH}_4\text{NO}_3(s) \xrightarrow[\text{shock}]{\text{heat or}} \text{N}_2\text{O}(g) + \text{2H}_2\text{O}(g) \nonumber$
ΔHm = –37 kJ mol–1
The reaction is exothermic and produces 3 mol of gaseous products for every mole of solid reactant. This causes a tremendous increase in pressure, and, if the reaction is rapid, an explosion.
The compound dinitrogen monoxide (nitrous oxide or laughing gas), produced by decomposition of NH4NO3, is a third important oxide of nitrogen (in addition to NO and NO2). N2O is produced during microbial decomposition of organic matter containing nitrogen. Because it is quite unreactive, it is the second most-concentrated nitrogen-containing substance in the atmosphere (after N2). It is used commercially as an anesthetic, is mildly intoxicating, and is poisonous in large doses.
The other two important oxides of nitrogen, NO and NO2, play a major role in an air-pollution problem known as photochemical smog (or Los Angeles smog). NO is formed by Equation $\ref{1}$ in automobile engines and other high-temperature combustion processes. At normal temperatures NO is oxidized to NO2 (Equation $\ref{2}$). Both these oxides are free radicals and are rather reactive. Moreover, brown-colored NO2 absorbs sunlight, and the energy of the absorbed photons breaks a nitrogen-oxygen bond:
The oxygen atoms produced are highly reactive. They combine with hydrocarbon molecules (from evaporated or unburned gasoline) to form aldehydes, ketones, and a number of other compounds which form an almost fog-like cloud and irritate the eyes, throat, and lungs. Photochemical smog is especially bad in cities like Los Angeles and Denver which have lots of sunshine and automobile traffic, but its effects have been observed in every large city in the United States.
Phosphorus
As in the case of carbon and silicon, there are major differences between the chemistries of nitrogen and phosphorus. The concentrations of phosphorus compounds in the earth’s atmosphere are so small as to be negligible, but phosphorus is more abundant than nitrogen in the solid crust. Here it is found as phosphate rock, which is mainly hydroxyapatite, Ca10(PO4)6(OH)2, or fluorapatite, Ca10(PO4)6F2. (These are the same substances involved in the discussion of dental decay in the section on group IVA elements.)
Phosphate rock is quite insoluble, and hence its phosphate ions cannot be assimilated by plants. Production of phosphate fertilizer requires treatment of apatite acid. This protonates the PO43 ions, converting them to H2PO4, whose calcium salt is much more soluble:
$\text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + \text{7H}_2\text{SO}_4 + \text{H}_2\text{O} \rightarrow \text{3Ca(H}_2\text{PO}_4)_2•\text{H}_2\text{O} + \text{7CaSO}_4 \nonumber$
$\text{Ca}_{10}(\text{PO}_4)_6(\text{OH})_2 + \text{14H}_3\text{PO}_4 \rightarrow \text{10Ca(H}_2\text{PO}_4)_2 + \text{2H}_2\text{O} \nonumber$
The compound Ca(H2PO4)•H2O is known as superphosphate, and Ca(H2PO4)2 is called triple superphosphate.
The phosphoric acid, H3PO4, used to make triple superphosphate is also obtained from phosphate rock. The first step is a reduction with carbon (coke) and silicon dioxide in an electric furnace:
$\text{2Ca}_{10}(\text{PO}_4)_6\text{(OH)}_2 + \text{18SiO}_2 + \text{30C} \rightarrow \text{3P}_4 + \text{30CO} + \text{2Ca(OH)}_2 + \text{18CaSiO}_3 \nonumber$
The phosphorus obtained this way is then oxidized to phosphorus pentoxide:
$\text{P}_4(s) + \text{5O}_2(g) \rightarrow \text{P}_4\text{O}_{10}(s) \nonumber$
(The name phosphorus pentoxide for P4O10 comes from the empirical formula P2O5 of this compound.) Phosphorus pentoxide is the acid anhydride of phosphoric acid:.
$\text{P}_4\text{O}_6(s) + \text{6H}_2\text{O}(l) \rightarrow \text{4H}_3\text{PO}_3(aq) \nonumber$
Although not a very strong acid, phosphoric acid is triprotic. Therefore, 1 mol of this acid can transfer 3 mol of protons to a strong base. There is another oxide of phosphorus, P4O6, which involves the + 3 oxidation state, corresponding to use of the 3p3, but not the 3s2, electrons for bonding. P4O6 is the acid anhydride of phosphorous acid, H3PO3:
$\text{P}_4\text{O}_6(s) + \text{6H}_2\text{O}(l) \rightarrow \text{4H}_3\text{PO}_3(aq) \nonumber$
Phosphorous acid is quite weak, and, contrary to what its formula might suggest, can only donate two protons. This is apparently because its Lewis structure is
Only the two protons bonded to highly electronegative oxygen atoms are expected to be acidic.
Another major commercial use of phosphates is in laundry detergents. The problem of precipitation of soap by hard-water ions such as Ca2+ was mentioned in the section on alkaline earth metals. This can be prevented, and the cleaning power of synthetic detergents can be improved, by adding phosphates. The compound usually used is sodium tripolyphosphate, whose anion is a condensation polymer of hydrogen phosphate and dihydrogen phosphate ions:
The tripolyphosphate ion has numerous O atoms whose lone pairs of electrons can form coordinate covalent bonds to metal ions like Ca2+:
The Ca2+ ions are effectively removed from solution (they are said to be sequestered) because they are bonded to the tripolyphosphate ion. Consequently Ca2+(aq) is not available to precipitate soap or detergent molecules.
The use of phosphates in detergents is responsible in part for an environmental problem known as accelerated eutrophication, or premature, aging of bodies of water. Over a period of many thousands of years, a lake or other body of water will slowly accumulate essential nutrient elements such as nitrogen or phosphorus because their compounds dissolve in streams that feed the lake. As the water becomes richer in nutrients, more plants and microorganisms can grow. Some of the organic matter which remains when these organisms die precipitates to the bottom of the lake and is not decomposed. Eventually the lake fills up with debris, becoming a swamp, and finally dry land.
This process of eutrophication can be greatly accelerated by human input of nutrients such as nitrogen or phosphorus fertilizers, or phosphates from detergents. Since reduction in the use of detergent phosphates would appear to have the least negative effects—people’s clothes might not look as clean—many have suggested that prohibiting or limiting phosphate content is the way to solve the problem. Many states have passed laws implementing such limitations or bans. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.06%3A_Group_VA_Elements.txt |
As we approach the right-hand side of the periodic table, similarities among the elements within a group become greater again. This is true of group VIA. Except polonium, which is radioactive and usually omitted from discussion, all members of the group form X2– ions when combined with highly electropositive metals. The tendency to be reduced to the –2 oxidation state decreases significantly from top to bottom of the group, however, and tellurium shows some metallic properties. The group VIA elements are called chalcogens because most ores of copper (Greek chalkos) are oxides or sulfides, and such ores contain traces of selenium and tellurium. Atomic properties of the chalcogens are summarized in the table.
Table $1$: Properties of the Group VIA Elements
Element Symbol Electron Configuration Usual Oxidation State Radius/pm
Covalent Ionic (X2-)
Oxygen O [He]2s22p4 -2 66 140
Sulfur S [Ne]3s23p4 +6, +4, -2 104 184
Selenium Se [Ar]4s23d104p4 +6, +4, -2 117 198
Tellurium Te [Kr]5s24d105p4 +6, +4, -2 135 221
Symbol Ionization Energy/MJ mol–1 Density/ g cm–3 Electro- negativity Melting Point (in °C)
First Second Third
O 1.320 3.395 5.307 1.43×10-3 3.5 -218
S 1.006 2.257 3.367 2.06 2.5 119
Se 0.947 2.051 2.980 4.82 2.4 217
Te 0.876 1.800 2.704 6.25 2.1 450
At ordinary temperatures and pressures, oxygen is a gas. It exists in either of two allotropic forms: O2, which makes up 21 percent of the earth’s atmosphere, or O3 (ozone), which slowly decomposes to O2. O3 can be prepared by passing an electrical discharge through O2 or air:
$\text{3O}_2(g) \xrightarrow{\text{electrical discharge}} \text{2O}_3(g) \nonumber$
This reaction occurs naturally as a result of lightning bolts. O3 is also produced by any device which produces electrical sparks. You may have noticed its distinctive odor in the vicinity of an electric motor, for example.
Ozone is formed in the earth’s stratosphere (between altitudes of 10 and. 50 km) by ultraviolet rays whose wavelengths are shorter than 250 nm:
$\text{O}_2 \xrightarrow{\text{ultraviolet light}} \text{2O} \nonumber$
$\text{O} + \text{O}_2 \rightarrow \text{O}_3\label{3}$
The ozone itself absorbs longer-wavelength ultraviolet radiation (up to 340 nm), preventing these harmful rays fom reaching the earth’s surface. Otherwise these rays would increase the incidence of human skin cancer and cause other environmental problems. In recent years convincing evidence has been obtained to show that nitrogen oxide emissions from supersonic transport (SST) airplanes (which fly in the stratosphere) can reduce the concentration of ozone. Similar conclusions have been drawn regarding chlorofluorocarbons(sometimes referred to as CFCs) used as propellants in aerosol hair sprays and deodorants. Once in the atmosphere, a photochemical reaction causes atomic chlorine to be broken off from CFCs. This atomic chlorine can then participate in a catalytic ozone depleting reaction:
$\text{Cl} + \text{O}_3 \rightarrow \text{ClO} + \text{O}_2 \nonumber$
$\text{ClO} + \text{O}_3 \rightarrow \text{Cl} + \text{2O}_2 \nonumber$
Atomic chlorine is regenerated, meaning that each CFC molecule has the potential to deplete large amounts of ozone. In the 1980s, it was determined that use of chemicals such as CFCs were thinning stratospheric ozone. This is also when the "ozone hole" over Antarctica was discovered. In response to the depletion of ozone, and the danger presented by it, the Montreal Protocol on Substances that Deplete the Ozone Layer was signed by leaders of multiple countries, with the goal to phase out production and use of CFCs and other chemicals harmful to the ozone layer. Today, 191 countries have signed the protocol, and while it is projected to take until 2075 for ozone levels to return to normal, the Montreal Protocol has so far proven a success[1].
O3 is also an important component of photochemical smog. It is produced when O atoms (formed by breaking N—O bonds in NO2) react with molecules according to Eq. $\ref{3}$. O3 is a stronger oxidizing agent than O2. It reacts with unsaturated hydrocarbons (alkenes) in evaporated gasoline to produce aldehydes and ketones which are eye irritants. Rubber is a polymeric material which contains bonds, and so it too reacts with O3. Further, ground level ozone and the accompanying smog has proven a significant health concern, irritating and damaging the respiratory system and also having links to asthma[2]. So ozone is beneficial when in the upper atmosphere, but has adverse effects when at ground level.
Sulfur occurs in a variety of allotropic forms. At room temperature the most stable form is rhombic sulfur. This yellow solid consists of S8 molecules (seen in the Jmol below) packed in a crystal lattice which belongs to the orthorhombic system (listed on the page discussing crystal systems).
Figure $1$ S8 molecule: The initial ball and stick model can be manipulated in three dimensional space. Click on VdW radii to see a space filling model of the same molecule.
When heated to 96°C, solid rhombic sulfur changes very slowly into monoclinic sulfur, in which one-third of the S8 molecules are randomly oriented in the crystal lattice. When either form of sulfur melts, the liquid is at first pale yellow and flows readily, but above 160°C it becomes increasingly viscous. Only near the boiling point of 444.6°C does it thin out again. This unusual change in viscosity with temperature is attributed to opening of the eight-membered ring of S8 and formation of long chains of sulfur atoms. These intertwine and prevent the liquid from flowing. This explanation is supported by the fact that if the viscous liquid is cooled rapidly by pouring it into water, the amorphous sulfur produced can be shown experimentally to consist of long chains of sulfur atoms.
Both selenium and tellurium have solid structures in which the atoms are bonded in long spiral chains. Both are semiconductors, and the electrical conductivity of selenium depends on the intensity of light falling on the element. This property is utilized in selenium photocells, which are often used in photographic exposure meters.
Selenium is also used in rectifiers to convert alternating electrical current to direct current. Compounds of selenium and tellurium are of little commercial importance, and they often are toxic. Moreover, many of them have foul odors, are taken up by the body, and are given off in perspiration and on the breath. These properties have inhibited study of tellurium and selenium compounds.
Chemical Reactions and Compounds
Oxygen Since oxygen has the second largest electronegativity among all the elements, it is found in the –2 oxidation state in most compounds. Important oxides have already been discussed in sections dealing with the elements from which they form, and so we will deal only with unusual oxidation states of oxygen here. One of these is the +2 state found in OF2, the most common compound in which oxygen is combined with the more electronegative fluorine. We have already mentioned the –½ and –1 states observed in alkali-metal superoxides and peroxides, but one important peroxide, hydrogen peroxide (H2O2), has not yet been discussed.
H2O2 can be prepared by electrolysis of solutions containing sulfate ions. H2O2 is a weak acid, and it can serve as an oxidizing agent (oxygen being reduced to the –2 state) or as a reducing agent (oxygen being oxidized to the 0 state). Like the peroxide ion, the H2O2 molecule contains an O—O single bond. This bond is rather weak compared with many other single bonds, and this contributes to the reactivity of H2O2. The compound decomposes easily, especially if exposed to light or contaminated with traces of transition metals. The decomposition
$\text{2H}_2\text{O}_2(l) \rightarrow \text{2H}_2\text{O}(l) + \text{O}_2(g) \nonumber$ can occur explosively in the case of the pure liquid. Sulfur Although this element is only sixteenth in abundance at the surface of the earth, it is one of the few that has been known and used throughout history. Deposits of elemental sulfur are not uncommon, and, because they were stones that would burn, were originally called brimstone. Burning sulfur produces sulfur dioxide,
$\text{S}_8(s) + \text{8O}_2(g) \rightarrow \text{8SO}_2(g) \nonumber$
This colorless gas has a choking odor and is more poisonous than carbon monoxide. It is the anhydride of sulfurous acid, a weak diprotic acid:
$\text{SO}_2(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_3(aq) \nonumber$
SO2 is also produced when almost any sulfur-containing substance is burned in air. Coal, for example, usually contains from 1 to 4% sulfur, and so burning coal releases SO2 to the atmosphere. Many metal ores are sulfides, and when they are heated in air, SO2 is produced. Copper, for example, may be obtained as the element by heating copper(I) sulfide:
$\text{Cu}_2\text{S}(s) + \text{O}_2(g) \xrightarrow{\Delta } \text{2Cu}(s) + \text{SO}_2(g) \nonumber$
Since SO2 is so poisonous, its release to the atmosphere is a major pollution problem. Once in the air, SO2 is slowly oxidized to sulfur trioxide, SO3:
$\text{2SO}_2(g) + \text{O}_2(g) \rightarrow \text{2SO}_3(g) \nonumber$
This compound is the anhydride of sulfuric acid, H2SO4:
$\text{SO}_3(g) + \text{H}_2\text{O}(l) \rightarrow \text{H}_2\text{SO}_4(aq) \nonumber$
Thus if air is polluted with SO2 and SO3, a fine mist of dilute droplets of can form. All three substances are very irritating to the throat and lungs and are responsible for considerable damage to human health.
The natural mechanism for removal of sulfur oxides from the air is solution in raindrops, followed by precipitation. This makes the rainwater more acidic than it would otherwise be, and acid rain is now common in industrialized areas of the United States and Europe. Acid rain can slowly dissolve limestone and marble, both of which consist of CaCO3:
$\text{CaCO}_3(s) + \text{H}_3\text{O}^{+}(aq) \rightarrow \text{Ca}^{2+}(aq) + \text{HCO}_3^{-}(aq) + \text{H}_2\text{O}(l) \nonumber$
Thus statues and buildings made of these materials may be damaged.
Despite the fact that a tremendous amount of sulfur is released to the environment by coal combustion and ore smelting, this element is not usually recovered from such processes. Instead it is obtained commercially from large deposits along the U.S. Gulf Coast and from refining of sour petroleum. Sour petroleum contains numerous sulfur compounds, including H2S, which smells like rotten eggs. The deposits of elemental sulfur in Texas and Louisiana are mined by the Frasch process. Water at 170°C is pumped down a pipe to melt the sulfur, and the latter is forced to the surface by compressed air. Most of the H2S or S8 obtained from these sources is oxidized to SO2, passed over a vanadium catalyst to make SO3, and dissolved in water to make H2SO4. In 2005 an estimated 190 billion kg of H2SO4 was produced in the world, making H2SO4 one of the most important industrial chemicals. About half of it is used in phosphate fertilizer production[3].
Pure H2SO4 is a liquid at room temperature and has a great affinity for H2O. This is apparently due to the reaction:
$\text{H}_2\text{SO}_4 + \text{H}_2\text{O} \rightarrow \text{H}_3\text{O}^{+} + \text{HSO}_4^{-} \nonumber$
Formation of H3O+ releases energy, and the reaction is exothermic. Concentrated H2SO4 is 93% H2SO4 and 7% H2O by mass, corresponding to more than twice as many H2SO4 as H2O molecules. Since many H2SO4 molecules still have protons to donate, concentrated H2SO4 also has a great affinity for H2O. It is often used as a drying agent and can be employed in condensation reactions which give off H2O.
1. ↑ "Montreal Protocol-One Page Factsheet." Environmental Protection Agency. 6 September 2007. http://www.epa.gov/ozone/downloads/MP20_FactSheet.pdf
2. ↑ Kenneth Olden. "Statement on Health Effects of Air Pollution before the Senate Committee on Health, Education, Labor and Pensions, Subcommittee on Health. Department of Health and Human Services. 3 September 2003.www.hhs.gov/asl/testify/t020903.html
3. ↑ Bala Suresh and Yoshio Inoguchi. "Sulfuric Acid". SRI Consulting. July 2006. http://www.sriconsulting.com/CEH/Pub...orts/781.5000/
12.07: Group VIA- Chalcogens
Hernan Cortes (1485-1547) was a Spanish conquistador who was responsible for the fall of the Aztec empire. In 1518, Cortes left with a fleet of 11 ships and 500 men to explore and secure inner Mexico for Spain. Cortes entered Aztec territory and was peacefully received by the emperor Moctezuma II. Moctezuma offered the Spaniards gold and took them into the
Figure 1. Hernan Cortes
heart of the city. This enticed the Spanish even more to take over the territory for what they believed was vast quantities of gold. Many battles took place with the Aztecs, and over time Cortes needed more gun powder for his army. Spain, being across the ocean, would not be a quick source that he needed to win this war. As a result, he had his men make the gun powder. The main ingredients of gunpowder are Carbon and Sulfur. Carbon was easy to come by, but Sulfur was not. Cortes sent an expedition of men on a treacherous mission to the top of Popocatépetl, an active volcano. Volcanoes are a good source of sulfur because they emit sulfur dioxide.
Sulfur has many similarities with the elements in its group. This group is VIA. Except polonium, which is radioactive and usually omitted from discussion, all members of the group form X2- ions when combined with highly electropositive metals. The tendency to be reduced to the –2 oxidation state decreases significantly from top to bottom of the group, however, and tellurium shows some metallic properties. The group VIA elements are called chalcogens because most ores of copper (Greek chalkos) are oxides or sulfides, and such ores contain traces of selenium and tellurium. Atomic properties of the chalcogens are summarized in the table.
Properties of the Group VIA Elements
Table 1.
Element Symbol
Electron
Configuration
Usual
Oxidation State
Radius/pm
Covalent
Ionic (X2-)
Oxygen O [He]2s22p4 -2 66
140
Sulfur S [Ne]3s23p4 +6, +4, -2 104
184
Selenium Se [Ar]4s23d104p4 +6, +4, -2 117
198
Tellurium Te [Kr]5s24d105p4 +6, +4, -2 135 221
Table 2.
Symbol Ionization Energy/MJ mol–1
Density/
g cm–3
Electro-
negativity
Melting
Point (in °C)
First Second Third
O 1.320 3.395 5.307 1.43×10-3 3.5
-218
S 1.006 2.257 3.367 2.06 2.5 119
Se 0.947 2.051 2.980 4.82 2.4 217
Te 0.876 1.800 2.704 6.25 2.1 450
Sulfur occurs in a variety of allotropic forms. At room temperature the most stable form is rhombic sulfur. This yellow solid consists of S8 molecules (seen in the Jmol below) packed in a crystal lattice which belongs to the orthorhombic system (listed on the page discussing crystal systems).
Figure 1 S8 molecule: The initial ball and stick model can be manipulated in three dimensional space. Click on VdW radii to see a space filling model of the same molecule.
When heated to 96°C, solid rhombic sulfur changes very slowly into monoclinic sulfur, in which one-third of the S8 molecules are randomly oriented in the crystal lattice. When either form of sulfur melts, the liquid is at first pale yellow and flows readily, but above 160°C it becomes increasingly viscous. Only near the boiling point of 444.6°C does it thin out again. This unusual change in viscosity with temperature is attributed to opening of the eight-membered ring of S8 and formation of long chains of sulfur atoms. These intertwine and prevent the liquid from flowing. This explanation is supported by the fact that if the viscous liquid is cooled rapidly by pouring it into water, the amorphous sulfur produced can be shown experimentally to consist of long chains of sulfur atoms.
Chemical Reactions and Compounds
Sulfur Although this element is only sixteenth in abundance at the surface of the earth, it is one of the few that has been known and used throughout history. Deposits of elemental sulfur are not uncommon, and, because they were stones that would burn, were originally called brimstone. Burning sulfur produces sulfur dioxide,
S8(s) + 8O2(g) → 8SO2(g) This colorless gas has a choking odor and is more poisonous than carbon monoxide. It is the anhydride of sulfurous acid, a weak diprotic acid: SO2(g) + H2O(l) → H2SO3(aq) SO2 is also produced when almost any sulfur-containing substance is burned in air. Coal, for example, usually contains from 1 to 4% sulfur, and so burning coal releases SO2 to the atmosphere. Many metal ores are sulfides, and when they are heated in air, SO2 is produced. Copper, for example, may be obtained as the element by heating copper(I) sulfide: Cu2S(s) + O2(g) \(\xrightarrow{\Delta }\) 2Cu(s) + SO2(g) Since SO2 is so poisonous, its release to the atmosphere is a major pollution problem. Once in the air, SO2 is slowly oxidized to sulfur trioxide, SO3: 2SO2(g) + O2(g) → 2SO3(g) This compound is the anhydride of sulfuric acid, H2SO4: SO3(g) + H2O(l) → H2SO4(aq) Thus if air is polluted with SO2 and SO3, a fine mist of dilute droplets of can form. All three substances are very irritating to the throat and lungs and are responsible for considerable damage to human health.
The natural mechanism for removal of sulfur oxides from the air is solution in raindrops, followed by precipitation. This makes the rainwater more acidic than it would otherwise be, and acid rain is now common in industrialized areas of the United States and Europe. Acid rain can slowly dissolve limestone and marble, both of which consist of CaCO3:
CaCO3(s) + H3O+(aq) → Ca2+(aq) + HCO3(aq) + H2O(l) Thus statues and buildings made of these materials may be damaged.
Despite the fact that a tremendous amount of sulfur is released to the environment by coal combustion and ore smelting, this element is not usually recovered from such processes. Instead it is obtained commercially from large deposits along the U.S. Gulf Coast and from refining of sour petroleum. Sour petroleum contains numerous sulfur compounds, including H2S, which smells like rotten eggs. The deposits of elemental sulfur in Texas and Louisiana are mined by the Frasch process. Water at 170°C is pumped down a pipe to melt the sulfur, and the latter is forced to the surface by compressed air. Most of the H2S or S8 obtained from these sources is oxidized to SO2, passed over a vanadium catalyst to make SO3, and dissolved in water to make H2SO4. In 2005 an estimated 190 billion kg of H2SO4 was produced in the world, making H2SO4 one of the most important industrial chemicals. About half of it is used in phosphate fertilizer production[1].
Pure H2SO4 is a liquid at room temperature and has a great affinity for H2O. This is apparently due to the reaction
H2SO4 + H2O → H3O+ + HSO4 Formation of H3O+ releases energy, and the reaction is exothermic. Concentrated H2SO4 is 93% H2SO4 and 7% H2O by mass, corresponding to more than twice as many H2SO4 as H2O molecules. Since many H2SO4 molecules still have protons to donate, concentrated H2SO4 also has a great affinity for H2O. It is often used as a drying agent and can be employed in condensation reactions which give off H2O.
References
1. Bala Suresh and Yoshio Inoguchi. "Sulfuric Acid". SRI Consulting. July 2006. http://www.sriconsulting.com/CEH/Pub...orts/781.5000/
2. en.Wikipedia.org/wiki/Hernán_Cortés | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.07%3A_Group_VIA-_Chalcogens/12.7.01%3A_Cultural_Connections-_Sulfur.txt |
The halogens include fluorine, chlorine, bromine, and iodine. Astatine is also in the group, but is radioactive and will not be considered here. A summary of atomic properties of the halogens is given in the following table. The free elemental halogens all consist of diatomic molecules X2, where X may be fluorine, chlorine, bromine, or iodine (recall the microscopic picture of bromine). They are strong oxidizing agents and are readily reduced to the X ions, and so the halogens form numerous ionic compounds. Fluorine, the most electronegative element, has no positive oxidation states, but the other halogens commonly exhibit +1, +3, +5, and +7 states. Most compounds containing halogens in positive oxidation states are good oxidizing agents, however, reflecting the strong tendency of these elements to gain electrons.
Table $1$ Properties of the Group VIIA Elements.
Element Symbol Electron Configuration Usual Oxidation State Radius/pm
Covalent Ionic (X-)
Fluorine F [He]2s22p5 -1 64 136
Chlorine Cl [Ne]3s23p5 +7, +5, +3, +1, -1 99 181
Bromine Br [Ar]4s23d104p5 +7, +5, +3, +1, -1 114 195
Iodine I [Kr]5s24d105p5 +7, +5, +3, +1, -1 133 216
Symbol Ionization Energy/MJ mol–1
Density/
g cm–3
Electro-
negativity
Melting
Point (in °C)
First Second Third
F 1.687 3.381 6.057 1.73×10-3 4.0
-220
Cl 1.257 2.303 3.828 3.17×10-3 3.0 -101
Br 1.146 2.113 3.471 3.14 2.8 -7
I 1.015 1.852 3.184 4.94 2.5 114
There is some variation among their physical properties and appearance. Fluorine and chlorine are both gases at room temperature, the former very pale yellow, and the latter yellow-green in color. Bromine is a red-brown liquid which vaporizes rather easily. Iodine forms shiny dark crystals and, when heated, sublimes (changes directly from solid to gas) to a beautiful violet vapor. All the gases produce a choking sensation when inhaled. Chlorine was used to poison soldiers on European battlefields in 1915 to 1918. Halogens are put to more humane uses such as to disinfect public water supplies by means of chlorination and to treat minor cuts by using an alcohol solution (tincture) of iodine. These applications depend on the ability of the halogens to destroy microorganisms which are harmful to humans.
All halogens are quite reactive, and in the natural world they always occur combined with other elements. Fluorine reacts so readily with almost any substance it contacts that chemists were not successful in isolating pure fluorine until 1886, although its existence in compounds had been known for many years. Chlorine, bromine, and iodine are progressively less reactive but still form compounds with most other elements, especially metals. A good example is mercury, whose reaction with bromine was discussed in the section covering macroscopic and microscopic views of a chemical reaction. Mercury reacts with other halogens in the same way:
$\text{Hg}(l) + \text{X}_2(g, l, or s) \rightarrow \text{HgX}_2 (s) \nonumber$ X = F, Cl, Br, or I
Already covered in the section on alkali metals, halogens react readily with alkali metals with the general form of:
$\text{2M} + \text{X}_2 \rightarrow \text{2MX} \nonumber$ M = Li, Na, K, Rb, or Cs and X = F, Cl, Br, I
Iodine combines less vigorously with alkali metals than other halogens, but its reactions are analogous to the reactions of alkali metals with florine, chlorine and bromine. Compounds of an alkali metal and a halogen, such as sodium chloride, potassium fluoride, lithium bromide, or cesium iodide, have closely related properties. (All taste salty, for example.) They belong to a general category called salts, all of whose members are similar to ordinary table salt, sodium chloride. The term halogen is derived from Greek words meaning “salt former.”
Halogens also react with alkaline-earth metals in the general reaction:
$\text{M} + \text{X}_2 \rightarrow \text{MCl}_2 \nonumber$ M = Be, Mg, Ca, Sr, Ba, or Ra and X = F, Cl, Br, I
Another vigorous reaction occurs when certain compounds containing carbon and hydrogen contact the halogens. Turpentine, C10H16, reacts quite violently. In the case of fluorine and chlorine the equation is
$\text{C}_{10}\text{H}_{16}(l) + \text{8X}_2(g) \rightarrow \text{10C}(s) + \text{16HX}(g) \nonumber$ X = F, Cl
but the products are different when bromine and iodine react. Before the advent of the automobile, veterinarians used solid iodine and turpentine to disinfect wounds in horses’ hooves. This may have been because of the superior antiseptic qualities of the mixture. However, a more likely reason is the profound impression made on the owner of the horse by the great clouds of violet iodine vapor which sublimed as a result of the increase in temperature when the reaction occurred! Below is a video of this impressive reaction:
The violent reaction is due to α-pinene in turpentine. The relief of ring strain is highly exothermic. This temperature increase causes the sublimation leading to the impressive violet iodine vapor.
The halogens also react directly with hydrogen, yielding the hydrogen halides:
$\text{H}_2 + \text{X}_2 \rightarrow \text{2HX} \nonumber$ X = F, Cl, Br, I
These compounds are all gases, are water soluble, and, except for HF, are strong acids in aqueous solution. They are conveniently prepared in the laboratory by acidifying the appropriate sodium or other halide:
$\text{NaCl}(s) + \text{H}_3\text{O}^{+}(aq) \xrightarrow{\Delta} \text{Na}^{+}(aq) + \text{H}_2\text{O}(l) + \text{HCl}(g) \label{6}$
The acid must be nonvolatile so that heating will drive off only the gaseous hydrogen halide. In the case of fluorides and chlorides, H2SO4 will do, but bromides and iodides are oxidized to Br2 or I2 by hot H2SO4 and so H3PO4 is used instead.
A reaction similar to Eq. $\ref{6}$ occurs when phosphate rock containing fluorapatite is treated with H2SO4 to make fertilizer:
$\text{Ca}_{10}(\text{PO}_4)_6\text{F}_2 + \text{7H}_2\text{SO}_4 + \text{3H}_2\text{O} \rightarrow \text{3Ca(H}_2\text{PO}_4)_2•\text{H}_2\text{O} + \text{7CaSO}_4 + \text{2HF} \nonumber$
The HF produced in this reaction can cause significant air-pollution problems. Fluorides are also emitted to the atmosphere in steelmaking and aluminum production. There is some evidence that fluorides, rather than sulfur dioxide, may have been responsible for human deaths in air-pollution episodes at Donora, Pennsylvania, and the Meuse Valley in Belgium.
The relative oxidizing strengths of the halogens can be illustrated nicely in the laboratory. If, for example, a solution of Cl2 in H2O is combined with a solution of NaI, the dark color of I2 can be observed, showing that the Cl2 has oxidized the I:
$\text{Cl}_2(aq) + \text{2I}^{-}(aq) \rightarrow \text{2Cl}^{-}(aq) + \text{I}_2(aq) \nonumber$
This very reaction is shown in the following video:
The video starts out with four solutions. The experimental solution is on the far left, and contain Cl2 in water, which is covered by a layer of hexane, a nonpolar solvent which is immiscible with H2O. The three other solutions, from left to right are a Cl2 solution, a Br2 solution, and an I2 solution. When a solution with iodide ions is added to the experimental solution, nonpolar I2 molecules are formed. They concentrate in the hexane layer, and a beautiful violet color can be observed, the same as I2 solution. From such experiments it can be shown that the strongest oxidizing agent is F2 (at the top of the group). F2 will react with Cl, Br, and I. The weakest oxidizing agent, I2, does not react with any of the halide ions.
The extremely high oxidizing power of F2 makes it the only element which can combine directly with a noble gas. The reactions
$\text{Xe}(g) + \text{F}_2(g) \rightarrow \text{XeF}_2(s)$
$\text{XeF}_2(s) + \text{F}_2(g) \rightarrow \text{XeF}_4(s)$
$\text{XeF}_4(s) + \text{F}_2(g) \rightarrow \text{XeF}_6(s)$
may be used to synthesize the three xenon fluorides, all of which are strong oxidizing agents. When an electrical discharge is passed through a mixture of Kr and F2 at a low temperature, KrF2 can be formed. This is the only compound of Kr, and it decomposes slowly at room temperature.
Fluorine is also set apart from the other halogens because of its ability to oxidize water:
$\text{3F}_2 + \text{6H}_2\text{O} \rightarrow \text{4H}_3\text{O}^{+} + \text{4F}^{-} + \text{O}_2 \nonumber$
Chlorine is also capable of oxidizing water, but it does so very slowly. Instead the reaction
$\text{Cl}_2 + \text{2H}_2\text{O} \rightleftharpoons \text{H}_3\text{O}^{+} + \text{Cl}^{-} + \text{HOCl} \nonumber$
goes partway to completion. Hypochlorous acid, HOCl, is a weak acid. Small concentrations of hypobromous and hypoiodous acids can also be obtained in this way. In basic solution the halogen is completely consumed, producing the hypohalite anion:
$\text{Cl}_2 + \text{2OH}^{-} \rightarrow \text{Cl}^{-} + \text{H}_2\text{O} + \text{OCl}^{-} \nonumber$
Since hypochlorite, OCl, could also be supplied from an ionic compound such as NaOCl, the latter is often used to chlorinate swimming pools.
Hypohalite ions disproportionate in aqueous solution:
$\text{3OCl}^{-} \rightarrow \text{2Cl}^{-} + \text{ClO}_3^{-} \nonumber$
This reaction is rather slow for hypochlorite unless the temperature is above 75°C, but OBr and OI are consumed immediately at room temperature. Chlorate, ClO3, bromate, BrO3, and iodate, IO3, salts can be precipitated from such solutions. All are good oxidizing agents. Potassium chlorate, KClO3, decomposes, giving O2 when heated in the presence of a catalyst:
$\text{2KClO}_{3} \xrightarrow[\text{MnO}_{\text{2}}\text{ catalyst}]{\Delta } \text{2KCl} + \text{3O}_{2} \nonumber$
This is a standard laboratory reaction for making O2.
If KClO2 is heated without a catalyst, potassium perchlorate, KClO4, may be formed. Perchlorates oxidize organic matter rapidly and often uncontrollably. They are notorious for exploding unexpectedly and should be handled with great care.
One other interesting group of compounds is the interhalogens, in which one halogen bonds to another. Some interhalogens, such as BrCl, are diatomic, but the larger halogen atoms have room for several smaller ones around them. Thus compounds such as ClF3, BrF3 and BrF5, and IF3, ICl3, IF5, and IF7 can be synthesized. Note that the largest halogen atom I can accommodate three chlorines and up to seven fluorines around it. The following video showcases a reaction which involves some of these interhalogens:
The video begins with a test tube containing a layer of KI aqueous solution on top of CCl4 below it. Chlorine is bubbled through the KI layer. As seen in the video on oxidizing strength of the halogens, Cl2 reacts with I- to form iodine, according to the reaction:
$\text{2I}^{-}(aq ) + \text{Cl}_2(aq ) \rightarrow \text{I}_2(aq) + \text{2Cl}^{-}(aq) \nonumber$
A brown triiodide ion is also formed in the aqueous layer, according to the reaction:
$\text{I}^{-}(aq ) + \text{I}_2(aq ) \rightarrow \text{I}_3^{-}(aq) \nonumber$
A purple solution begins to form in the CCl4 layer, as iodine dissolves in it. The iodine in the aqueous layer also reacts with the excess Cl2 to form the red ICl, according to the following reaction:
$\text{I}_2(aq ) + \text{Cl}_2(aq ) \rightarrow \text{2ICl}(aq) \nonumber$
The final reaction takes place as more Cl2 is added, which reacts with ICl, to form the yellow ICl3. This reaction causes the aqueous solution to decolorize. This goes according to the reaction:
$\text{ICl}(aq ) + \text{Cl}_2(aq ) \rightarrow \text{ICl}_3(aq) \nonumber$
At the end of the video, the layers have decolorized, with a red portion in the CCl4 which is, due to its color, most likely remaining ICl.
12.09: Group VIIIA- Noble Gases
The properties of the noble gases are summarized in the table below. The noble gases have a complete octet of electrons ns2np6 or just ns2 for helium, leaving them with little chemical reactivity. Sure enough,the ionization energies of He and Ne are greater than 2000 kJ mol–1, and it is unlikely that these noble gases will ever be induced to form chemical bonds. The same probably applies to Ar. Kr and especially Xe do form compounds though, which was discussed in the halogens section, and Rn might be expected to be even more reactive. Rn is radioactive, however, and study of its chemistry is difficult.
Table \(1\) Properties of the Group VIII Elements.
Element Symbol
Electron
Configuration
Usual
Oxidation State
Radius/pm - Covalent
Helium He 1s2 0
...
Neon Ne [He]2s22p6 0
...
Argon Ar [Ne]3s23p6 0
...
Krypton Kr [Ar]4s23d104p6 +2
110
Xenon Xe [Kr]5s24d105p6 +8, +6, +4, +2 130
Symbol Ionization Energy/MJ mol–1
Density/
g cm–3
Electro-
negativity
Melting
Point (in °C)
First Second
He 2.379 5.257 0.179 ...
-272
Ne 2.087 3.959 0.901 ... -249
Ar 1.527 2.672 1.78 ... -190
Kr 1.357 2.374 3.74 2.6 -157
Xe 1.177 2.053 5.86 2.4 -112
Because of the lack of reactivity of the noble gases, they are often used when an nonreactive atmosphere is needed, such as in welding. Due to their low boiling points, noble gases are also cryogens in their liquid forms. One familiar use of helium is in balloons and blimps, since it is buoyant in the atmosphere, and unlike hydrogen, nonreactive. Another familiar use is as lighting in gas discharge lamps. Referred to popularly as neon lights, they can contain other noble gases, or mixtures of the gases. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/12%3A_Chemistry_of_the_Representative_Elements/12.08%3A_Group_VIIA-_Halogens.txt |
• 13.1: Prelude to Equilibria
The term chemical equilibrium is used to describe a chemical reaction in which the concentrations of the substances involved remain constant. Read on to learn how chemical equilibrium is defined.
• 13.2: The Equilibrium State
The term chemical equilibrium is used to describe a chemical reaction in which the concentrations of the substances involved remain constant. Read on to learn how chemical equilibrium is defined.
• 13.3: The Equilibrium Constant
The equilibrium constant represents the constant ratio between reactants and products when a reaction has reached equilibrium. Read on to find out more about how this ratio is calculated.
• 13.4: The Law of Chemical Equilibrium
Chemical equilibrium is attained when the concentration of reactants and products stops changing. Since reactants and products eventually reach a constant value (given constant temperature and pressure) a ratio called the rate constant can be used to describe the equilibrium. This section unpacks this ratio and how it is calculated.
• 13.5: The Equilibrium Constant in Terms of Pressure
In its most familiar form, the equilibrium constant is described in terms of the concentration of products and reactants. However, for gases it is often more convenient to relate the equilibrium constant in terms of pressure. Read on to find out more about expressing the equilibrium constant in terms of pressure.
• 13.6: Calculating the Extent of a Reaction
In chemistry, it's often convenient to predict what the outcome of a reaction will be in numerical terms. This section teaches you how to calculate the extent of a reaction - how much product will be formed.
• 13.7: Successive Approximation
An approximation is often useful even when it is not a very good one, because we can use the initial inaccurate approximation to calculate a better one. With practice, using this method of successive approximations is much faster than using the quadratic formula. It also has the advantage of being self-checking.
• 13.8: Predicting the Direction of a Reaction
Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibrium. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium. A useful tool for making such predictions is the reaction quotient, Q. Q has the same mathematical form as the equilibrium-constant expression, but Q is a ratio of the actual concentrations (not the equilibrium concentrations).
• 13.9: Le Chatelier’s Principle
Le Chatelier’s principle states that if a system is in equilibrium and some factor in the equilibrium conditions is altered, then the system will (if possible) adjust to a new equilibrium state so as to counteract this alteration to some degree.
• 13.10: The Effect of a Change in Pressure
In general, whenever a gaseous equilibrium involves a change in the number of molecules (Δn ≠ 0), increasing the pressure by reducing the volume will shift the equilibrium in the direction of fewer molecules. This applies even if pure liquids or solids are involved in the reaction.
• 13.11: The Effect of a Change in Temperature
Similar to a change in volume, a change in temperature forces a reaction to change in order to offset it's effect.
• 13.12: Effect of Adding a Reactant or Product
Just as varying temperature or volume can affect equilibrium, so can adding/subtracting a reaction/product. Read on to learn the specifics.
• 13.13: The Molecular View of Equilibrium
Chemical equilibrium can seem to be an unchanging phenomenon from a macroscopic perspective. Diving into the microscopic perspective, we find a different story. Read on to find out more.
13: Chemical Equilibrium
Chemical equilibrium is an essential concept in chemistry, allowing for an accurate understanding of the molecular processes behind those symbols in chemical equations. However, chemical equilibrium is easily misunderstood. To dispel any misconceptions before we begin our investigation of chemical equilibrium, we will first define what chemical equilibrium is not.
Chemical equilibrium is not when there are equal amounts or concentrations of reactants and products. Instead, chemical equilibrium is defined as the state at which the concentration of the reactants and products are constant, not necessarily equal.
Chemical equilibrium is not when the reaction "stops". Rather chemical equilibrium is achieved when the forward and reverse rates of a reaction are equal. We will learn more about that further along in the chapter, but before then it is essential to understand that chemical equilibrium is not a reaction coming to a stop.
Now that we have alerted you to the two biggest misconceptions that pop up when learning about chemical equilibrium, we can proceed to learning more about what exactly chemical equilibrium is and why it's important. For a preview to what we will be learning in the coming chapter, check out the video below or move on to the next section. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.01%3A_Prelude_to_Equilibria.txt |
A simple and instructive example of a chemical equilibrium is provided by the interconversion of the cis and trans isomers of difluoroethene:
${cis-}\text{C}_{2}\text{H}_{2}\text{F}{2}\rightleftharpoons {trans-}\text{C}_{2}\text{H}_{2}\text{F}_{2} \nonumber$
The two molecules involved in this equilibrium were illustrated in Figure $1$. The only difference between them is that in the cis isomer the two fluorine atoms are on the same side of the molecule, while in the trans isomer they are on opposite sides of the molecule. Although their molecules are so similar, these two isomers of difluoroethene are distinct chemical substances. They both condense to liquids at low temperatures, but these liquids have different boiling points. At room temperature both are gases, but they may be separated from each other and analyzed quantitatively by the technique of gas chromatography.
Other sections discuss that the barrier to free rotation about the bond prevents cis-C2H2F2 from changing rapidly into trans-C2H2F2. The same applies to the reverse reaction, conversion of the trans isomer to the cis. These reactions occur very slowly at higher temperatures, and even at 700 K (427°C), several weeks are required before equilibrium is reached and the concentrations of cis and trans species no longer vary with time. To study the reaction conveniently, a catalyst, such as I2(g) is added, speeding the reaction so that equilibrium is reached in a few minutes. When this is done, we always end up with a mixture which is slightly richer in the cis isomer. Furthermore, at a given temperature, the ratio of concentrations of the two isomers is always the same. For example, at 623 K the ratio
$\frac{\text{Equilibrium concentration of trans}}{\text{Equilibrium concentration of cis}}=\frac{[trans\text{-C}_{2}\text{H}_{2}\text{F}_{2}]}{[cis\text{-C}_{2}\text{H}_{2}\text{F}_{\text{2}}]}=\text{0.50 }\label{2}$
[In the second ratio in Eq. $\ref{2}$ square brackets are used to indicate the concentrations of trans C2H2F2 and cis-C2H2F2 once equilibrium has been reached.]
Apart from a change in temperature, nothing will alter this equilibrium ratio from 0.50. Whether we start with the pure cis isomer, the pure trans isomer, or even a mixture of isomers, the same ratio is obtained (see Figure $2$ ). Other variations, such as starting with half the amount of either isomer, changing the volume of the container, or heating the mixture to 1000 K and then cooling it to 623 K, are likewise without effect. Even adding a catalyst has no effect on the equilibrium ratio. If we heat cis-C2H2F2 and trans-C2H2F2 to 623 K with iodine added as a catalyst, the only difference is that equilibrium is achieved in a few minutes instead of a few weeks. The final composition is the same as in the uncatalyzed case.
We have described this equilibrium between the cis and trans isomers of difluoroethene because it demonstrates very clearly the four features which are characteristic of any chemical situation in which appreciable concentrations of reactants and products are in equilibrium with each other. These four features are
1. Even though the attainment of an equilibrium may be slow, once equilibrium has been achieved, the concentrations of all species participating in the equilibrium remain constant.
2. A chemical equilibrium can always be attained by approaching the equilibrium from more than one direction. We can begin with pure products or with pure reactants. Alternatively we can approach the equilibrium from either a higher or lower temperature.
3. The final equilibrium concentrations of the reactants are unaffected by the presence or absence of a catalyst.
4. There is always some mathematical relationship connecting the concentrations of the various species involved in the equilibrium mixture at a constant temperature. In the example just discussed, the concentrations were in a constant ratio. Usually the relationship is more complex, as we will see in the next section.
It is not always easy to tell when a chemical system is in a genuine equilibrium state. We often find mixtures of substances whose compositions do not change with time but which are not really in equilibrium with each other. A mixture of hydrogen and oxygen gas at room temperature is a good example. Although hydrogen and oxygen do react with each other to form water at room temperature, this reaction is so slow that no detectable change is apparent even after a few years. However, if an appropriate catalyst is added, the two gases react explosively and are converted completely to water according to the equation
$\text{2H}_{2}\text{ }({g}) + \text{O}_{2}\text{ } ({g})\rightarrow \text{2H}_{2}\text{O} \text{ }({l}) \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.02%3A_The_Equilibrium_State.txt |
The constancy of the ratio of the equilibrium concentration of one isomer to the concentration of the other at a given temperature is characteristic of all gaseous equilibria between isomers, i.e.,of all reactions of the general type
$\text{A}\text{ }({g})\rightleftharpoons \text{B}\text{ }({g})\label{1}$
The constant ratio of concentrations is called the equilibrium constant and is given the symbol $K_c$. For reactions of the type given by Equation $\ref{1}$ the equilibrium constant is thus described by the equation
$K_{c}=\frac{[\text{ B }]}{[\text{ A }]}\label{2}$
where, by convention, the concentration of the product B appears in the numerator of the ratio. If, for some reason, we wish to look at this reaction in reverse,
$\text{B}\text{ }({g})\rightleftharpoons \text{A}\text{ }({g}) \nonumber$
then the equilibrium constant is denoted as the reciprocal of the constant given in Equation $\ref{2}$.
$K_{c}=\frac{[\text{ A }]}{[\text{ B }]} \nonumber$
In general the equilibrium constant $K_c$ varies with temperature and also differs based on the substances involved. Examples illustrating this behavior are given in Table $1$ where the experimentally determined equilibrium constants for various cis-trans isomerization equilibria are recorded at various temperatures. Note that the equilibrium changes as the temperature and the composition of the molecule changes.
TABLE $1$ The Equilibrium Constant Kc for some Cis-Trans Interconversions.
When we turn our attention to more complex equilibrium reactions, we find that the relationship between the concentrations of the various species is no longer a simple ratio. A good demonstration of this fact is provided by the dissociation of dinitrogen tetroxide, N2O4. This compound is a colorless gas, but even at room temperature it dissociates partly into a vivid red-brown gas, NO2, according to the equation
$\text{N}_{2}\text{O}_{4}\text{ }({g})\rightleftharpoons \text{2NO}_{2} \text{ }({g})\label{5}$
If 1 mol N2O4 contained in a flask of volume 1 L is heated to 407.2 K, exactly one-half of it dissociates into NO2. If the volume is now increased, the ratio of [NO2] to [N2O4] does not remain constant but increases as more dissociates. As shown in Table $2$ , if we increase the volume still further, even more dissociation occurs. By the time we have increased the volume to 10 L, the fraction of N2O4 molecules dissociated has increased to 0.854 (i.e., to 85.4 percent).
Obviously the situation is now not quite so straightforward as in the previous example. Nevertheless there is a simple relationship between the equilibrium concentrations of the reactant and product in this case too. We find that it is the quantity
$\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]} \nonumber$
rather than the simple ratio of concentrations, which is now constant. Accordingly we also call this quantity an equilibrium constant and give it the symbol Kc. Thus Kc for Equation $\ref{5}$ is given by the relationship
$K_{c}=\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]} \nonumber$
where again by convention the product appears in the numerator. It is easy to check that Kc actually is a constant quantity with the value 2.00 mol/L from the data given in Table $2$ . Thus if we take the result from line d, we find that when 1 mol N2O4 is placed in a 10 L flask at 407 K, 0.854 mol dissociate.
TABLE $2$ The Dissociation of 1 mol $N_2O_4$ into NO2 at 407.2 K (134°C) and Various Volumes.
Amount of N2O4 Added
mol
Volume of Flask
L
Fraction N2O4 Dissociated
Amount N2O4 at Equilibrium
mol
Amount NO2 at Equilibrium
mol
Concentration N2O4 at Equilibrium
mol/L
Concentration NO2 at Equilibrium
mol/L
Equilibrium Constant Kc
mol/L
a 1 1 0.500 0.500 1.000 0.500 1.000 2.000
b 1 2 0.618 0.382 1.236 0.191 0.618 2.000
c 1 5 0.766 0.234 1.532 0.0468 0.3064 2.006
d 1 10 0.854 0.146 1.708 0.0146 0.1708 1.998
Since from Equation $\ref{5}$ each mole which dissociates yields 2 mol NO2,there will be
$\text{0.854 mol N}_{2}\text{O}_{4}\times \frac{\text{2 mol NO}_{2}}{\text{1 mol N}_{2}\text{O}_{4}}=\text{1.708 mol NO}_{2} \nonumber$
present in the reaction vessel. There will also be (1 – 0.854) mol = 0.146 mol N2O4 left undissociated in the flask. Since the total volume is 10 L, the equilibrium concentrations are
$[\text{ NO}_{2}]=\frac{\text{1.708 mol}}{\text{10 L}}=\text{0.1708 mol/L} \nonumber$
and
$[\text{ N}_{2}\text{O}_{4}]=\frac{\text{0.146 mol}}{\text{10 L}}=\text{0.0146 mol/L} \nonumber$
Accordingly
$K_{c}=\frac{\text{1.708 mol/L}\times \text{ 0.1708 mol/L}}{\text{0.0146 mol/L}}=\text{2.00 mol/L} \nonumber$
In exactly the same way, if we use the data from line a in Table $2$ , we find
$K_{c}=\frac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]}=\frac{\text{(1.00 mol/L}\text{)}^{2}}{\text{0.5 mol/L}}=\text{2.00 mol/L} \nonumber$
You can check for yourself that lines b and c also yield the same value forKc.
Example $1$ : Equilibrium Constant
When 2 mol N2O4 gas is heated to 407 K in a vessel of volume 5 dm3, it is found that 0.656 of the molecules dissociate into NO2. Show that these data are in agreement with the value for Kc of 2.00 mol dm–3 given in the text.
Solution
Many equilibrium problems can be solved in a fairly standardized fashion in three stages.
a) Calculate the amount of each substance transformed by the reaction as it comes to equilibrium, i.e., the amount of each reactant consumed by the reaction and the amount of each product produced by the reaction. Stoichiometric ratios derived from the equation must always be used in these calculations.
In this particular example we note that 0.656 of the original N2O4 dissociates. Since 2 mol was used, a total of 0.656 × 2 = 1.312 mol N2O4 is consumed. The amount of NO2 produced is accordingly
$n_{\text{NO}_{2}}=\text{1.312 mol N}_{2}\text{O}_{4}\times \dfrac{\text{2 mol NO}_{2}}{\text{1 mol N}_{2}\text{O}_{4}}=\text{2.624 mol NO}_{2}$
b) Use the amounts calculated in the first stage to calculate the amount of each substance present at equilibrium. Dividing by the volume, we can obtain the equilibrium concentrations.
Since 1.312 mol N2O4 dissociated out of an original 2 mol, we have (2 – 1.312) mol = 0.688 mol N2O4 left. The equilibrium concentration of N2O4 is thus
$[\text{ N}_{2}\text{O}_{4}] = \dfrac{0.688\text{ mol N}_{2}\text{O}_{4}}{\text{ 5.00 dm}^{3}} = \text{0.1376 mol dm}^{-3}$
Since no NO2 was originally present, the amount of NO2 present at equilibrium is the amount produced by the dissociation, namely, 2.624 mol NO2. Thus
$[\text{ NO}_{2}]=\dfrac{\text{2.624 mol NO}_{2}}{\text{5.00 dm}^{3}}=\text{0.525 mol dm}^{-3}$
It is usually worthwhile tabulating these calculations, particularly in more complex examples.
Note that a negative quantity in the column headed Amount Produced indicates that a given substance (such as N2O4 in this example) has been consumed. There is less of that substance when equilibrium is reached than was present initially.
Substance
Initial Amount
mol
Amount Produced
mol
Equilibrium Amount
mol
Equilibrium Concentration
mol dm-3
N2O4 2.00 -1.312 0.688 0.688/5
NO2 0.00 2.624 2.624 2.624/5
c) In the third stage we insert the equilibrium concentrations in an expression for the equilibrium constant:
$K_{c}=\dfrac{[\text{ NO}_{2}]^{2}}{[\text{ N}_{2}\text{O}_{4}]}=\dfrac{\text{0.525 mol dm}^{-3}\times \text{ 0.525 mol dm}^{-3}}{\text{0.1376 mol dm}^{-3}}=\text{2.00 mol dm}^{-3}$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.03%3A_The_Equilibrium_Constant.txt |
Two examples of an equilibrium constant dealt with in other sections, namely:
$K_{c}=\frac{[trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]}{[cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]} \nonumber$
for the reaction
$\text{cis-C}_2\text{H}_2\text{F}_2 \rightleftharpoons \text{trans-C}_2\text{H}_2\text{F}_2 \nonumber$
In the equation above, notice that both sides have the same molecular formula, yet in the image below, the elements involved are arranged differently. These molecules are called isomers and to learn more about them, check out the following site: https://www.thoughtco.com/geometric-isomerism-cis-and-trans-608702.
and
$K_{c}=\frac{[\text{ NO}_{\text{2}}]^{\text{2}}}{[\text{ N}_{\text{2}}\text{O}_{\text{4}}]} \nonumber$
for the reaction
$\text{N}_2\text{O}_4 \rightleftharpoons \text{2NO}_2 \nonumber$
In the equation above, we see the decomposition of N2O4 (the clear gas pictured below) into NO2, which is the brownish red gas seen below. As you can see in the molecular structure above, this decomposition of N2O4 occurs when the middle bond holding N2O4 together is broken, leaving two NO2 molecules. For more information on this specific equilibrium, check out the page on The Effect of a Change in Temperature.
Both the trans-cis transition of C2H2F2 and the decomposition of N2O4 are particular examples of a more general law governing chemical equilibrium in gases. If we write an equation for a gaseous equilibrium in general in the form
$aA(g) + bB(g) \rightleftharpoons cC(g) + dD(g) \label{1}$
then the equilibrium constant defined by the equation
$K_{c}=\frac{[\text{ C }]^{c}[\text{ D }]^{d}}{[\text{ A }]^{a}[\text{ B }]^{b}} \label{2}$
is found to be a constant quantity depending only on the temperature and the nature of the reaction. This general result is called the law of chemical equilibrium, or the law of mass action.
Example $1$: Equilibrium Constant
Write expressions for the equilibrium constant for the following reactions:
1. $\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber$
2. $\text{N}_2(g) + \text{3H}_2(g) \rightleftharpoons \text{2NH}_3(g) \nonumber$
3. $\text{O}_2(g) + \text{4HCl}(g) \rightleftharpoons \text{2H}_2\text{O}(g) + \text{2Cl}_2(g) \nonumber$
Solution
1. $K_{c}=\frac{[\text{ H}_{\text{2}}]\text{ }[\text{ I}_{\text{2}}]}{[\text{ HI }]^{\text{2}}} \nonumber$
2. $K_{c}=\frac{[\text{ NH}_{\text{3}}]^{\text{2}}}{[\text{ N}_{\text{2}}]\text{ }[\text{ H}_{\text{2}}]^{3}} \nonumber$
3. $K_{c}=\frac{[\text{ H}_{\text{2}}\text{O }]^{\text{2}}[\text{ Cl}_{\text{2}}]^{\text{2}}}{[\text{ O}_{\text{2}}]\text{ }[\text{ HCl }]^{\text{4}}} \nonumber$
Example $2$: Methane and Steam Equilibrium Constant
A mixture containing equal concentrations of methane and steam is passed over a nickel catalyst at 1000 K. The emerging gas has the composition [CO] = 0.1027 mol dm–3, [H2] = 0.3080 mol dm–3, and [CH4] = [H2O] = 0.8973 mol dm–3. Assuming this mixture is at equilibrium, calculate the equilibrium constant Kc for the reaction
$\text{CH}_4(g) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{3H}_2(g) \nonumber$
Solution
The equilibrium constant is given by the following equation:
$K_{c}=\frac{[\text{ CO }]\text{ }[\text{ H}_{\text{2}}]^{\text{3}}}{[\text{ CH}_{\text{4}}]\text{ }[\text{ H}_{\text{2}}\text{O }]}=\frac{\text{0}\text{.1027}^{-\text{3}}\times \text{ (0}\text{.3080}^{-\text{3}}\text{)}^{\text{3}}}{\text{0}\text{.8973}^{-\text{3}}\times \text{ 0}\text{.8973}^{-\text{3}}} \nonumber$
$=\text{3}\text{.727 }\times \text{ 10}^{-\text{3}} \nonumber$
Note: The yield of H2 at this temperature is quite poor. In the commercial production of H2 from natural gas, the reaction is run at a somewhat higher temperature where the value Kc is larger.
As the above example shows, the equilibrium constant Kc is a dimensionless quantity. This lack of units is a result of the derivation of Kc from Keq which uses the activities of the reactants and products in the equilibrium system instead of their concentrations. The activity of a substance is a measure of its effective concentration under specified conditions. While a detailed discussion of this important quantity is beyond the scope of an introductory text, it is necessary to be aware of a few important aspects:
• Activities are dimensionless (unitless) quantities and are in essence “adjusted” concentrations.
• For relatively dilute solutions, a substance's activity and its molar concentration are roughly equal. However, to obtain a unitless quantity, each molar concentration is divided by a standard state defined as 1 $\dfrac{mole}{dm^{3}}$. By using this standard state, the numerical value of the concentration remains unchanged, but the units cancel out.
• Activities for pure condensed phases (solids and liquids) are equal to 1.
• Activities for solvents in dilute solutions are equal to 1.
Thus, when we apply the equilibrium law to reactions which involve pure solids and pure liquids, we find in such cases that as long as some solid or liquid is present, the actual amount does not affect the position of equilibrium. Accordingly, only the molar concentrations of gaseous species are explicitly written in the expression for the equilibrium constant. For example, the equilibrium constant for the reaction
$\text{CaCO}_3(s) \rightleftharpoons \text{CaO}(s) + \text{CO}_2(g) \label{3}$
Notice below the molecular representation of the above reaction, with the ionic structure of CaCO3 'breaking apart' to form the simpler ionic structure CaO as well as gaseous CO2.
is given by the expression
$\text{K}_c = [\text{CO}_2] \label{4}$
in which only the concentration of the gas appears explicitly because the two solids each have an activity with a value equal to 1. Equation $\ref{4}$ suggests that if we heat CaCO3 to a high temperature so that some of it decomposes, the concentration of CO2 at equilibrium will depend only on the temperature and will not change if the ratio of amount of solid CaCO3 to amount of solid CaO is altered. Experimentally this is what is observed.
Example $3$ : Equilibrium Constant Expression
Write expressions for the equilibrium constants for the following reactions:
1. $\text{C}(s) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{H}_2(g) \nonumber$
2. $\text{C}(s) + \text{CO}_2(g) \rightleftharpoons \text{2CO}(g) \nonumber$
3. $\text{Fe}_3\text{O}_4(s) + \text{H}_2(g) \rightleftharpoons \text{3FeO}(s) + \text{H}_2\text{O}(g) \nonumber$
Solution
Since only gaseous species need be included, we obtain
1. $K_{c}=\frac{[\text{ CO }]\text{ }[\text{ H}_{\text{2}}]}{[\text{ H}_{\text{2}}\text{O }]} \nonumber$
2. $K_{c}=\frac{[\text{ CO }]^{\text{2}}}{[\text{ CO}_{\text{2}}]} \nonumber$
3. $K_{c}=\frac{[\text{ H}_{\text{2}}\text{O }]}{[\text{ H}_{\text{2}}]} \nonumber$
The equilibrium law can be shown experimentally to apply to dilute liquid solutions as well as to mixtures of gases, and the equilibrium-constant expression for a solution reaction can be obtained in the same way as for a gas-phase reaction.
Acetic acid, for example, reacts as follows when it dissolves in water:
$\text{CH}_3\text{COOH}(aq) + \text{H}_2\text{O}(l) \rightleftharpoons \text{CH}_3\text{COO}^{-}(aq) + \text{H}_3\text{O}^{+} (aq)\label{5}$
Notice in the figure below how the white hydrogen attached to the red oxygen on the acetic acid molecule is transferred to water, forming acetate (acetic acid minus a hydrogen) and a hydronium ion. Symbolically in the equation above, we can see this same transfer, except hydrogen is represented as an H rather than a white ball.
In solution only the concentrations of the solute species are explicitly shown in the equilibrium law. Even though the solvent may be a reactant or product, as long as the solution is dilute, any small amount of water produced or used by the chemical reaction will not change the amount of water to an appreciable extent because it is the solvent. As the solvent, water is assigned an activity equal to 1. Thus the activity of water is conventionally not explicitly written as part of the equilibrium law for a reaction in aqueous solution. Since it applies to a weak acid, Ka is called an acid constant. (The a stands for acid.) Other equilibrium constants in which the activity of the solvent water has a value of 1 and is thus not explicitly included in the equilibrium law are the base constant, Kb, for ionization of a weak base and the solubility product constant, Ksp, for dissolution of a slightly soluble compound.
Definition: Activity of the Solvent
The activity of the solvent in a dilute solution has a defined value of 1. This activity is part of the equilibrium law, but it is conventionally not written in the equilibrium law because its numerical value of 1 does not change the numerical value of the equilibrium constant.
Note
It is a common error to claim that the molar concentration of the solvent is in someway involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Kc = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Kc = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
Example $4$: Ionic Equilibria
Write out expressions for the equilibrium constants for the following ionic equilibria in dilute aqueous solution:
1. $\text{HF}(aq) + \text{H}_2\text{O} \rightleftharpoons \text{F}^{-}(aq) + \text{H}_3\text{O}^{+}(aq) \nonumber$
2. $\text{H}_2\text{O} + \text{NH}_3(aq) \rightleftharpoons \text{OH}^{-}(aq) + \text{NH}_4^{+}(aq) \nonumber$
3. $\text{H}_2\text{O} + \text{CO}_3^{2-}(aq) \rightleftharpoons \text{HCO}_3^{-}(aq) + \text{OH}^{-}(aq) \nonumber$
4. $\text{BaSO}_4(s) \rightleftharpoons \text{Ba}^{2+}(aq) + \text{SO}_4^{2-}(aq) \nonumber$
Solution
We do not explicitly include the activity of H2O in the first three examples and the activity of solid BaSO4 in the fourth because in all of these cases, the acivity of the specific species is equal to 1, and thus would not change the numerical value of the equilibrium constant..
1. $\text{K}_a = \text{K}_c = \frac{[\text{ H}_{\text{3}}\text{O}^{\text{+}}]\text{ }[\text{ F}^{-}]}{[\text{ HF }]} \nonumber$
2. $\text{K}_b = \text{K}_c = \frac{[\text{ NH}_{\text{4}}^{\text{+}}]\text{ }[\text{ OH}^{-}]}{[\text{ NH}_{\text{3}}]} \nonumber$
3. $\text{K}_b =\text{K}_c = \frac{[\text{ HCO}_{\text{3}}^{-}]\text{ }[\text{ OH}^{-}]}{[\text{ CO}_{\text{3}}^{\text{2}-}]} \nonumber$
4. $\text{K}_{sp} = \text{K}_c = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \nonumber$
Example $5$ : Acetic Acid
Measurements of the conductivities of acetic acid solutions indicate that the fraction of acetic acid molecules converted to acetate and hydronium ions is
1. 0.0296 at a concentration of 0.020 00 mol dm–3
2. 0.5385 at a concentration of 2.801 × 10–5 mol dm–3
Use these data to calculate the equilibrium constant for Equation (5) at each concentration.
Solution
Consider first 1 dm3 of solution a. This originally contained 0.02 mol CH3COOH of which the fraction 0.0296 has ionized. Thus (1 – 0.0296) × 0.02 mol undissociated CH3COOH is left, while 0.0296 × 0.02 mol H3O+ and CH3COO have been produced. In tabular form
Substance Original Amount Amount Produced Equilibrium Amount Equilibrium Concentration
CH3COOH 0.02 mol -0.0296×0.02 mol (0.02-0.000 592) mol 0.0194 mol dm-3
H3+ 0 mol +0.0296×0.02mol 0.000 592 mol 5.92×10-4 mol dm-3
CH3COO- 0 mol +0.0296×0.02 mol 0.000 592 mol 5.92×10-4 mol dm-3
Substituting into the expression for Ka gives
$K_{a}=\frac{[\text{ CH}_{\text{3}}\text{COO}^{-}]\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}]}{[\text{ CH}_{\text{3}}\text{COOH }]}=\frac{\text{(5}\text{.92 }\times \text{ 10}^{-\text{4}}\text{)}^{\text{2}}}{\text{0}\text{.0194}}=\text{1}\text{.81 }\times \text{ 10}^{-\text{5}} \nonumber$
A similar calculation on the second solution yields
$K_{a}=\frac{\text{(1}\text{.5083 }\times \text{ 10}^{-\text{5}}\text{)}^{\text{2}}}{\text{1}\text{.2926 }\times \text{ 10}^{-\text{5}}}=\text{1}\text{.760 }\times \text{ 10}^{-\text{5}} \nonumber$
Note
The two values of the equilibrium constant are only in approximate agreement. In more concentrated solutions the agreement is worse. If the concentration is 1 mol dm–3, for instance, Ka has the value 1.41 × 10–5. This is the reason for our statement that the equilibrium law applies to dilute solutions.
Acknowledgements
molview.org was used to create the molecular representations found on this webpage. Check out the site at the following link: Molview. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.04%3A_The_Law_of_Chemical_Equilibrium.txt |
Some equilibria involve physical instead of chemical processes. One example is the equilibrium between liquid and vapor in a closed container. In other sections we stated that the vapor pressure of a liquid was always the same at a given temperature, regardless of how much liquid was present. This can be seen to be a consequence of the equilibrium law if we recognize that the pressure of a gas is related to its concentration through the ideal gas law. Rearranging PV = nRT we obtain
$P=\frac{n}{V}RT=cRT\label{1}$
since c = amount of substance/volume = n/V. Thus if the vapor pressure is constant at a given temperature, the concentration must be constant also. Equation $\ref{1}$ also allows us to relate the equilibrium constant to the vapor pressure. In the case of water, for example, the equilibrium reaction and Kc are given by:
$\text{H}_2\text{O}(l) \rightleftharpoons \text{H}_2\text{O}(g) \nonumber$ $\text{K}_c = [\text{H}_2\text{O}(g)] \nonumber$
Substituting for the concentration of water vapor from Equation $\ref{1}$, we obtain
$K_{c}=\frac{P_{\text{H}_{\text{2}}\text{O}}}{RT} \nonumber$
At 25°C for example, the vapor pressure of water is 17.5 mmHg (2.33 kPa), and so we can calculate
$K_{c}=\frac{\text{2}\text{.33 kPa}}{\text{(8}\text{.314 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\text{)(298}\text{.15 K)}} \nonumber$
$=\text{9}\text{.40 }\times \text{ 10}^{-\text{4}}\text{ mol/L} \nonumber$
For some purposes it is actually more useful to express the equilibrium law for gases in terms of partial pressures rather than in terms of concentrations. In the general case:
$a\text{A}(g) + b\text{B}(g) \rightleftharpoons c\text{C}(g) + d\text{D}(g) \nonumber$
The pressure-equilibrium constant Kp is defined by the relationship:
$K_{p}=\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}} \nonumber$
where pA is the partial pressure of component A, pB of component B, and so on. Since pA = [A] × RT, pB = [B] × RT, and so on, we can also write as follows:
\begin{align} K_{p} & =\frac{p_{\text{C}}^{c}p_{\text{D}}^{d}}{p_{\text{A}}^{a}p_{\text{B}}^{b}}=\frac{\text{( }\!\![\!\!\text{ C }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{c}\text{( }\!\![\!\!\text{ D }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{d}}{\text{( }\!\![\!\!\text{ A }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{a}\text{( }\!\![\!\!\text{ B }\!\!]\!\!\text{ }\times \text{ }RT\text{)}^{b}} \ & =\frac{\text{ }\!\![\!\!\text{ C }\!\!]\!\!\text{ }^{c}\text{ }\!\![\!\!\text{ D }\!\!]\!\!\text{ }^{d}}{\text{ }\!\![\!\!\text{ A }\!\!]\!\!\text{ }^{a}\text{ }\!\![\!\!\text{ B }\!\!]\!\!\text{ }^{b}}\text{ }\times \text{ }\frac{\text{(}RT\text{)}^{c}\text{(}RT\text{)}^{d}}{\text{(}RT\text{)}^{a}\text{(}RT\text{)}^{b}} \ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\text{(}c\text{ + }d\text{ }-\text{ }a\text{ }-\text{ }b\text{ )}} \ & =K_{c}\text{ }\times \text{ (}RT\text{)}^{\Delta n} \end{align} \nonumber
Again $Δn$ is the increase in the number of gaseous molecules represented in the equilibrium equation. If the number of gaseous molecules does not change, Δn = 0, Kp = Kc, and both equilibrium constants are dimensionless quantities.
Example $1$ : SI Units
In what SI units will the equilibrium constant Kc be measured for the following reactions? Also predict for which reactions Kc = Kp.
1. $\text{2NOBr}(g) \rightleftharpoons \text{2NO}(g) + \text{Br}_2(g)$
2. $\text{H}_2\text{O}(g) + \text{C}(s) \rightleftharpoons \text{CO}(g) + \text{H}_2(g)$
3. $\text{N}_2(g) + \text{3H}_2(g) \rightleftharpoons \text{2NH}_3(g)$
4. $\text{H}_2(g) + \text{I}_2(g) \rightleftharpoons \text{2HI}(g)$
Solution
We apply the rule that the units are given by (mol dm–3)Δn.
1. Since Δn = 1, units are moles per cubic decimeter.
2. Since Δn = 1, units are moles per cubic decimeter (the solid is ignored).
3. Here Δn = -2 since two gas molecules are produced from four. Accordingly the units are mol–2 dm6
4. Since Δn = 0, Kc is a pure number. In this case also Kc = Kp | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.05%3A_The_Equilibrium_Constant_in_Terms_of_Pressure.txt |
Once we know the equilibrium constant for a reaction, we can calculate what will happen when any arbitrary mixture of reactants and products is allowed to come to equilibrium. To take a simple case: What would happen if we mixed 1 mol cis isomer with 1 mol trans isomer of difluoroethene in a 10-dm3 flask at 623 K? From what we know of this reaction, it is easy to guess that some of the trans isomer will be converted to cis isomer, since an equilibrium mixture of the isomers always contains more of the cis than of the trans form. If we want to know the exact amount of trans isomer converted in this way, we can use the equilibrium constant to calculate it.
In all calculations of this sort we need to concentrate on the increase in the amount of one of the products or the reactants. Since this quantity is unknown, we label it algebraically, calling it x mol, where x indicates a pure number.1 In the case under consideration, we would label the amount of cis isomer produced as the system moves to equilibrium as x mol. Once this step has been taken, the amount of each of the other products and reactants transformed by the reaction can be deduced from the equation and the appropriate stoichiometric factors. In our current case the equation is
$cis-\text{C}_2\text{H}_2\text{F}_2 \rightleftharpoons trans-\text{C}_2\text{H}_2\text{F}_2 \nonumber$
and it is obvious that if x mol cis isomer have been produced, x mol trans isomer have been consumed.
1If x is used to indicate a quantity rather than a number, the units become much more difficult to handle in the algebra which follows.
We can now construct a table showing the initial amounts, the amounts transformed by reaction, the amounts present at equilibrium, and finally the equilibrium concentrations of each product and reagent:
Substance Initial Amount Amount Produced Equilibrium Amount Equilibrium Concentration
cis-C2H2F2 1 mol x mol (1 + x) mol ((1 + x)/10) mol/L
trans-C2H2F2 1 mol -x mol (1 - x) mol ((1 - x)/10) mol/L
Once the final concentration of each species has been obtained in this way, an algebraic equation can be set up linking the equilibrium concentrations to the value of the equilibrium constant:
\begin{align}K_{c}=\text{0}\text{.500}=\frac{\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}=\frac{{\text{((1}-x\text{)}}/{\text{10) mol/L}}\;}{{\text{((1 + }x\text{)}}/{\text{10) mol/L}}\;} \ \text{0}\text{.500}=\frac{\text{1}-x}{\text{1 + }x} \ \end{align}
At this stage we are left with an algebraic equation to solve for x. Inevitably this equation involves only numbers. If any units remain, a mistake must have been made. From
by cross-multiplying, we have
$\text{1 – x} = \text{0.500 + 0.500x}$
which rearranges to
$\text{0.500} = \text{1.500x}$
so that $x=\dfrac{\text{0}\text{.500}}{\text{1}\text{.500}x}=\text{0}\text{.333}$
We thus conclude that 0.333 mol trans isomer is converted to the cis form when the original mixture is allowed to equilibrate. The final equilibrium concentrations are as follows:
$\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1 + }x}{\text{10}}\text{ mol/L}=\frac{\text{1}\text{.333}}{\text{10}}\text{ mol/L} \nonumber$
$[cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}]=\text{0}\text{.1333 mol/L}\, \nonumber$
While
$\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{\text{1}-x}{\text{10}}\text{ mol/L}=\text{0}\text{.0667 mol/L} \nonumber$
We can easily cross check that the ratio of these two concentrations is actually equal to the equilibrium constant, that is, to 0.5.
In calculating the extent of a chemical reaction from an equilibrium constant, it is often useful to realize that if the equilibrium constant is very small, the reaction proceeds to only a limited extent, while if it is very large, the reaction goes almost to completion. This point is easiest to see in the case of an equilibrium between two isomers of the type
$\text{A} \rightleftharpoons \text{B} \nonumber$
If Kc for this reaction is very small, say 10–6, then the ratio [B]/[A] = 10–6. There will thus be a million times more molecules of the A isomer than of the B isomer in the equilibrium mixture. For most purposes we can regard the equilibrium mixture as being pure A. Conversely if Kc has a very large value like 106, the very opposite is true. In an equilibrium mixture governed by this second constant, there would be a million times more B isomer than A isomer, and for most purposes the equilibrium mixture could be regarded as pure B.
The realization that an equilibrium mixture can contain only small concentrations of some of the reactants or products is often very useful in solving equilibrium problems, as the 3rd example below demonstrates.
Example $1$: Concentration
When colorless hydrogen iodide gas is heated, a beautiful purple color appears, indicating that some iodine gas has been produced and that the compound has decomposed partially into its elements according to the equation
$\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber$
At 745 K (471.8°C), Kc for this reaction has the value 0.0200. Calculate the concentration of I2 produced when 1.00 mol HI is heated to this temperature in a flask of volume 10.0 dm3. Also calculate the fraction of the HI which has dissociated.
Solution
Let us denote the amount of I2 produced by the reaction as x mol. The equation then tells us that the amount of H2 produced will also be x mol, while the amount of HI consumed by the decomposition will be 2x mol. The initial amount of HI, 1 mol, will thus be reduced to(1 – 2x) mol at equilibrium. Dividing the above amounts by the volume 10 dm³, we easily obtain the equilibrium concentrations
Substance Initial Amount Amount Produced Equilibrium Amount Equilibrium Concentration
I2 0.00 mol x mol x mol (x/10) mol dm-3
H2 0.00 mol x mol x mol (x/10) mol dm-3
HI 1.00 mol (-2)x mol (1 - 2x) mol ((1 - 2x)/10) mol dm-3
We can now write an expression for the equilibrium constant
$\text{K}_{c}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}}\;}\;} \nonumber$
or
$\text{0}\text{.0200}=\frac{x^{\text{2}}}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber$
which is the required algebraic expression, free of units. This equation is easily solved by taking the square root of both sides.
$\sqrt{\text{2 }\times \text{ 10}^{-\text{2}}}=\sqrt{\text{2}}\text{ }\times \text{ 10}^{-\text{1}}=\text{0}\text{.1414}=\frac{x^{\text{2}}}{\text{1 }-\text{ 2}x}$
Thus $\text{0.1414 – 0.2828x} = \text{x}$
or $\text{1.2828x} = \text{0.1414}$
so that $x=\frac{\text{0}\text{.1414}}{\text{1}\text{.2828}}=\text{0}\text{.110}$
Thus $\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{x}{\text{10}}\text{mol dm}^{-\text{3}}=\text{1}\text{.10 }\times \text{ 10}^{-\text{2}}\text{ mol dm}^{-\text{3}}$
Since 2x mol HI dissociated and 1 mol HI was originally present, we concluded that the fraction of HI which dissociated is
$\frac{\text{2}x\text{ mol}}{\text{1 mol}}=\text{0}\text{.220} \nonumber$
It is wise at this point to check the answer. We found x = 0.110. If this is the correct value, we should also find that
$\frac{x^{\text{2}}}{\text{(1}-\text{2}x\text{)}^{\text{2}}}=\text{0}\text{.0200} \nonumber$
The value obtained using a calculator is 0.019 888 2 (which rounds to 0.0199 to three significant figures). The difference is due to errors introduced by rounding off during the calculation.
Example $2$ : Equilibrium Mixture
A mixture of 1.00 mol HI pas and 1.00 mol H2 gas is heated in a 10.0-dm³ flask to 745 K. Calculate the concentration of I2 produced in the equilibrium mixture and also the fraction of HI which dissociates.
Solution:
Again we let x mol represent the amount of I2 produced. Our table then becomes
Substance Initial Amount Amount Produced Equilibrium Amount Equilibrium Concentration
I2 0.00 mol x mol x mol (x/10) mol dm-3
H2 1.00 mol x mol (1 + x) mol ((1 + x)/10) mol dm-3
HI 1.00 mol (-2x) mol (1 - 2x) mol ((1 - 2x)/10) mol dm-3
Substituting the final concentrations in an expression for the equilibrium constant, we then have
$K_{c}=\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}\text{ }\text{ }]\text{ }\!\!\!\!\text{ }\text{ }\!\!][\!\!\text{ I }_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{{\text{(1 + }x}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(}x}/{\text{10) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{((1}-\text{2}x\text{)}}/{\text{10) mol dm}^{-\text{3}}}\;}\;} \nonumber$
or $\text{0}\text{.0200}=\frac{\text{(1 + }x\text{)}x}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber$
Because of the added H2, it is no longer possible to take a square root as in the previous example. Instead we need to multiply out and rearrange in order to obtain a quadratic equation of the form ax2 + bx + c = 0. Accordingly we have
$\text{0.0200(1 – 2x)}^{2} = \text{(1 + x)x}$
or $\text{0.0200(1 – 4x + 4x}^{2}) = \text{x} + \text{x}^{2}$
multiplying both sides by 50, we obtain
$\text{1 – 4x} + \text{4x}^{2} = \text{50x} + \text{50x}^{2}$
which on rearrangement has the required form
$\text{46x}^{2} + \text{54x – 1} = \text{0} \nonumber$
where
• a = 46
• b = 54
• c = – 1
We can now use the conventional quadratic formula
$x=\frac{-\text{b }\!\!\pm\!\!\text{ }\sqrt{\text{b}^{\text{2}}-\text{4}ac}}{\text{2}a}=\frac{-\text{54 }\!\!\pm\!\!\text{ }\sqrt{\text{54}^{\text{2}}\text{ + 4 }\times \text{ 46 }\times \text{ 1}}}{\text{2 }\times \text{ 46}}$
$=\frac{-\text{54 }\!\!\pm\!\!\text{ 55}\text{.678}}{\text{92}}=\text{0}\text{.0182 or }-\text{1}\text{.192}$
The negative root, x = – 1.192, implies that 1.192 mol I2 was consumed. Since no I2 was present to begin with, this is impossible. We conclude that the positive root, namely x = 0.0182, is the correct one. Thus
$\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }=\frac{x}{\text{10}}\text{mol dm}^{-\text{3}}=\text{1}\text{.82 }\times \text{ 10}^{-\text{3}}\text{ mol dm}^{-\text{3}} \nonumber$
Again the fraction dissociated is given by 2x mol/1 mol and is thus equal to 0.0364. To check this solution, we can substitute x = 0.0182 in the expression
$\frac{\text{(1 + }x\text{)}x}{\text{(1}-\text{2}x\text{)}^{\text{2}}} \nonumber$
We then obtain the value 0.019 96, which rounds to the correct value of 0.0200.
Note:
The inclusion of one of the products (H2 gas) in the mixture reduces the extent to which the hydrogen iodide dissociates quite appreciably. We will explore this phenomenon more extensively in the next section.
Example $3$ : Fraction of Dissociation
The equilibrium constant Kc for the dissociation of dinitrogen tetroxide according to the equation
$\text{N}_2\text{O}_4(g) \rightleftharpoons \text{2NO}_2(g) \nonumber$
changes from a very small to a very large value as the temperature is increased, as shown in the table. Calculate the fraction of N2O4 dissociated at
Temperature/K Kc/mol×dm-3
200 1.09×10-7
400 1.505
600 1.675×103
each temperature if 1.00 mol N2O4 is sealed in a container of volume 4.00 dm3.
Solution
Let the amount of N2O4 dissociated at the temperature under consideration be x mol. From the equation, 2x mol NO2 will be produced. In tabular form we then have
Substance Initial Amount Amount Produced Equilibrium Amount Equilibrium Concentratioin
N2O4 1.00 mol -x mol (1-x) mol ((1-x)/4) mol×dm-3
NO2 0.00 mol +2x mol 2x mol (2x/4) mol×dm-3
We thus have
$\text{K}_{c}=\frac{\text{ }\!\![\!\!\text{ NO}_{\text{2}}\text{ }\!\!]\!\!\text{ }^{\text{2}}}{\text{ }\!\![\!\!\text{ N}_{\text{2}}\text{O}_{\text{4}}\text{ }\!\!]\!\!\text{ }}=\frac{{\text{(2}x}/{\text{4) mol dm}^{-\text{3}}\text{ }\times \text{ }{\text{(2}x}/{\text{4) mol dm}^{-\text{3}}\text{ }}\;}\;}{{\text{((1}-x\text{)}}/{\text{4) mol dm}^{-\text{3}}}\;} \nonumber$
$=\frac{\text{4}x^{\text{2}}}{\text{16}}\text{ }\times \text{ }\frac{\text{4}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}} \nonumber$
a) At 200 K, Kc = 1.09 × 10–7 mol dm–3, so that
$\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}\text{ mol dm}^{-\text{3}}$
or $\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}$
Since Kc is so small, we guess that very little N2O4 is dissociated at this temperature. This means that x is very small, and it is probably valid to make the approximation
1 – x ≈ 1
(The symbol ≈ means approximately equal.) With this approximation our equation becomes:
$\text{x}^{2} = \text{1.09 × 10}^{-7} mol$
or $x =\sqrt{\text{1}\text{.09 }\times \text{ 10}^{-\text{8}}}\text{ mol}=\sqrt{\text{1}\text{.09}}\text{ }\times \text{ 10}^{-\text{4}}=\text{3}\text{.30 }\times \text{ 10}^{-\text{4}} \nonumber$
Our guess about x was thus correct. It is a small number, especially in comparison with 1. The approximation 1 – x ≈ 1 is valid to three decimal-place accuracy since 1 – x = 0.9997.
Since x mol has dissociated, the fraction of N2O4 dissociated is given by x mol/1 mol = x. The fraction dissociated is thus 3.30 × 10-4.
b) At 400 K, Kc = 1.505 mol dm-3, so that:
$\frac{x^{\text{2}}}{\text{1}-x}\text{ mol dm}^{-\text{3}}=\text{1}\text{.505 mol dm}^{-\text{3}}$
or $\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.505}$
Since Kc is not small, we can no longer use the approximation:
1 – x ≈ 1
Indeed, such an assumption leads to a ridiculous conclusion, since if 1 – x ≈ 1, then x2 = 1.505, or x = 1.227. This result is impossible since it tells us that more N2O4 has dissociated (1.227 mol) than was originally present (1 mol).
Accordingly we return to the orthodox method for solving quadratic equations. Multiplying out our equation, we obtain:
$\text{x}^{2} = \text{1.505 – 1.505x}$
or $\text{x}^{2} + \text{1.505x – 1.505} = \text{0}$
so that $x=\frac{-\text{1}\text{.505 }\!\!\pm\!\!\text{ }\sqrt{\text{(1}\text{.505)}^{\text{2}}\text{ + 4(1}\text{.505)}}}{\text{2}}=\frac{-\text{1}\text{.505 }\!\!\pm\!\!\text{ 2}\text{.878}}{\text{2}} \nonumber$
= − 2.19 or 0.6865
Since a negative result has no physical meaning, we conclude that 0.6865 is the correct answer. (As a cross check we can feed this result back into our original equation.) The fraction of N2O4 dissociated at this temperature is accordingly 0.6865.
c) At 600 K,Kc = 1.675 × 103 mol dm–3, so that:
$\frac{x^{\text{2}}}{\text{1}-x}=\text{1}\text{.675 }\times \text{ 10}^{\text{3}}$
Since Kc is fairly large, we guess that almost all the N2O4 has dissociated at this temperature and that x is accordingly close to 1. A valid approximation for these circumstances is then x ≈1. With this approximation our equation becomes:
$\frac{1}{\text{1}-x}=\text{1}\text{.675 }\times \text{ 10}^{\text{3}}$
or $\text{1-x}=\frac{1}{\text{1}\text{.675 }\times \text{ 10}^{\text{3}}}=\text{5}\text{.97 }\times \text{ 10}^{-\text{4}}$
Thus $\text{x} = \text{1 – 5.97 × 10}^{-4} = \text{0.999 40} \nonumber$
and our approximation is a good one. Since the fraction dissociated is x, we conclude that 0.9994 of the original N2O4 has dissociated.
Note:
We did not use the approximation x ≈ 1 to say that 1 – x ≈ 0.This is not a productive way to solve the problem because we would end up dividing by zero. Common sense tells us that it will not work. Similarly, in part a of this example we used the fact that x was very small to say that 1 – x ≈ 1. Saying that x2 ≈ 0 would not make sense because it would lead to the algebraic equation
$\frac{0}{\text{1}-x}=\text{1}\text{.09 }\times \text{ 10}^{-\text{7}}$
that is, that 0 = 1.09 × 10–7 – (1.09 × 10–7)x. This would lead to the solution x = 1 which is rather far from the original assumption. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.06%3A_Calculating_the_Extent_of_a_Reaction.txt |
An approximation is often useful even when it is not a very good one, because we can use the initial inaccurate approximation to calculate a better one. A good example of this occurs in Example 2 from Calculating the Extent of Reactions where it was necessary to solve the equation
$\text{0}\text{.0200}=\frac{\text{(1 + }x\text{)}x}{\text{(1 }-\text{ 2}x\text{)}^{\text{2}}} \nonumber$
The conditions of the problem suggest that x is so small that it can be ignored.
1. Accordingly we can approximate
$1 + x \approx 1 \approx 1 – 2x \nonumber$
from which we obtain the approximate result (called the first approximation), x1:
$\text{0}\text{.0200 }\approx \text{ }\frac{\text{1}x_{\text{1}}}{\text{1}^{\text{2}}} ~~~ \text{or} ~~~ x_{1} \approx 0.0200 \nonumber$
Although for some purposes this is a sufficiently accurate result, a much better approximation can be obtained by feeding this one back into the formula. If we write the formula as
$x=\frac{\text{0}\text{.0200 (1 - 2}x\text{)}^{\text{2}}}{\text{1 + }x} \nonumber$
we can now substitute x1 = 0.0200 on the right-hand side, giving the second approximation:
$x_{\text{2}}=\frac{\text{0}\text{.0200 (1 - 2 }\times \text{ 0}\text{.0200)}^{\text{2}}}{\text{1 + 0}\text{.0200}}=\frac{\text{0}\text{.0200 }\times \text{ 0}\text{.96}^{\text{2}}}{\text{1}\text{.0200}}=\text{0}\text{.0181} \nonumber$
If we repeat this process, a third approximation is obtained:
$x_{3} \approx 0.0182 \nonumber$
in exact agreement with the accurate result obtained from the quadratic formula in the example.
With practice, using this method of successive approximations is much faster than using the quadratic formula. It also has the advantage of being self-checking. A mistake in any of the calculations almost always leads to an obviously worse approximation. In general, if the last approximation for x differs from the next to last by less than 5 percent, it can be assumed to be accurate, and the successive-approximation procedure can be stopped. In the example just given,
$\frac{x_{\text{2}}\text{ }-\text{ }x_{\text{1}}}{x_{\text{1}}}=\frac{\text{0}\text{.0181 - 0}\text{.0200}}{\text{0}\text{.0200}}=\frac{-\text{0}\text{.0019}}{\text{0}\text{.0200}}\approx -\text{0}\text{.10}=-\text{10 percent} \nonumber$
so a third approximation was calculated. This third approximation was almost identical to the second and so was taken as the final result.
13.08: Predicting the Direction of a Reaction
Often you will know the concentrations of reactants and products for a particular reaction and want to know whether the system is at equilibrium. If it is not, it is useful to predict how those concentrations will change as the reaction approaches equilibrium. A useful tool for making such predictions is the reaction quotient, Q. Q has the same mathematical form as the equilibrium-constant expression, but Q is a ratio of the actual concentrations (not a ratio of equilibrium concentrations).
For example, suppose you are interested in the reaction
$2 \text{SO}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{SO}_{3} (g) \nonumber$
$K_{\text{c}} = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = 245 \text{ mol/L} \text{(at 1000 K)} \nonumber$
and you have added 0.10 mol of each gas to a container with volume 10.0 L. Is the system at equilibrium? If not, will the concentration of SO3 be greater than or less than 0.010 mol/L when equilibrium is reached? You can answer these questions by calculating Q and comparing it with Kc. There are three possibilities:
• If Q = Kc then the actual concentrations of products (and of reactants) are equal to the equilibrium concentrations and the system is at equilibrium.
• If Q < Kc then the actual concentrations of products are less than the equilibrium concentrations; the forward reaction will occur and more products will be formed.
• If Q > Kc then the actual concentrations of products are greater than the equilibrium concentrations; the reverse reaction will occur and more reactants will be formed.
For the reaction given above,
$Q = \frac{[\text{SO}_{3}]^{2}}{[\text{SO}_{2}]^{2} [\text{O}_{2}]} = \frac{(\frac{0.1 mol}{10 L})^{2}}{(\frac{0.1 mol}{10 L})^{2} (\frac{0.1 mol}{10 L})} = 100 \frac{\text{mol}}{\text{L}} \nonumber$
(In the expression for Q each actual concentration is enclosed in braces {curly brackets} in order to distinguish it from the equilibrium concentrations, which, in the Kc expression, are enclosed in [square brackets].) In this case Q = 100 mol/L. This is less than Kc, which has the value 245 mol/L. This implies that the concentrations of products are less than the equilibrium concentrations (and the concentrations of reactants are greater than the equilibrium concentrations). Therefore the reaction will proceed in the forward direction, producing more products, until the concentrations reach their equilibrium values.
Example $1$: Equilibrium
At 2300 K, the equilibrium constant, Kc, is 1.7 x 10-3 for the reaction
$\text{N}_{2} (g) + \text{O}_{2} (g) \rightleftharpoons 2 \text{NO} (g) \nonumber$
A mixture of the three gases at 2300 K has these concentrations, [N2] = 0.17 mol dm-3, [O2] = 0.17 mol dm-3, and [NO] = 0.034 mol dm-3.
(a) Is the system at equilibrium?
(b) In which direction must the reaction occur to reach equilibrium?
(c) What are the equilibrium concentrations of N2, O2, and NO?
Solution
Use the known concentrations to calculate Q. Compare Q with Kc to answer questions (a) and (b). Use an ICE table to answer part (c).
$Q = \frac{\{\text{NO\}}^{2}}{\{\text{N}_{2}\}\{\text{O}_{2}\}} = \frac{(0.034 \text{mol dm}^{-3})^{2}}{(0.17 \text{mol dm}^{-3})(0.17 \text{mol dm}^{-3})} = 4.0 \times 10^{-2} \nonumber$
1. Q is larger than Kc, so the reaction is not at equilibrium.
2. Because Q is larger than Kc, the concentration of the product, NO, is larger than its equilibrium concentration and the concentrations of the reactants, N2 and O2, are smaller than their equilibrium concentrations. Therefore some of the product, NO, will be consumed and more of the reactants, N2 and O2, will be formed.
3. Use the given concentrations as the initial concentrations of reactants and product. Enter these into an ICE table. Let x be the increase in the concentration of N2 as the system reacts to equilibrium. The ICE table looks like this:
N2 O2 NO
Initial concentration/mol dm-3 0.17 0.17 0.034
Change in concentration/mol dm-3 x x -2x
Equilibrium concentration/mol dm-3 0.17 + x 0.17 + x 0.034 - 2x
Next, substitute the equilibrium concentrations into the Kc expression and solve for x
$K_{\text{c}} = 1.7 \times {10^{ - 3}} = \frac{(0.034 - 2x)^{2}}{(0.17 + x)(0.1 + x)} \nonumber$
Now take the square root of both sides of this equation. This gives
$\sqrt {1.7 \times {10^{ - 3}}} = 0.0412 = \dfrac{0.034 - 2x}{0.17 + x}$
Multiplying both sides by 0.17 + x gives
$0.0070 + 0.412 x = 0.034 - 2 x$
$2 x + 0.0412 x =0.034 - 0.0070$
$x = \dfrac{0.0270}{2.0412} = 0.0132$
$\text{[N}_{2} ] = \text{[O}_{2} ] = 0.17 + 0.013 = 0.183 \text{ mol dm}^{-3}$
$\text{[NO] } = 0.034 - 2(0.0132) = 0.0076 \text{ mol dm}^{-3}$
Check the result by substituting these concentrations into the equilibrium constant expression.
$K_{\text{c}} = \frac{(0.0076)^{2}}{(0.18)(0.18)} = 1.8 \times {10^{ - 3}} \nonumber$
This agrees to two significant figures with the Kc value of 1.7 x 10-3. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.07%3A_Successive_Approximation.txt |
Often it is useful to predict qualitatively (without doing calculations such as those just described) what will happen to a system at equilibrium when conditions such as temperature or volume change or when a reactant or product is added or removed from the reaction mixture. Fortunately a simple rule, Le Chatelier’s principle, enables us to make such qualitative predictions. This rule states that if a system is in equilibrium and some factor in the equilibrium conditions is altered, then the system will (if possible) adjust to a new equilibrium state so as to counteract this alteration to some degree.
13.10: The Effect of a Change in Pressure
As an example of the application of Le Chatelier's principle, consider the effect of tripling the pressure on an equilibrium mixture of NO2 and N2O4:
$\text{N}_2\text{O}_4(g) \rightleftharpoons \text{2NO}_2(g) \label{1}$
This could be done using the piston and cylinder shown in Figure $1$, in which case tripling the pressure would be expected to reduce the volume of the mixture to one-third its former value. Under the new conditions, however, Le Chatelier’s principle tells us that a new equilibrium will be achieved which counteracts the alteration of pressure. That is, the concentrations of N2O4 and NO2 should change in such a way as to lessen the pressure increase. This can happen if some of the NO2 reacts to form N2O4 because two molecules of NO2 are consumed for every one molecule of N2O4 produced. This reduction in the number of gas molecules will reduce the pressure at the new volume. Thus Le Chatelier’s principle predicts that the reverse of Equation $\ref{1}$ will occur, producing more N2O4 and using up some NO2. We say that increasing the pressure on the N2O4and NO2 causes the equilibrium to shift to the left. This agrees with the experimental data on this equilibrium, already given in Table 2 in The Equilibrium Constant. Note that the value of the equilibrium constant remains the same, even though the equilibrium shifts.
Notice that the effect just described occurs because the gas volume decreased and the concentrations of NO2 and N2O4 both increased. If we had increased the total pressure on the equilibrium system by pumping in an inert gas such as N2(g), the volume would have remained the same, as would the partial pressures and concentrations of NO2 and N2O4. In such a case no shift in the equilibrium would be expected. Note also the words if possible in the statement of Le Chatelier’s principle. If the equilibrium reaction had not involved a change in the number of molecules in the gas phase, no shift in the concentrations could have made any difference in the pressure. Thus for the reaction
$\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber$
changing the pressure by changing the size of the container would have no effect. You can check this for yourself by redoing Example 1 from Calculating the Extent of a Reaction using several different volumes and the same initial amount of HI.
In general, whenever a gaseous equilibrium involves a change in the number of molecules (Δn ≠ 0), increasing the pressure by reducing the volume will shift the equilibrium in the direction of fewer molecules. This applies even if pure liquids or solids are involved in the reaction. An example is the reaction
$\text{C}(s) + \text{H}_2\text{O}(g) \rightleftharpoons \text{CO}(g) + \text{H}_2(g) \nonumber$
in which superheated steam is passed over carbon obtained from coal to produce carbon monoxide and hydrogen. Since the volume of solid carbon is negligible compared with the volumes of the gases, we need consider only the latter. Hence Δn = 1 (two gas molecules on the right for every one on the left), and an increase in pressure should favor the reverse reaction. This reaction is an important industrial process, and for the reason we have just outlined, is carried out at a low pressure.
The video below demonstrates what this process looks like, using the reaction of N2O4 and NO2 once again. Here we see visual confirmation of the fact that an increase in the volume of a container causes a shift in the equilibrium based on the number of moles of gas, according to Le Chatelier's principle.
13.11: The Effect of a Change in Temperature
In a chemical equilibrium there is almost always a difference in energy, and hence in enthalpy, between the reactants and the products. The thermochemical equation for dissociation of N2O4 for example, is
$\text{N}_2\text{O}_4(g) \rightleftharpoons \text{2NO}_2(g) \nonumber$
ΔHm = 54.8 kJ mol–1
Because of this enthalpy difference, any shift in the equilibrium toward further dissociation will result in the absorption of heat energy and a momentary decrease in temperature. Conversely, a shift in the reverse direction will cause a small rise in temperature. If we increase the temperature of a mixture of N2O4 and NO2, the mixture should respond in such a way as to oppose the rise in temperature.
This can happen if some N2O4 in the mixture dissociates, since the resulting absorption of energy will produce a cooling effect. We would therefore expect that by raising the temperature of the equilibrium mixture, we would shift the equilibrium in favor of dissociation. Indeed we see in the result of Example 3 from Calculating the Extent of a Reaction that raising the temperature from 200 to 600 K changes an equilibrium mixture which is almost pure N2O4 into an equilibrium mixture which is almost pure NO2.
In the general case, if we raise the temperature of any mixture of species which are in chemical equilibrium with each other, Le Chatelier's principle tells us that we will shift the equilibrium in the direction of those species with the higher energy. Thus, if the reaction is endothermic, as in the dissociation just discussed, raising the temperature will swing the equilibrium toward the products, and the value of the equilibrium constant Kc will increase with temperature.
Conversely, if the reaction is exothermic, a rise in temperature will favor the reactants, and Kc will get smaller as the temperature increases. We can also turn the argument around. If we find a reaction for which Kc increases with temperature, we know immediately that the reaction must be endothermic. Conversely, if Kc decreases as temperature increases, the reaction must be exothermic. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.09%3A_Le_Chateliers_Principle.txt |
If we have a system which is already in equilibrium, addition of an extra amount of one of the reactants or one of the products throws the system out of equilibrium. Either the forward or the reverse reaction will then occur in order to restore equilibrium conditions. We can easily tell which of these two possibilities will happen from Le Chatelier's principle. If we add more of one of the products, the system will adjust in order to offset the gain in concentration of this component. The reverse reaction will occur to a limited extent so that some of the added product can be consumed. Conversely, if one of the reactants is added, the system will adjust by allowing the forward reaction to occur to some extent. In either case some of the added component will be consumed.
We see this principle in operation in the case of the decomposition of HI at high temperatures:
$\text{2HI}(g) \rightleftharpoons \text{H}_2(g) + \text{I}_2(g) \nonumber$
In Example 1 from Calculating the Extent of a Reaction we saw that if 1 mol HI is heated to 745 K in a 10 L flask, some of the HI will decompose, producing an equilibrium mixture of composition 1 $\text{[HI]} = \text{0.0780 mol/L}$ ; $\text{[I}_2] = \text{0.0110 mol/L}$; $\text{[H}_2] = \text{0.0110 mol/L}$ This is a genuine equilibrium mixture since it satisfies the equilibrium law
If an extra mole of H2 is added to this mixture, the concentrations become 2 $\text{[HI]} = \text{0.0780 mol/L}$ ; $\text{[I}_2] = \text{0.0110 mol/L}$ ; $\text{[H}_2] = \text{0.111 mol/L}$ The system is no longer in equilibrium (hence the lack of square brackets to denote equilibrium concentrations) as we can easily check from the equilibrium law
The addition of H2 has increased the concentration of this component. Accordingly, Le Chatelier’s principle predicts that the system will achieve a new equilibrium in such a way as to reduce this concentration. The reverse reaction occurs to a limited extent. This not only reduces the concentration of H2 but the concentration of I2 as well. At the same time the concentration of HI is increased. The system finally ends up with the concentrations calculated in Example 2 from Calculating the Extent of a Reaction, namely,
$\text{H}_2+ \text{I}_2 \rightarrow \text{2HI} \nonumber$
3 $\text{[HI]} = \text{0.0963 mol/L}$ ;$\text{[I}_2] = \text{0.001 82 mol/L}$ ; $\text{[H}_2] = \text{0.1018 mol/L}$ This is again an equilibrium situation since it conforms to the equilibrium law
$\frac{\text{ }\!\![\!\!\text{ H}_{\text{2}}]\text{ }\!\!]\!\!\text{ }\!\![\!\!\text{ I}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ HI }\!\!]\!\!\text{ }^{\text{2}}}=\frac{\text{0}\text{.1018 mol L}^{-\text{1}}\text{ }\times \text{ 0}\text{.001 82 mol L}^{-\text{1}}}{\text{(0}\text{.0963 mol L}^{-\text{1}}\text{)}^{\text{2}}\text{ }}=\text{0}\text{.02}=K_{c} \nonumber$
The way in which this system responds to the addition of H2 is also illustrated schematically in Figure 1. The actual extent of the change is exaggerated in this figure for diagrammatic effect.
Le Chatelier's principle can also be applied to cases where one of the components is removed. In such a case the system responds by producing more of the component removed. Consider, for example, the ionization of the weak diprotic acid H2S:
$\ce{H_{2}S + 2H_{2}O \rightleftharpoons 2H_{3}O^{+} + S^{2-}} \nonumber$
Since H2S is a weak acid, very few S2– ions are produced, but a much larger concentration of S2–ions can be obtained by adding a strong base. The base will consume most of the H3O+ ions. As a result, more H2S will react with H2O in order to make up the deficiency of H3O+, and more S2–ions will also be produced. This trick of removing one of the products in order to increase the concentration of another product is often used by chemists, and also by living systems.
Previously, we've investigated the effect of adding/subtracting a product/reactant in mathematical terms. The video below allows you to visually see the changes the equilibrium shifts that occur upon the addition of a reactant or product, courtesy of the North Caroline School of Science and Mathematics.
Example $1$ : Yield
When a mixture of 1 mol N2 and 3 mol H2 is brought to equilibrium over a catalyst at 773 K (500°C) and 10 atm (1.01 MPa), the mixture reacts to form NH3 according to the equation
$\text{N}_2(g) + \text{3H}_2(g) \rightleftharpoons \text{2NH}_3(g)$ ΔHm = – 94.3 kJ The yield of NH3, however, is quite small; only about 2.5 percent of the reactants are converted. Suggest how this yield could be improved (a) by altering the pressure; (b) by altering the temperature; (c) by removing a component; (d) by finding a better catalyst.
Solution:
a) Increasing the pressure will drive the reaction in the direction of fewer molecules. Since Δn = - 2, the forward reaction will be encouraged, increasing the yield of NH3. b) Increasing the temperature will drive the reaction in an endothermic direction, in this case in the reverse direction. In order to increase the yield, therefore, we need to lower the temperature. c) Removing the product NH3 will shift the reaction to the right. This is usually done by cooling the reaction mixture so that NH3(l) condenses out. Then more N2(g) and H2(g) are added, and the reaction mixture is recycled to a condition of sufficiently high temperature that the rate becomes appreciable. d) While a better catalyst would speed up the attainment of equilibrium, it would not affect the position of equilibrium. It would therefore have no effect on the yield.
Note: As mentioned in Chaps. 3 and 12, NH3 is an important chemical because of its use in fertilizers. In the design of a Haber-process plant to manufacture ammonia, attempts are made to use as high a pressure and as low a temperature as possible. The pressure is usually of the order of 150 atm (15 MPa), while the temperature is not usually below 750 K. Although a lower temperature would give a higher yield, the reaction would go too slowly to be economical, at least with present-day catalysts. The discoverer of a better catalyst for this reaction would certainly become a millionaire over-night.
Acknowledgements:
Two of the equations in this article were created using help from https://www.latex4technics.com/. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.12%3A_Effect_of_Adding_a_Reactant_or_Product.txt |
In other sections, we show that equilibrium constants exist, how they can be determined, and how they can be used, but we have not explained why they are sometimes large and sometimes small. In order to answer this and related questions we must switch our point of view from the macroscopic to the microscopic level.
The cis-trans isomerism of difluoroethene provides a simple example for examination on the molecular level. As we saw earlier, if 1 mol cis isomer of this compound is heated to 623 K, it is gradually transformed into the equilibrium mixture of ⅔ mol cis isomer and ⅓ mol trans isomer. From a macroscopic point of view, we might conclude that the reaction stops once equilibrium is attained, but on the microscopic level, such a statement is obviously untrue.
As cis molecules move around a container at 623 K, they occasionally suffer a collision with sufficient energy to flip them over to the trans conformation ( for a microscopic view of this process, see Figure $1$ ). This conversion process does not suddenly stop when equilibrium has been attained. There are plenty of cis molecules present in the equilibrium mixture, and they continue to collide with their fellow molecules and hence to flip over from the cis to the trans conformation. Thus the forward reaction
$cis-\text{C}_2\text{H}_2\text{F}_2 \rightarrow trans-\text{C}_2\text{H}_2\text{F}_2 \nonumber$
must continue to occur in the equilibrium mixture.
If this were the only reaction taking place in the flask, the eventual result would be that all the cis isomer would be converted to the trans isomer and no equilibrium mixture would result. This does not happen because the reverse reaction
$trans-\ce{C_{2}H_{2}F_{2} \rightarrow} cis-\ce{C_{2}H_{2}F_{2}} \nonumber$
is also occurring at the same time. The trans molecules in the equilibrium mixture are also continually being bombarded by their fellow molecules. Many of these collisions have sufficient energy to flip the trans molecule back into the cis form. Thus cis molecules are not only being consumed but also being produced in the flask. The concentration of the cis molecules remains constant because the rate at which they are being consumed is exactly balanced by the rate at which they are being produced.
In other words, for every cis molecule which flips to the trans conformation in one part of the container, there will be on the average a molecule of trans isomer flipping in the reverse direction in another part of the container. This constant reshuffle of molecules between the cis and the trans forms can then continue indefinitely without any net change in the concentration of either species.
This molecular interpretation of equilibrium is not confined to the particular example of cis-trans isomerism or even just to chemical reactions. In the section on phase transitions we describe vapor-liquid equilibria in which the condensation of vapor just balanced the evaporation of liquid. Any state of equilibrium corresponds to a situation in which the rate at which the forward reaction is occurring is exactly equal to the rate at which the reverse reaction is occurring. Since these rates are equal, there is no net change in the concentration of any of the reactants or products with time. Because an equilibrium corresponds to a balance between the forward and the reverse reaction in this way, we use double arrows ($\rightleftharpoons$) in the equation.
As another example of chemical equilibrium, let us take a 0.001 M solution of acetic acid at 25°C. As seen in the section on weak acids, slightly more than 10 percent of the acetic acid molecules in this solution have ionized according to the equation
$\text{CH}_3\text{COOH} + \text{H}_2\text{O} \rightleftharpoons \text{CH}_3\text{COO}^{-} + \text{H}_3\text{O}^{+} \nonumber$
Although the concentrations of all species in this solution remain constant with time, this does not mean that acetic acid molecules have stopped transferring protons to water molecules. The concentrations remain constant because the reverse reaction (transfer of a proton from a hydronium ion to an acetate ion) is occurring at the same rate as the forward reaction. For every hydronium ion produced by a proton transfer somewhere in the solution, another hydronium ion is losing its proton somewhere else. The net result is that the concentration of the hydronium ion remains constant, and with it the concentrations of the other species involved.
The microscopic view of equilibrium as a constant reshuffle of chemical species is often given a special name and referred to as a dynamic equilibrium. It is important to realize that once a dynamic equilibrium has been set up, a particular atom will sometimes turn up as part of one of the product molecules and sometimes as part of one of the reactant molecules.
In the ionization equilibrium of acetic acid just considered, suppose for the sake of argument that we could identify a particular oxygen atom in some way. If we could now follow the history of this atom, we would find that it would sometimes form part of an acetic acid molecule, CH3CO*OH and sometimes part of an acetate ion, CH3CO*O. Since the acid is about 10 percent ionized, we would find on the average that our labeled oxygen would spend only 10 percent of its time as part of an acetate ion and the remaining 90 percent as part of an acetic acid molecule.
Figure $1$ gives a computer simulation of this dynamic view of equilibrium for the cis-trans isomerization equilibrium of difluoroethene at 623 K. Suppose that it were possible to color one of the molecules in the equilibrium mixture and to photograph it at regular intervals thereafter under a high-power microscope. The figure shows the kind of results we could expect. In each “photograph” the molecule appears at a different orientation― between photos it has been bounced around quite a bit by collisions with other molecules. A careful inspection reveals a more important fact. Three of the nine photographs show the screened molecule in the trans conformation, while in the other six the molecule has the cis conformation. If we went on taking a large number of photographs of this molecule, we would find on the average that this ratio would be preserved. On the average one-third of the photographs would show the trans conformation, while the other two-thirds would show the cis form. The situation is much as though we had a conventional six-sided cubic die with the word trans printed on two sides and the word cis printed on the other four. If we roll this die a sufficient number of times, it will come up cis twice as often as trans.
We can now begin to see why it is that the equilibrium constant for this reaction is 0.5 at 623 K. Not only the screened molecule, but every other molecule in the equilibrium mixture as well, spends one-third of its time in the trans conformation and two-thirds in the cis conformation. At any given time, therefore, we will find a mixture in which one-third of the molecules are trans while the other two-thirds are cis. The ratio of trans to cis will be 1:2, or 0.5, as found experimentally.
Such an explanation is still not the whole story. We still need to explain why the dice are loaded against the trans conformation of a molecule of difluoroethene. Why is it that this isomer is less likely to occur in the constant reshuffle between the two conformations caused by the continual random collisions of molecules with each other in the equilibrium mixture? The reason is that the trans isomer is slightly higher in energy than the cis form. The enthalpy change ΔHm for the reaction
$cis-\text{C}_2\text{H}_2\text{F}_2 \rightarrow trans-\text{C}_2\text{H}_2\text{F}_2 \nonumber$
has the small, but nevertheless positive, value of +3.88 kJ mol–1.
In any collection of molecules, energy is constantly being transferred from one molecule to another as they collide. In this continual interchange of energy, inevitably some molecules acquire more kinetic or potential energy than their fellow molecules. However, when considering the distribution of kinetic energy, the probability of a molecule’s acquiring a given energy depends on the magnitude of that energy. The higher the energy, the less likely it is to occur. In the constant reshuffle of energies, molecules which are even slightly higher in energy than their fellows occur slightly less often. In the case under consideration, since a molecule of trans-difluoroethene is 3.88 kJ mol–1 higher in energy than the cis form, it will occur less often than the cis form in an equilibrium mixture in which each molecule is constantly flipping back and forth between the two forms. In other words, the die is loaded against the trans configuration because it is higher in energy.
We can now also understand why the position of equilibrium for this reaction varies with temperature. At a very low temperature it will be a rare occurrence for a molecule to acquire the extra 3.88 kJ mol–1 needed if it is to have the trans configuration. Accordingly, we expect the equilibrium mixture to contain many fewer trans molecules than cis molecules and the equilibrium constant
$K_{c}=\frac{\text{ }\!\![\!\!\text{ }trans\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ }cis\text{-C}_{\text{2}}\text{H}_{\text{2}}\text{F}_{\text{2}}\text{ }\!\!]\!\!\text{ }} \nonumber$
to be quite small. Indeed, if it were possible to take the system down to absolute zero, no trans molecules would be possible and the value of Kc would be zero. As the temperature is increased, the probability of a molecule acquiring enough energy to assume the trans configuration increases, and increases along with it. Kc cannot increase indefinitely, though.
If the temperature is high enough, the energy difference between the cis and trans configurations becomes insignificant compared with the average kinetic energy of a molecule. Both forms will then be almost equally probable, and the value of Kc will be very slightly less than 1.
Very few chemical equilibria are as simple as the cis-trans isomerism of difluoroethene. In almost all other cases the position of equilibrium is governed not only by the energy difference between reactants and products but also by a probability factor which is independent of the energy. A simple example of such a probability factor is provided by the equilibrium between the ring and chain forms of 1,4-butanediol. The projection formula for this alcohol is
When a dilute solution (about 0.04 M) of this compound is prepared in carbon tetrachloride, the two ends of the molecule hydrogen-bond together to form a ring containing seven atoms, and an equilibrium between the ring form and the chain form is set up:
At 70°C (343 K) the equilibrium constant for this reaction is found by experiment to be
$K_{c}=\frac{\text{ }\!\![\!\!\text{ chain }\!\!]\!\!\text{ }}{\text{ }\!\![\!\!\text{ ring }\!\!]\!\!\text{ }}=\text{1}\text{.22} \nonumber$
So that there are about 22 percent more chain molecules than ring molecules in the equilibrium mixture at this temperature. This is contrary to what one would expect if energy were the only factor involved in determining the position of equilibrium. The chain form of the molecule is obviously higher in energy than the ring form, since energy must be supplied in order to break the hydrogen bond. Indeed ΔHm for Eq. (1) is found to be +13.5 kJ mol–1. On the basis of our previous argument, if the chain form is higher in energy, it should be less probable than the ring form and Kc should be less than 1. The experimental fact that Kc is greater than 1 means that there is some other factor making the chain form more probable than would be the case if energy alone were responsible.
In order to see what this other factor is, let us suppose that we can somehow investigate our system at a temperature which is so high that the energy difference between the ring and the chain form is insignificant. (In practice, the carbon tetrachloride solvent would boil off long before this temperature was obtained.) At such a temperature, if the energy difference between the two forms were the only factor determining their relative probability, we would expect the ring and chain forms (seen below) to be equally probable. It is easy to see, though, why this would not actually be the case.
The chain form of 1,4-butanediol is a very flexible molecule because of the free rotation which is possible around the three carbon-carbon bonds and the two carbon-oxygen bonds. As it collides with its fellow molecules, it can adopt a very large number of different conformations. The figure below illustrates some of these conformations, selected at random by a computer. Again we can take these as representing a series of photographs of the same molecule taken at regular intervals, say every 5 min. For the sake of clarity, though, we have kept the position of two of the carbon atoms the same for each photograph. In actual fact the molecule would be rotating as well as flexing, and the orientation of the bond joining these two carbon atoms would be continually changing. We should also remember that the molecule has undergone very many collisions between successive photographs.
Figure $2$ The chain form of butanediol. Because various segments of the molecule are free to rotate about carbon-carbon and carbon-oxygen bonds, the molecule can adopt a very large number of conformations. A random selection of nine of these is shown. Note that in none of these are the two oxygen atoms close enough to hydrogen bond together and form a ring. The ring form of butanediol. is at the bottom right. The two oxygen atoms at the top of the ring are hydrogen bonded together. From a strictly geometrical point of view this is a rather improbable conformation for the molecule. It is only because of the lowering in energy caused by the formation of a hydrogen bond that this ring form turns out to be slightly more probable in the equilibrium mixture than the chain form at low temperatures.
The important thing to notice about this random selection of nine conformations shown in Figure $2$ is that in none of them are the two oxygen atoms close enough to allow the formation of a hydrogen bond. At this high temperature, because of the large number of ways in which a butanediol molecule can arrange itself in space, it will only occasionally adopt a ring conformation like that shown in Figure $3$ in which the two oxygen atoms are close enough to hydrogen bond to each other. In the equilibrium mixture at high temperature, therefore, there will be many more chain molecules than ring molecules and Kc will be much larger than 1.
This situation begins to change as the temperature is reduced. With decreasing temperature, a smaller and smaller fraction of the molecules will have enough energy to break open a hydrogen bond. In consequence the ring form will become more probable. Once having formed, a ring will be able to survive quite a few collisions with other molecules before it is broken open again. Indeed, if we could reduce the temperature of our system close to absolute zero without the solvent freezing, then we would find that virtually all the butanediol molecules were in the ring form. Once formed, a ring would almost never experience a collision which was sufficiently energetic to break it open. At such a low temperature the value of Kc would be very close to zero.
The two equilibria we have just described have both been very simple examples, especially since they each involved only one reactant and one product. Nevertheless the principles we have discovered in discussing them apply to all chemical equilibria. We can always interpret the equilibrium constant of a chemical reaction as the product of two factors, each of which has a meaning on the microscopic level:
K = energy factor × probability factor (2)
The energy factor takes account of the fact that either the products or the reactants are higher in energy and hence less probable in the constant reshuffling of energy that goes on between molecules. The probability factor takes account of the fact that even if there were no energy difference between products and reactants, one or the other would still be more probable. This greater probability arises because of the larger number of ways in which the products (or reactants) can be realized on the molecular level.
In general the energy factor has the greatest effect on the value of the equilibrium constant at low temperatures. The lower the temperature, the very much lower the probability of a high-energy species occurring by chance in the constant energy reshuffle. At higher temperatures the effect of the energy factor becomes less pronounced and the probability factor becomes more important. If the temperature can be raised high enough, the probability factor eventually predominates. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/13%3A_Chemical_Equilibrium/13.13%3A_The_Molecular_View_of_Equilibrium.txt |
Many reactions in aqueous solutions involve weak acids or bases or slightly soluble substances, and in such cases one or more equilibria are achieved in solution. Furthermore, the equilibrium state is usually reached almost instantaneously, and so we can use the equilibrium law to calculate the concentrations and amounts of substance of different species in solution. Such information enables us to understand, predict, and control what will happen in solution, and it has numerous practical applications. Equilibrium constants may be used to obtain information about reactions in solution, and in many cases the results of equilibrium calculations will be applied to practical problems.
14: Ionic Equilibria in Aqueous Solutions
We have already noted the importance of reactions in aqueous solutions in the chemical laboratory, in the natural environment, and in the human body. Many reactions in aqueous solutions involve weak acids or bases or slightly soluble substances, and in such cases one or more equilibria are achieved in solution.
Furthermore, the equilibrium state is usually reached almost instantaneously, and so we can use the equilibrium law to calculate the concentrations and amounts of substance of different species in solution. Such information enables us to understand, predict, and control what will happen in solution, and it has numerous practical applications. Equilibrium constants may be used to obtain information about reactions in solution, and in many cases the results of equilibrium calculations will be applied to practical problems.
Acid-base reactions in aqueous solutions are intimately related to water’s ability to act as both a weak acid and a weak base, producing H3O+ and OH by proton transfer. In any aqueous solution at 25°C, the equilibrium constant of water (kw) is as follows:
$\text{K}_w={[\text{ H}_{3}\text{O}^{+}][\text{OH}^{-}]}=\frac{1.00 \times{10}^{-14}}{\text{1 mol L}^{-\text{1}}} \nonumber$
$[\text{H}_{3}\text{O}^{+}][\text{OH}^{-}]={K_{w}}=\text{1.00}\times\text{10}^{-14} \text{ mol}^{2} \text{ L}^{-2} \nonumber$
and concentrations of H3O+ and OH can vary from roughly 100 to 10–14 mol/L. This makes it convenient to define pH and pOH as:
\begin{align}\text{pH}=-\text{log}\frac{[\text{ H}_{3}\text{O}^{+}]}{\text{1 mol L}^{1}}\text{ } \\\text{pOH}=-\text{log}\frac{[\text{ OH}^{-}]}{\text{1 mol L}^{-1}} \\end{align} \nonumber
Since molecules of a strong acid transfer their protons to water molecules completely, [H3O+] (and hence pH) can be obtained directly from the stoichiometric concentration of the solution. Similarly [OH] and pOH may be obtained from the stoichiometric concentration of a strong base. In the case of weak acids and weak bases, proton-transfer reactions proceed to only a limited extent and a dynamic equilibrium is set up. In such cases an acid constant Ka (equilibrium constant for acids) or a base constant Kb as well as the stoichiometric concentration of weak acid or base are required to calculate [H3O+], [OH], pH, or pOH. Ka and Kb for a conjugate acid-base pair are related, and their product is always Kw
Often it is necessary or desirable to restrict the pH of an aqueous solution to a narrow range. This can be accomplished by means of a buffer solution―one which contains a conjugate weak acid-weak base pair. If a small amount of strong base is added to a buffer, the OH ions are consumed by the conjugate weak acid, so they have little influence on pH. Similarly, a small amount of strong acid can be consumed by the conjugate weak base in a buffer. To a good approximation the [H3O+] in a buffer solution depends only on Ka for the weak acid and the stoichiometric concentrations of the weak acid and weak base.
Indicators for acid-base on are conjugate acid-base pairs, each member of which is a different color. An indicator changes from the color of the conjugate acid to the color of the conjugate base as pH increases from approximately pKIn – 1 to pKIn + 1. For titrations involving only strong acids and strong bases, several indicators are usually capable of signaling the endpoint because there is a large jump in within ± 0.05 L of the exact stoichiometric volume of titrant. In the case of titrations which involve a weak acid or a weak base, a buffer solution is involved and the jump in pH is smaller. Consequently greater care is required in selection of an appropriate indicator.
A dynamic equilibrium is set up when a solid compound is in contact with a saturated solution. In the case of an ionic solid, the equilibrium constant for such a process is called the solubility product. Ksp can be determined by measurement of the solubility of a compound, and it is useful in predicting whether the compound will precipitate when ionic solutions are mixed. The common-ion effect, in which an increase in the concentration of one ion decreases the concentration of the other ion of an insoluble compound, can be interpreted quantitatively using solubility products. It is also true that removal of one ion of an insoluble compound from solution will increase the concentration of the other ion, and hence the solubility. It is for this reason that salts of weak acids often dissolve in acidic solutions―protonation of the anion effectively reduces its concentration to the point where the solubility product is not exceeded. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.01%3A_Prelude_to_Ionization_of_Water.txt |
In the section on amphiprotic species, we saw that water can act as a very weak acid and a very weak base, donating protons to itself to a limited extent:
$\text{2H}_{2}\text{O}\text{ }({l}) \rightleftharpoons \text{H}_{3}\text{O}^{+} ({aq}) + \text{OH}^{-} ({aq}) \nonumber$
The equilibrium constant $K$ for this reaction can be written as follows:
$K_{a}=\dfrac{a_{H_3O^+}·a_{OH^-}}{a_{H_2O}^2} \approx \frac{[H_{3}O^{+}][HO^{-}]}{(1)^{2}}=[H_{3}O^{+}][HO^{-}] \label{16.3.4}$
where $a$ is the activity of a species. Because water is the solvent, and the solution is assumed to be dilute, the activity of the water is approximated by the activity of pure water, which is defined as having a value of 1. The activity of each solute is approximated by the molarity of the solute.
Note
It is a common error to claim that the molar concentration of the solvent is in some way involved in the equilibrium law. This error is a result of a misunderstanding of solution thermodynamics. For example, it is often claimed that Ka = Keq[H2O] for aqueous solutions. This equation is incorrect because it is an erroneous interpretation of the correct equation Ka = Keq($\textit{a}_{H_2O}$). Because $\textit{a}_{H_2O}$ = 1 for a dilute solution, Ka = Keq(1), or Ka = Keq.
In this reaction, one water molecule acts as an acid and one water molecule acts as a base. Thus, this reaction actually can be designated as the $K_a$ of water and as the $K_b$ of water. It is most common, however, to designate this reaction and the associated law of mass action as the $K_w$ of water:
$K_{w}=[H_{3}O^{+}][HO^{-}] \label{16.3.5}$
Measurements of the electrical conductivity of carefully purified water indicate that at 25°C [H3O+] = [OH] = 1.00 × 10–7 mol/L, so that
\begin{align}K_{w}={1.00}\times{10}^{-7}\text{ mol L}^{-1}\times{1.00}\times{10}^{-7}\text{ mol L}^{-1} \nonumber \\text{ }\\text{ }={1.00}\times{10}^{-14}\text{ mol}^2\text{L}^{-2} \nonumber \end{align} \nonumber
(Since the equilibrium law is not obeyed exactly, even in dilute solutions, results of most equilibrium calculations are rounded to three significant figures. Hence the value of Kw = 1.00 × 10–14 mol2/L2 is sufficiently accurate for all such calculations.)
The equilibrium constant Kw applies not only to pure water but to any aqueous solution at 25°C. Thus, for example, if we add 1.00 mol of the strong acid HNO3 to H2O to make a total volume of 1 L, essentially all the HNO3 molecules donate their protons to H2O:
$\text{HNO}_{3} + \text{H}_{2}\text{O} \rightarrow \text{NO}_{3}^{-} + \text{H}_{3}\text{O}^{+} \nonumber$
and a solution in which [H3O+] = 1.00 mol/L is obtained. Although this solution is very acidic, there are still hydroxide ions present. We can calculate their concentration by rearranging Eq. $\ref{3}$:
\begin{align}\text{ }[\text{OH}^{-}]=\dfrac{K_{w}}{[\text{ H}_{3}\text{O}^{+}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{1.00 mol L}^{-1}}\\text{ }\\text{ }=\text{1.00 }\times \text{ 10}^{-14}\text{ mol L}^{-1}\end{align} \nonumber
The addition of the HNO3 to H2O not only increases the hydronium-ion concentration but also reduces the hydroxide-ion concentration from an initially minute 10–7 mol/L to an even more minute 10–14 mol/L.
Example $2$: Ion Concentration
Calculate the hydronium-ion concentration in a solution of 0.306 M Ba(OH)2.
Solution
Since 1 mol Ba(OH)2 produces 2 mol OH in solution, we have
$[OH^-] = 2 \times 0.306 \dfrac{mol}{L} = 0.612 \dfrac{mol}{L} \nonumber$
Then
\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=\dfrac{K_{w}}{[\text{OH}^{-}]}=\dfrac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{0.612 mol L}^{-1}} \nonumber \\text{ }\\text{ }=\text{1.63 }\times \text{ 10}^{-14}\text{ mol L}^{-1} \nonumber \end{align} \nonumber
Note
Note that since strong acids like HNO3 are completely converted to H3O+ in aqueous solution, it is a simple matter to determine [H3O+], and from it, [OH]. Similarly, when a strong base dissolves in H2O it is entirely converted to OH, so that [OH], and from it [H3O+] are easily obtained. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.02%3A_Ionization_of_Water.txt |
The calculations we have just done show that the concentrations of hydronium and hydroxide ions in aqueous solution can vary from about 1 mol L-1down to about 1 × 10–14 mol L–1, and perhaps over an even wider range. The numbers used to express [H3O+] and [OH] in the units mole per liter will often include large negative powers of 10. Consequently it is convenient to define the following:
$\text{pH}=-\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{1 mol L}^{-\text{1}}}\\text{ } \ \text{pOH}=-\text{log}\frac{\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{1 mol L}^{-\text{1}}}\text{ } \ \nonumber$
Note carefully what these equations tell us to do. To obtain pH, for example, we divide [H3O+] by the units mole per liter. This gives a pure number, and so we can take its logarithm. (It does not make sense to take the logarithm of a unit, such as mole per cubic decimeter.) The minus sign insures that we will obtain a positive result most of the time.
The logarithm of a number is the power to which 10 must be raised to give the number itself. Therefore the definitions of pH and pOH mean that we can deal with powers of 10 rather than numerical values. Since the numbers needed to express [H3O+] and [OH] are usually between 1 and 10-14 pH and pOH values are usually between 0 and 14.
Example $1$: pH & pOH
Calculate the pH and the pOH of each of the following aqueous solutions: (a) 1.00 M HNO3; 0.306 M Ba(OH)2 .
Solution
a) Our previous discussion showed that for this solution [H3O+] = 1.00 mol/L and [OH] = 1.00 x 10-14. Applying the definitions of pH and pOH, we have
\begin{align}\text{pH}=-\text{log}\frac{\text{1}\text{.00 mol L}^{-\text{1}}}{\text{1 mol L}^{-\text{1}}}=-\text{log(10 }^{\text{0}}\text{)}\\text{ }=-\text{(0)}=\text{0}\text{.00}\\text{ }\\text{pOH}=-\text{log}\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol L}^{-\text{1}}}{\text{1 mol L}^{-\text{1}}}=-\text{log(10}^{-\text{14}}\text{)}\\text{ }=-\text{(}-\text{14)}=14.\text{00}\end{align} \nonumber
b)In the example in the section on ionization of water, we found for this solution [H3O+] = 1.63 × 10–14 mol/L and [OH] = 6.12 × 10–1 mol/L. Thus
\begin{align}\text{pH} = -\text{log}{ 1.63}\times{10^{-14}}=-({-13.788})=13.788\\text{ }\\text{pOH} = -\text{log}{ 6.12}\times{10^{-1}}=-({-0.213})=0.213\end{align} \nonumber
In the laboratory it is convenient to measure the pH of a solution using a pH meter. Such a device works on a different principle from the conductivity measurements we have already mentioned, and an accurate explanation of how it works is beyond the scope of the present discussion. While greater accuracy can be obtained when great care and special instruments are used, pH is usually measured to an accuracy of ± 0.01. Therefore pH values are usually rounded to the second decimal place; the results of Example 1b would commonly be rounded to pH = 13.79 and pOH = 0.21.
Because pH measurements are so easily made, it is essential that you be able to convert from pH to [H3O+]. This is the reverse of finding pH from [H3O+]. Consequently it involves antilogs instead of logs. From the definition
$\text{pH}=-\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber$
we have
$\text{pH}=\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber$
Taking the antilog of both sides, we have
$\text{antilog}\left( -\text{pH} \right)\text{ = antilog}-\text{pH}=\left\{ \text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \right\} \nonumber$
so that
$\text{antilog}\left( -\text{pH} \right)=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber$
remembering that antilog x = 10x, we can write this expression as
$\text{10}^{-\text{pH}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}} \nonumber$
or
$\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^\text{-pH}\text{ mol}\text{ L}^{-1}\label{15}$
An alternative method of writing this equation is
$\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }=\frac{\text{1}}{\text{10}^{\text{pH}}}\text{ mol L}^{-\text{1}} \nonumber$
Example $2$: Hydronium Concentration
The pH of a solution is found to be 3.40. Find the hydronium-ion concentration of the solution.
Solution
If you have a calculator which has an antilog or 10x button, the problem is very simple. You enter – 3.40 and hit the button. The number thus obtained, 3.9822 x 10–4 is the number of moles of hydronium ion per liter. This follows from Eq. $\ref{15}$:
$\left[\text{H}_{3}\text{O}^{+}\right]=\text{10}^{-\text{pH}}\text{ mol}\text{ L}^{-1}=\text{10}^{-3.4}\text{ mol}\text{ L}^{-1}=\text{3.98}\times\text{10}^{-4}\text{ mol}\text{ L}^{-1}$
The same result is almost as easy to find using Eq. (1b).
$\text{10}^{-\text{pH}}=\text{antilog}\left(\text{pH}\right)=\text{antilog 3.40}=\text{antilog 3}\times \text{antilog 0.40}=\text{10}^{3}\times \text{2.51}$
Thus
$\text{10}^{-\text{pH}}=\frac{\text{1}}{\text{10}^{\text{pH}}}=\frac{\text{1}}{\text{2}\text{.51 }\times \text{ 10}^{\text{3}}}=\text{3}\text{.98 }\times \text{ 10}^{-\text{4}}$
in other words,
$\left[\text{H}_{3}\text{O}^{+}\right]=\text{3.98}\times\text{10}^{-4}\text{ mol}\text{ L}^{-1}$
There is a very simple relationship between the pH and the pOH of an aqueous solution at 25°C. We know that at this temperature
${K}_{w}={K}_{c}\left(\text{55.5 mol}\text{ L}^{-1}\right)^{2}\left[\text{H}_{3}\text{O}^{+}\right] \left[\text{OH}^{-}\right]={K}_{w}=\text{ 10}^{-14} \text{mol}^{2}\text{L}^{-2} \nonumber$
Dividing both sides by mol2 L–2, we obtain
$\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}}\text{ }\times \text{ }\frac{[\text{OH}^{-}]}{\text{mol L}^{-\text{1}}}=\text{10}^{-\text{14}} \nonumber$
Taking logs and multiplying both by – 1, we then have
$\text{log}\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }}{\text{mol L}^{-\text{1}}}\text{ }-\text{ }\log \text{ }\frac{[\text{OH}^{-}]}{\text{mol L}^{-\text{1}}}=-\text{log}(\text{10}^{-\text{14}}\text{)} \nonumber$
or
$\text{pH} + \text{pOH}= \text{14.00} \nonumber$
This simple relationship is often useful in finding the pH of solutions containing bases, as the following example shows.
Example $3$: pH of a Solution
If 3.53 g of pure NaOH is dissolved in 10 L of H2O find the pH of the resulting solution.
Solution
We first calculate the concentration of the NaOH.
$n_{\text{NaOH}}=\text{3}\text{.53 }\times \text{ }\frac{\text{1 mol}^{-\text{1}}}{\text{40}\text{.0 g}}=\text{0}\text{.088 25 mol}$
so that
$c_{\text{NaOH}}=\frac{n_{\text{NaOH}}}{V}=\frac{\text{0}\text{.088 25 mol}}{\text{10 L}^{\text{1}}}=\text{8}\text{.82 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}}\text{ }$
Since NaOH is a strong base, each mole of NaOH dissolved produces 1 mol OH ions, so that
$\left[\text{OH}^{-}\right]=\text{8.82 }\times \text{10}^{-3}\text{mol L}^{-1}$
Thus
$\text{pOH}=-\text{log }\left(\text{8.82}\times \text{10}^{-3}\right)=-\left(\text{0.95}-\text{3.00}\right)=+\text{2.05}$
From which
$\text{pH} =\text{14.00 - pOH} =\text{11.95}$
While the ability to calculate the pH of a solution from the hydronium-ion concentration and vice versa is useful, it is not the only thing we need to understand about pH. If someone gives you a solution whose pH is 14.74, it is true that the hydronium-ion concentration must be 1.82 × 10–15 mol L–1 but it is perhaps more important to know that the solution is corrosively basic. In general, then, we need not only to be able to calculate a pH but also to have some realization of what kind of solutions have what kind of pH, as displayed in the following table. This table is part of our collection of acid and base constants.
Table $2$: The pH Scale
Substance pH [H3O+] [OH-] pOH Strength
Battery acid
0 1 10-14 14 Strongly acidic
Stomach acid
Lemon juice
1 10-1 10-13 13
2 10-2 10-12 12
3 10-3 10-11 11 Weakly acidic
Soda water 4 10-4 10-10 10
Black coffee 5 10-5 10-9 9 Barely acidic
6 10-6 10-8 8
Pure water 7 10-7 10-7 7 Neutral
Seawater 8 10-8 10-6 6 Barely basic
Baking soda 9 10-9 10-5 5
Toilette soap 10 10-10 10-4 4 Mildly basic
Laundry water 11 10-11 10-3 3
Household ammonia 12 10-12 10-2 2 Very basic
13 10-13 10-1 1
Drain cleaner 14 10-14 1 0
In pure water at 25°C the hydronium-ion concentration is close to 1.00 × 10–7 mol/L, so that the pH is 7. In consequence any solution, not only pure water, which has a pH of 7 is described as being neutral. An acidic solution, as we know, is one in which the hydronium-ion concentration is greater than that of pure water, i.e., greater than 10–7 mol/L. In pH terms this translates into a pH which is less than 7 (because the pH is a negative logarithm). Small pH values are thus characteristic of acidic solutions; the smaller the pH, the more acidic the solution.
By contrast, a basic solution is one in which the hydroxide-ion concentration is greater than 10–7 mol/L. In such a solution the hydronium-ion concentration is less than 10–7 mol/L, so that the pH of a basic solution is greater than 7. Large pH values are thus characteristic of basic solutions. The larger the pH, the more basic the solution. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.03%3A_pH_and_pOH.txt |
When any weak acid, which we will denote by the general formula HA, is dissolved in water, the reaction
$\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}\nonumber$
proceeds to only a limited extent, and we must allow for this in calculating the hydronium-ion concentration and hence the pH of such a solution. In general the pH of a solution of a weak acid depends on only two factors, the concentration of the acid, ca, and the magnitude of an equilibrium constant Ka, called the acid constant, which measures the strength of the acid. The acid constant is defined by the relationship:
$K_{c} \times \text{ 1}=K_{a}=\frac{[\text{ H}_{3}\text{O}^{+}][\text{ A}^{-}]}{[\text{ HA }] }\nonumber$
(The acid constant is $K_c$ multiplied by the activity of the water, which has a value of 1 because water is the solvent as already defined in the section on the law of chemical equilibrium.)
In Example 5 on The Law of Chemical Equilibrium we showed how measurements of the conductivity of acetic acid solutions could be used to find the acid constant for acetic acid, Ka(CH3COOH). You will recall from that example that Ka was only approximately a constant, varying from a value of 1.81 × 10–5 mol/L in very dilute solutions to a value of 1.41 × 10–7 mol/L in a 1M solution. A similar variation is found for other weak acids, so that most of the calculations we do using Ka are only approximate. Only two or possibly three significant figures should be retained.
The table below gives the Ka values for a few selected acids arranged in order of their strength. This table is part of our larger collection of acid and base resources. It is at once apparent from this table that the larger the Ka value, the stronger the acid. The strongest acids, like HCl and H2SO4 have Ka values which are too large to measure, while another strong acid, HNO3, has Ka value close to 20 mol/L. Typical weak acids such as HF and CH3COOH have acid constants with a value of 10–4 or 10–5 mol/L. Acids like the ammonium ion, NH4+, and hydrogen cyanide, HCN, for which Ka is less than 10–9 mol/L are very weakly acidic.
You may notice that some acids have multiple Ka values. This is due to the fact they are polyprotic, which means they are able to give up more than one proton. However, due to molecular forces, the ease with which each proton gives up its protons varies, giving different K values for each proton given up.
Table $2$ The Acid Constants for Some Acids at 25°C.
Acid Formula and Ionization Equation Ka Molecular Shape
$\text{Acetic}$
$\ce{CH3COOH + H2O <=> H3O^+ +CH3COO^-}$
$1.8 \times 10^{-5}$
$\text{Aluminum ion}$ $\ce{ Al(H2O)6^{3+} + H2O <=> H3O^+ + Al(H2O)5OH^{2+} }$ $7.2 \times 10^{-6}$
$\text{Ammonium ion}$ $\ce{NH4^+ + H2O <=> H3O^+ + NH3}$ $5.6 \times 10^{-10}$
$\text{Hydrochloric}$ $\ce{HCl + H2O -> H3O+ + Cl^-}$ $\text{very large}$
$\text{Hydrogen peroxide}$ $\ce{H2O2 + H2O <=> H3O^+ + HO2^-}$ $2.1 \times10^{-12}$
$\text{Phosphoric}$
$\ce{H3PO4 + H2O <=> H3O^+ + H2PO4^-}$
$\ce{H2PO4^- + H2O <=> H3O^+ + HPO4^{2-}}$
$\ce{HPO4^{2-} + H2O <=> H3O^+ + PO4^{3-}}$
$K_1 = 7.2 \times10^{-3}$
$K_2 = 6.3 \times10^{-8}$
$K_3 = 4.6 \times10^{-13}$
$\text{Sulfuric}$
$\ce{H2SO4 + H20 <=> H3O^+ + HSO4^-}$
$\ce{HSO4^- + H20 <=> H3O^+ +SO4^{2-}}$
$K_1 = \text{very large}$
$K_2 = 1.1\times 10^{-2}$
• Taken from Hogfelt, E. Perrin, D. D. Stability Constants of Metal Ion Complexes, 1st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on Equilibrium. ISBN: 008020958
Before we can go on to discuss how the hydronium-ion concentration and the pH of a solution of a weak acid depend on the concentration of the acid, we need to clarify a point of terminology. In order to do this let us take as an example a 0.0010 M solution of acetic acid. Conductivity measurements discussed in the section on weak acids in reactions in aqueous solutions show that only about 10 percent of the acid molecules have donated protons to water at any given time. We thus have a situation which can be summarized schematically in the following way:
$\underset{\text{0.0009 mol L}^{-1}}{\overset{\text{90 }\% }{\mathop{\text{CH}_{3}\text{COOH}}}}\,\text{ + H}_{2}\text{O } \rightleftharpoons \underset{\text{0.0001 mol L}^{-1}}{\overset{\text{10 }\% }{\mathop{\text{CH}_{3}\text{COO}^{-}}}}\,\text{ + }\underset{\text{0.0001 mol L}^{-1}}{\mathop{\text{H}_{3}\text{O}^{\text{+}}}} ~~~~~~~ \nonumber$
In such a solution there is some ambiguity as to what we mean by the phrase concentration of acetic acid. Do we mean 0.0010 mol L–1, or do we mean 90 percent of this value, namely, 0.0009 mol L–1? In order to resolve this difficulty, we will use the term stoichiometric concentration of acid and the symbol ca to indicate the quantity 0.0010 mol L–1, that is, to indicate the total amount of acetic acid originally added per unit volume of solution. On the other hand we will use the term equilibrium concentration
and the symbol [CH3COOH] indicate the quantity 0.0009 mol L–1 that is, the final concentration of this species in the equilibrium mixture.
Let us now consider the general problem of finding [H3O+]in a solution of a weak acid HA whose acid constant is Ka and whose stoichiometric concentration is ca. According to the equation for the equilibrium,
$\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{ A}^{-}\nonumber$
for every mole of H3O+ produced, there must also be a mole of A produced. At the same time a mole of HA and a mole of H2O must be consumed. Since the volume which all these species occupy is the same, any increase in [H3O+] must be accompanied by an equal increase in [A] and an equal decrease in [HA]. Consequently we can draw up the following table (in which equilibrium concentrations of all species have been expressed in terms of [H3O+]:
Species HA H3O+ A
Initial concentration (mol/L) ca 10–7 0
Change in concentrationa (mol/L)
–[H3O+] [H3O+] [H3O+]
Equilibrium concentration (mol/L)
ca –[H3O+] [H3O+] [H3O+]
aThe hydronium-ion concentration actually increases from 107 mol/L to the equilibrium concentration, and so the change in each of the concentrations is ± ([H3O+ ] –10-7 mol/L). However, the concentration of hydronium ions produced by the weak acid is usually so much larger than –10-7 mol L-1 that the latter quantity can be ignored. In the case of 0.0010 M acetic acid, for example, [H3O+ ] ≈ 1 × 10-4 mol/L. Subtracting gives:
\begin{align*} & \text{ 0.000 100 0 mol L}^{-1} \ - & \underline{\text{0.000 000 1 mol L}^{-1}} \ & \text{0.000 099 9 mol L}^{-1} \end{align*} \nonumber
which is very close to 1 × 10-4 mol/L.
We can now substitute the equilibrium concentrations into the expression
$K_{a}=\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] [ \text{ A}^{-} ] }{ [ \text{ HA } ] }=\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] ^{\text{2}}}{c_{a}- [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] } \label{8}$
This could be solved for [H3O+] by means of the quadratic formula, but in most cases a quicker approximate method is available. Since the acid is weak, only a small fraction of the HA molecules will have donated protons to form H3O+ ions. Therefore [H3O+] is only a small fraction of ca and can be ignored when we calculate ca – [H3O+]. That is,
$c_{a} – \text{[H}_{3} \text{O}^{+} ] \approx c_{a} ~~~~~~~~~ \label{5}$
Equation $\ref{8}$ then becomes
$K_{a}= \frac{\text{ [} \text{ H}_{\text{3}}\text{O}^{\text{+}} ] ^{\text{2}}}{c_{a}}\nonumber$
which rearranges to
$\text{[H}_{3} \text{O}^{+}]^{2} \approx K_{a}c_{a} \nonumber$
Taking the square root of both sides gives an important approximate formula:
$\text{[H}_{3} \text{O}^{+} ] \approx \sqrt{K_{a}c_{a}} ~~~~~~~\label{12}$
Example $1$: pH Calculation
Use Equation $\ref{12}$ to calculate the pH of a 0.0200-M solution of acetic acid. Compare this with the pH obtained using the [H3O+] of 5.92 × 10–4 mol L–1 derived from accurate conductivity measurements.
Solution
From first table above, Ka = 1.8 × 10–5 mol L–1. Since ca = 2.00 × 10–2 mol L–1, we have
\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{K_{a}c_{a}} \[4pt] &=\sqrt{\text{(1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}}\text{)(2}\text{.00 }\times \text{ 10}^{-\text{2}}\text{ mol L}^{-\text{1}}\text{)}} \[4pt] &=\sqrt{\text{3}\text{.6 }\times \text{ 10}^{-\text{7}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \[4pt] &=\sqrt{\text{36 }\times \text{ 10}^{-8}\text{ mol L}^{-\text{1}}} \[4pt] [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\text{6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}} \\text{ pH} &=-\text{log(6}\text{.0 }\times \text{ 10}^{-\text{4}}\text{)} \[4pt] &=-\text{(0}\text{0.78}-\text{4)} \[4pt] &=\text{3}\text{.22} \end{align*} \nonumber
Using the accurate [H3O+] from conductivity measurements,
\begin{align*} \text{pH} &= – \text{log} (5.92 \times 10^{-4}) \[4pt] &= –(0.772 – 4) \[4pt] &= 3.228 \end{align*} \nonumber
Note that the approximate equation gives an [H3O+] which differs by 1 in the second significant digit from the accurate value. The calculated pH differs by 1 in the second place to the right of the decimal―roughly the same as the accuracy of simple pH measurements.
In a few cases, if the acid is quite strong or the solution very dilute, it turns out that Equation $\ref{12}$ is too gross an approximation. A convenient rule of thumb for determining when this is the case is to take the ratio [H3O+]/ ca. If this is larger than 5% or so, we need to make a second approximation, and then the rules for successive approximations can be applied. Equation $\ref{8}$ can be converted to a convenient form for successive approximations by multiplying both sides by ca – [H3O+]:
$\text{[H}_{3} \text{O}^{+}]^{2} = K_{a}(c_{a} - \text{[H}_{3} \text{O}^{+}]) \nonumber$
or
$\text{[H}_{3} \text{O}^{+} ] = \sqrt{K_{\text{a}}\text{(}c_{\text{a}}- [\text{ H}_{\text{3}}\text{O}^{\text{+}} ]\text{ )}} \nonumber$
To get a second approximation for [H3O+] we can feed the first approximation into the right side of this equation. The exact procedure is detailed in the following example.
Example $2$: pH of a Solution
Find the pH of 0.0200 M $\ce{HF}$ using the table of Acid Constants on this page.
Solution
Using Equation $\ref{12}$ we find
\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] =\sqrt{K_{a}c_{a}} =\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.0200 mol L}^{-\text{1}}} =\text{3}\text{.69 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \\end{align*} \nonumber
Checking we find
$\frac{ [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] }{c_{a}}=\frac{\text{0}\text{.003 69}}{\text{0}\text{0.0200}}= 0.185 \nonumber$
that is 18.5 percent.
A second approximation is thus necessary. We feed our first approximation of [H3O+]= 3.69 × 10–3 mol L–1 into Equation \ref{12}:
\begin{align*} [ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{K_{a}c_{a}} \[5pt] &=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{(0}\text{0.02}-\text{0}\text{0.003 69) mol L}^{-\text{1}}} \[4pt] &=\text{3}\text{.33 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \end{align*} \nonumber
Taking a third approximation we find
\begin{align*}[ \text{ H}_{\text{3}}\text{O}^{\text{+}} ] &=\sqrt{\text{6}\text{.8 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{(0}\text{.02}-\text{0}\text{.003 33) mol L}^{-\text{1}}} \[4pt] &=\text{3}\text{.37 }\times \text{ 10}^{-3}\text{ mol L}^{-\text{1}} \end{align*} \nonumber
Since this differs from the second approximation by less than 5 percent, we take it as the final result. The pH is
$\mathrm{pH}=-\log \left(3.37 \times 10^{-3}\right)=2.47 \nonumber$
Cross check: Since $\ce{HF}$ is a stronger acid than acetic acid, we expect this solution to have a lower pH than 0.02 M $\ce{CH3COOH}$ and indeed it does. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.04%3A_The_pH_of_Solutions_of_Weak_Acids.txt |
The pH of a solution of a weak base can be calculated in a way which is very similar to that used for a weak acid. Instead of an acid constant Ka, a base constant Kb must be used. If a weak base B accepts protons from water according to the equation
$\text{B} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons\text{BH}^{+} + \text{OH}^{-} \label{1}$
then the base constant is defined by the expression
$K_{b}=\dfrac{ \text{ BH}^{\text{+}} \text{ OH}^{-} }{ \text{ B } } \label{2}$
A list of Kb values for selected bases arranged in order of strength is given in the table below. This table is part of our larger collection of acid-base resources.
Table $1$: The Base Constants for Some Bases at 25°C. Taken from Hogfelt, E. Perrin, D. D. Stability Constants of Metal Ion Complexes, 1st ed. Oxford; New Pergamon, 1979-1982. International Union of Pure and Applied Chemistry, Commission on Equilibrium. ISBN: 0080209580
Base Formula and Ionization Equation Kb Molecular Shape
Ammonia $NH_3 + H_2O \rightleftharpoons NH^+_4 + OH^–$ 1.77 × 10–5
Aniline $C_6H_5NH_2 + H_2O \rightleftharpoons C_6H_5NH^+_3 + OH^–$ 3.9 × 10–10
Carbonate ion $CO_3^{2–} + H_2O \rightleftharpoons HCO^-_3 + OH^–$ 2.1 × 10–4
Hydrazine $N_2H_4 + H_2O \rightleftharpoons N_2H^+_5 + OH^–$
$N_2H^+_5 + H_2O \rightleftharpoons N_2H_6^{2+} + OH^–$
K1 = 1.2 × 10–6
K2 = 1.3 × 10–15
Hydride ion $H^– + H_2O \rightarrow H_2 + OH^–$ 1.0
Phosphate ion $PO_4^{3–} + H_2O \rightleftharpoons HPO^{2-}_4 + OH^–$ 5.9 × 10–3
Pyridine $C_5H_5N + H_2O \rightleftharpoons C_5H_5NH^+ + OH^–$ 1.6 × 10–9
To find the pH we follow the same general procedure as in the case of a weak acid. If the stoichiometric concentration of the base is indicated by cb, the result is entirely analogous to equation 4 in the section on the pH of weak acids; namely,
$K_{b}=\dfrac{ [\text{OH}^{-}]^2}{c_{b}- [\text{ OH}^{-}] } \label{3}$
Under most circumstances we can make the approximation
$c_b – [OH^–] \approx c_b \nonumber$
in which case Equation \ref{3} reduces to the approximation
$[OH^–] ≈ \sqrt{K_{b}c_{b}} \label{4}$
which is identical to the expression obtained in the acid case (approximation shown in equation 6 in the section on the pH of weak acids) except that OH replaces H3O+ and b replaces a. Once we have found the hydroxide-ion concentration from this approximation, we can then easily find the pOH, and from it the pH.
Example $1$: pH using Kb
Using the value for Kb listed in the table, find the pH of 0.100 M NH3.
Solution
It is not a bad idea to guess an approximate pH before embarking on the calculation. Since we have a dilute solution of a weak base, we expect the solution to be only mildly basic. A pH of 13 or 14 would be too basic, while a pH of 8 or 9 is too close to neutral. A pH of 10 or 11 seems reasonable. Using Equation \ref{4} we have
\begin{align*} [\text{ OH}^{-}] &=\sqrt{K_{b}c_{b}} \[4pt] & =\sqrt{\text{1.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}} \times \text{ 0.100 mol L}^{-\text{1}}} \[4pt] &=\sqrt{\text{1.8 }\times \text{ 10}^{-\text{6}}\text{ mol}^{\text{2}}\text{ L}^{-2}} \[4pt] &=\text{1.34 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \end{align*} \nonumber
Checking the accuracy of the approximation, we find
$\dfrac{ [\text{ OH}^{-} ]}{c_{\text{b}}}=\dfrac{\text{1.34 }\times \text{ 10}^{-\text{3}}}{\text{0.1}}\approx \text{1 percent}$
The approximation is valid, and we thus proceed to find the pOH.
$\text{pOH}=-\text{log}\dfrac{ [\text{ OH}^{-} ]}{\text{mol L}^{-\text{1}}}=-\text{log(1.34 }\times \text{ 10}^{-\text{3}}\text{)}=\text{2.87}$
From which
$pH = 14.00 – pOH = 14.00 – 2.87 = 11.13 \nonumber$
This calculated value checks well with our initial guess.
Occasionally we will find that the approximation
$c_b – [OH^{–}] ≈ c_b \nonumber$
is not valid, in which case we must use a series of successive approximations similar to that outlined above for acids. The appropriate formula can be derived from Equation \ref{3} and reads
$[OH^{-}] \approx \sqrt{K_{b} ( c_b - [OH^{-}] )} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.05%3A_The_pH_of_Solutions_of_Weak_Bases.txt |
Whereas monoprotic acids (like $\ce{HCl}$ pictured below) only have one hydrogen ion (aka proton - pictured below in white) to donate, polyprotic acids can donate multiple protons (see the H2SO4 molecule pictured below). Following a similar logic, polyprotic bases are bases that can accept multiple protons. With the basics of polyprotic acids and bases covered, let's dive into some of the details of how these special acids and bases work.
In the case of polyprotic acids and bases we can write down an equilibrium constant for each proton lost or gained. These constants are subscripted 1, 2, etc., to distinguish them. For sulfurous acid, a diprotic acid, we can, for example, write
Step 1
$\ce { H2SO3 + H2O \rightleftharpoons H3O^{+} + HSO3^{-}} \nonumber$
with appropriate acid constant
$K_{a\text{1}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ HSO}_{\text{3}}^{-}\text{ }]\text{ }}{\text{ }[\text{ H}_{\text{2}}\text{SO}_{\text{3}}\text{ }]\text{ }}=\text{1}\text{.7 }\times \text{ 10}^{-\text{2}}\text{ mol L}^{-\text{1}} \label{step1}$
and
Step 2
$\text{H}\text{SO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{SO}_{3}^{2-} \nonumber$
with appropriate acid constant
$K_{a\text{2}}=\frac{\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }[\text{ SO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}{\text{ }[\text{ HSO}_{\text{3}}\text{ }]\text{ }}=\text{5}\text{.6 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}} \label{step2}$
The carbonate ion, $\ce{CO3^{2–}}$, is an example of a diprotic base.
Step 1
$\text{CO}_{3}^{2-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HCO}_{3}^{-} + \text{OH}^{-} \nonumber$
with appropriate base constant
$K_{b\text{1}}=\frac{\text{ }[\text{ HCO}_{\text{3}}^{-}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{\text{2}-}\text{ }]\text{ }}=\text{2}\text{.1 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}} \label{step1C}$
Step 2
$\text{HCO}_{3}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{H}_{2}\text{CO}_{3}^{} + \text{OH}^{-} \nonumber$
with appropriate base constant
$K_{b\text{2}}=\frac{\text{ }[\text{ H}_{\text{2}}\text{CO}_{\text{3}}\text{ }]\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{\text{ }[\text{ CO}_{\text{3}}^{-}\text{ }]\text{ }}=\text{2}\text{.4 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}} \label{step2C}$
A general treatment of the pH of solutions of polyprotic species is beyond our intended scope, but it is worth noting that in many cases we can treat polyprotic species as monoprotic. In the case of \ce{H2SO3}\), for example, $K_{a1} \gg K_{a2}$ indicating that $\ce{H2SO3}$ is a very much stronger acid than $\ce{HSO3^{–}}$. This means that when $\ce{H2SO3}$ is dissolved in water, we can treat it as a monoprotic acid and ignore the possible loss of a second proton. Solutions of salts containing the carbonate ion, such as $\ce{Na2CO3}$ or $\ce{K2CO3}$ can be treated similarly.
Example $1$: pH Calculation
Find the pH of a 0.100-M solution of sodium carbonate, $\ce{Na2CO3}$. Use the base constant $K_{b1} = 2.10 \times 10^{–4}\, mol\, L^{–1}$ (Equation \ref{step1C}).
Solution
We ignore the acceptance of a second proton and treat the carbonate ion as a monoprotic base. We then have
\begin{align}\text{ }[\text{ OH}^{-}\text{ }]\text{ }=\sqrt{K_{b}c_{b}}=\sqrt{\text{2}\text{.10 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\text{ }\times \text{ 0}\text{.100 mol L}^{-\text{1}}} \\text{ }=\text{4}\text{.58 }\times \text{ 10}^{-\text{3}}\text{ mol L}^{-\text{1}} \\end{align}
Checking, we find that
$\frac{\text{ }[\text{ OH}^{-}\text{ }]\text{ }}{c_{b}}=\frac{\text{4}\text{0.58 }\times\text{ 10}^{-\text{3}}}{\text{0}\text{.1}}\approx \text{4}\text{.6 percent} \nonumber$
so that our approximation is only just valid.
We now find
$\text{pOH}=-\text{log}\left(\text{4.58}\times\text{10}^{-3}\right)= \text{2.34} \nonumber$
while
$\text{pH}={14}-\text{pOH}={14}-{2.34}={11.6} \nonumber$
Since the carbonate ion is a somewhat stronger base than NH3, we expect a 0.1-M solution to be somewhat more basic, as actually found.
A glance at the Ka and and Kb tables (Tables E1 and E2) reveals that most acid and base constants involve numbers having negative powers of 10. As in the case of [H3O+] and [OH], then, it is convenient to define
$\text{p}K_{a}=-\text{log}\frac{K_{a}}{\text{mol L}^{-\text{1}}} \nonumber$]
${p}K_{b}=-\text{log}\frac{K_{b}}{\text{mol L}^{-\text{1}}} \nonumber$
Using these definitions, the larger Ka or Kb is (i.e., the stronger an acid or base, respectively), the smaller pKaor pKb will be. For a strong acid like HNO3, Ka = 20 mol L–1 and
$\text{p}K_{a} = -\text{log}\text{ 20}=-\left(\text{1.30}\right)=-\text{1.30} \nonumber$
Thus for very strong acids or bases pK values can even be negative. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.06%3A_Polyprotic_Acids_and_Bases.txt |
One of the more useful aspects of the Brönsted-Lowry definition of acids and bases in helping us deal with the pH of solutions is the concept of the conjugate acid-base pair. We argued qualitatively in the section on conjugate acid-base pairs in aqueous reactions that the strength of an acid and its conjugate base are inversely related. The stronger one is, the weaker the other will be. This relationship can be expressed quantitatively in terms of a very simple mathematical equation involving the appropriate acid and base constants.
Suppose in the general case we have a weak acid HA whose conjugate base is A. If either or both of these species are dissolved in H2O we will have both the following equilibria set up simultaneously.
$\text{HA} + \text{ H}_{\text{2}}\text{O} \rightleftharpoons$ $\text{ H}_{\text{3}}\text{O}^{\text{+}}+ \text{ A}^{-}$ in which HA acts as acid
and $\text{ A}^{-} + \text{ H}_{\text{2}}\text{O}\rightleftharpoons$ $\text{HA} + \text{ OH}^{-}$ in which A acts as base
To the first of these equilibria we can apply the equilibrium constant Ka(HA):
$K_{a}\text{(HA)}=\frac{[\text{ H}_{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\text{ }]}{[\text{ HA }]} \nonumber$
while to the second we can apply the equilibrium constant Kb(A):
$K_{b}\text{(A}^{-}\text{)}=\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]}{[\text{ A}^{-}\text{ }]\text{ }} \nonumber$
Multiplying these two constants together, we obtain a simple relationship between them.
\begin{align}K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=\frac{[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{A}^{-}]}{[\text{HA }]}\times \frac{[\text{HA}][\text{OH}^{-}]}{[\text{A}^{-}]}\\\text{ }=[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{ OH}^{-}]\end{align}
$K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=K_{w}\label{6}$
If we divide both sides of this equation by the units and take negative logarithms of both sides, we obtain
\begin{align}\text{p}K_{a}=-\text{log}\frac{K_{a}\text{(HA)}}{\text{mol L}^{-\text{1}}}-\text{log}\frac{K_{b}\text{(A}^{-}\text{)}}{\text{mol L}^{-\text{1}}}\\text{ }\\text{ }=-\text{log}\frac{\text{10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{mol}^{\text{2}}\text{ L}^{-2}}\end{align} \nonumber
$\text{p}K_{a}\left(\text{HA}\right) + \text{ }\text{p}K_{b}\text{(A}^{-}\text{)}=\text{p}K_{w}\label{10}$
Thus the product of the acid constant for a weak acid and the base constant for the conjugate base must be Kw, and the sum of pKa and pKb for a conjugate acid-base pair is 14.
Equation $\ref{6}$ or $\ref{10}$ enables us to calculate the base constant of a conjugate base from the acid constant of the acid, and vice versa. Given the acid constant for a weak acid like HOCl, for instance, we are able to calculate not only the pH of HOCl solutions but also the pH of solutions of salts like NaOCl or KOCl which are, in effect, solutions of the conjugate base of HOCl, namely, the hypochlorite ion, OCl.
Example $1$: pH Calculations with Ka
Find the pH of (a) 0.1 M HOCl (hypochlorous acid) and (b) 0.1 M NaOCl (sodium hypochlorite) from the value for Ka given in the table of Ka values.
Solution
a) For 0.1 M HOCl, we find in the usual way that
\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]=\sqrt{K_{a}c_{a}}\\text{ }=\sqrt{\text{3}\text{.1 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.1 mol}^{\text{2}}\text{ L}^{-\text{2}}}\\text{ }=\text{5}\text{.57 }\times \text{ 10}^{-5}\text{ mol L}^{-\text{1}}\end{align}
so that pH = 4.25
b) For 0.1 M NaOCl we must first calculate Kb:
\begin{align}K_{b}\text{(OCl}^{-}\text{)}=\frac{K_{w}}{K_{a}\text{(HOCl)}}\\text{ }=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{3}\text{.1 }\times \text{ 10}^{-\text{8}}\text{ mol L}^{-\text{1}}}\\text{ }=\text{3}\text{.22 }\times \text{ 10}^{-\text{7}}\text{ mol L}^{-\text{1}}\end{align}
Thus
\begin{align}\left[\text{OH}^{-}\right]=\sqrt{K_{b}c_{b}}\\text{ }=\sqrt{\text{3}\text{.22 }\times \text{ 10}^{-\text{7}}\times \text{ 0}\text{.1 mol}^{\text{2}}\text{ L}^{-\text{1}}}\\text{ }=\text{1}\text{.79 }\times \text{ 10}^{-\text{4}}\text{ mol L}^{-\text{1}}\end{align}
From which
pOH = 3.75
and pH = 14.00 – pOH = 10.25
In this, as in all pH problems, it is worth checking that the answers obtained are not wildly unreasonable. A pH of 4 for a weak acid is reasonable, though a little high, but then HOCl is among the weaker acids in table. A pH of 10 corresponds to a mildly basic solution―reasonable enough, for a weak base like OCl.
Not only can we use Eq. $\ref{6}$ to find the value of Kb for the base conjugate to a given acid, we can also employ it in the reverse sense to find the value of Ka for the acid conjugate to a given base, as the following example shows.
Example $2$: pH Calculation with Kb
Find the pH of 0.05 M NH4Cl (ammonium chloride), using the value Kb(NH3) = 1.8 × 10–5 mol L–1.
Solution
We regard this solution as a solution of the weak acid NH4+ and start by finding Ka for this species:
\begin{align}K_{a}\text{(NH}_{\text{4}}^{\text{+}}\text{)}=\frac{K_{w}}{K_{b}\text{(NH}_{\text{3}}\text{)}}=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol L}^{-\text{1}}}\;\text{ }=\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol L}^{-\text{1}}\end{align}
We can now evaluate the hydronium-ion concentration with the usual approximation:
\begin{align}\left[\text{ H}_{3}\text{O}^{+}\right]=\sqrt{K_{a}c_{a}}\\text{ }=\sqrt{\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol L}^{-\text{1}}\times \text{ 0}\text{.05 mol L}^{-\text{1}}}\\text{ }=\text{5}\text{.27 }\times \text{ 10}^{-6}\text{ mol L}^{-\text{1}}\end{align}
whence pH = –log(5.27 × 10–6) = 5.28
Note
The ammonium ion is a very weak acid (as seen in the Tables of Ka and and Kb values). A solution of NH4+ ions will thus not produce a very acidic solution. A pH of 5 is about the same pH as that of black coffee, not very acidic.
Before the Brönsted-Lowry definition of acids and bases and the idea of conjugate acid-base pairs became generally accepted, the interpretation of acid-base behavior revolved very much around the equation
Acid + base → salt + water
In consequence the idea prevailed that when an acid reacted with a base, the resultant salt should be neither acidic or basic, but neutral. In order to explain why a solution of sodium acetate was basic or a solution of ammonium chloride was acidic, a special term called hydrolysis had to be invoked. Thus, for instance, sodium acetate was said to be hydrolyzed because the acetate ion reacted with water according to the reaction
CH3COO- + H2O $\rightleftharpoons$ CH3COOH + OH-
From the Brönsted-Lowry point of view there is, of course, nothing special about such a hydrolysis. It is a regular proton transfer. Nevertheless you should be aware of the existence of the term hydrolysis since it is still often used in this context.
Because the Brönsted-Lowry definition is so successful at explaining why some salt solutions are acidic and some basic, one must beware of making the mistake of assuming that no salt solutions are neutral. Many are. A good example is 0.10 M NaNO3. This solution is neutral because neither the Na+ ion nor the NO3 ion shows any appreciable acidic or basic properties. Since NO3 is the conjugate base of HNO3 we might expect it to produce a basic solution, but NO3 is such a weak base that it is almost impossible to detect such an effect. Just how weak a base NO3 is can be demonstrated using the value of Ka (HNO3) = 20 mol L–1 obtained from the Tables of Ka and and Kb values.
\begin{align}K_{b}\text{(NO}_{\text{3}}^{-}\text{)};=\frac{K_{w}}{K_{a}\text{(HNO}_{\text{3}}\text{)}}\\text{ }=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ L}^{-\text{2}}}{\text{20 mol L}^{-\text{1}}}\\text{ }=\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\text{ mol L}^{-\text{1}}\end{align}
If we now apply the conventional formula from equation 4 from the section on the pH of weak base solutions to calculate [OH] in 0.10 M NaNO3, we obtain
\begin{align}\left[\text{OH}^{-}\right]=\sqrt{K_{b}c_{b}}\\text{ }=\sqrt{\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\times \text{ 1}\text{.0 }\times \text{ 10}^{-\text{1}}\text{ mol L}^{-\text{1}}}\\text{ }=\text{7}\text{.1 }\times \text{ 10}^{-\text{9}}\text{ mol L}^{-\text{1}}\end{align}
But this is less than one-tenth the concentration of OH ion which would have been present in pure H2O, with no added NaNO3. Essentially all the OH ions are produced by H2O, and the pH turns out to be only slightly above 7.00. (Note also that the derivation of equation 4 from the pH of weak base solutions section assumed that the [OH] produced by H2O was negligible. To get an accurate result in this case requires a completely different equation.)
In general all salts in which group I and group II cations are combined with anions which are the conjugate bases of strong acids yield neutral solutions when dissolved in water. Examples are CaI2, LiNO3, KCl, Mg(ClO4)2.
There is only one exception to this rule. The hydrated beryllium ion, Be(H2O)42+, is a weak acid (Ka = 3.2 × 10–7 mol L–1) so that solutions of beryllium salts are acidic.
Table $1$ The Acid-Base Properties of Some Common Ions
Cations Anions
Acidic
Cr3+, Fe3+, Al3+
Hg2+, Be2+
NH4+, H3O+
HSO4
Neutral
Mg2+, Ca2+, Sr2+, Ba2+
Li+, Na+, K+
Ag+
NO3, ClO4
Cl, Br, I
SO42 (very weakly basic)
Basic None
PO43, CO32, SO32
F, CN, OH, S2–
CH3COO, HCO3
The table lists the acid-base properties of some of the more frequently encountered ions and provides a quick reference for deciding whether a given salt will be acidic, basic, or neutral in solution. Note that the table tells us nothing about the strength of any acid or base. If we need to know more about the pH, other than whether it is above, below, or equal to 7, we need information about the actual value of the acid or base constant. The table also lists the SO42ion as neutral, though classifying it as very feebly basic would be more accurate.
Example $3$: Acid, Base, or Neutral
Classify the following solutions as acidic, basic, or neutral: (a) 1 M KBr; (b) 1 M calcium acetate; (c)1 M MgF2; (d) 1 M Al(NO3)3; (f) 1 M KHSO4; (f) 1 M NH4I.
Solution
1. Both cation and anion are neutral: neutral.
2. Cation is neutral but anion basic: basic.
3. Cation is neutral but anion basic: basic.
4. Cation is acidic and anion neutral: acidic.
5. Cation is neutral but anion acidic: acidic.
6. Cation is acidic but anion neutral: acidic.
Example $4$: pH Matching
Without actually doing any calculations, match the following solutions and pH values, using the Tables of Ka and and Kb values, and the table on this page.
Aqueous Solution pH
1 M NH4NO3 8.0
1 M KCN 11.7
1 M Ca(NO3)2 9.4
1 M MgSO4 7.0
1 M CH3COONa(sodium acetate) 1.0
1 M KHSO4 4.6
Solution
The pH of 7.0 is easiest to pick. Only one of the salt solutions given has both a neutral anion and a neutral cation. This is Ca(NO3)2. In the case of MgSO4 the Mg2+ ion is neutral but theSO42 ion is very feebly basic; this would agree with a pH of 8.0, only slightly basic. The SO42 ion is such a feeble base because its conjugate acid, HSO4, is quite a strong acid, certainly the most acidic of all the ions featured. Accordingly we expect 1 M KHSO4 to correspond to the lowest pH, namely, 1.0. The only other acidic solution is 4.6, and this must correspond to 1 M NH4NO3 since NO4+ is the only other acidic ion present. Among basic ions the cyanide ion, CN, is the strongest. The most basic pH, 11.7, thus corresponds to 1 M KCN. Only one solution is left: 1 M CH3COONa. This should be feebly basic and so matches the remaining pH of 9.4 rather well.
14.07: Conjugate Acid-Base Pairs and pH
How are conjugate acid-base pairs related to microorganisms, imitation fish products, and bread baking?
Pseudomonas aeruginosa
Surimi
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Yeast leavened bread
One of the more useful aspects of the Brönsted-Lowry definition of acids and bases in helping us deal with the pH of solutions is the concept of the conjugate acid-base pair. The strength of an acid and its conjugate base are inversely related. The stronger one is, the weaker the other will be. The acids presented in Weak acids in foods - pH and beyond generate relatively strong bases with respect to the strength of their conjugate acid. The function of these acids and bases in foods and other biological systems largely depends on the pH of the medium. Another important conjugate acid-base pair related to foods is hypochlorous acid and hypochlorite (HClO/ClO-). Yes, even though this pair does not occur naturally in foods, it is of great importance in food processing. NaOCl is an ideal disinfectant and possesses excellent cleaning action making it the most widely used disinfectant for processing equipment, finished products, ingredients, and worker's hands in the food industry . A dynamic pair: HClO/ClO-
Sodium hypochlorite solutions are obtained by the absorption of gaseous chlorine in a sodium hydroxide solution,
$\text{Cl}_{2} + \text{2NaOH} \rightleftharpoons \text{NaOCL} + \text{NaCl} + \text{H}_{2}\text{O}$ The effectiveness of the cleaning and disinfecting activity of sodium hypochlorite depends not only on its concentration but also on the the pH of the solution.[1] In solution, the hipochlorite ion will form hypochlorous acid and hydroxide ions $\text{ClO}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{HOCL} + \text{OH}^{-}$ generating basic solutions. As the pH of a sodium hypochlorite solution decreases, between 4 and 6, HOCl becomes the predominant species. At lower pH, around 4, HOCl will be converted to chlorine gas (Cl2). $\text{HOCL} + \text{H}^{+} +\text{Cl}^{-}\rightleftharpoons \text{Cl}_{2} + \text{H}_{2}\text{O}$ Thus, depending on the pH, chlorine can exists in three different forms in aqueous solutions: Cl2, HOCl, and ClO-. Chlorine gas is extremely toxic, reason why it is so important not to mix sodium hypochlorite solutions with acidic cleaning products. Commercial sodium hypochlorite solutions can be found in different concentrations ranging from 5 to 40% to which sodium chloride and alkali (usually NaOH), are added in order to ensure a solution with basic pH and reduce the corrosive effect of ClO-. Commercial solutions between 5-15% NaClO contain 0.25 - 0.35% free alkali and 0.5 - 1.5% NaCl.[2] What is the relationship between conjugate acid-base pairs? How can we calculate the pH of their solutions? The relationship between the strength of an acid and its conjugate base can be expressed quantitatively in terms of a very simple mathematical equation involving the appropriate acid and base constants. Suppose that we have a weak acid HA whose conjugate base is A. If either or both of these species are dissolved in H2O, the following equilibria will occur simultaneously. $\text{HA} + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{A}^{-}$ in which HA acts as acid and $\text{A}^{-} + \text{H}_{2}\text{O}\rightleftharpoons \text{HA} + \text{OH}^{-}$ in which A acts as base To the first of these equilibria we can apply the equilibrium constant Ka(HA): $K_{a}\text{(HA)}=\frac{[\text{ H}_{\text{3}}\text{O}^{\text{+}}][\text{ A}^{-}\text{ }]}{[\text{ HA }]}$ while to the second we can apply the equilibrium constant Kb(A): $K_{b}\text{(A}^{-}\text{)}=\frac{[\text{ HA }][\text{ OH}^{-}\text{ }]}{[\text{ A}^{-}\text{ }]\text{ }}$ Multiplying these two constants together, we obtain a simple relationship between them. \begin{align} K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}&=\frac{[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{A}^{-}]}{[\text{HA }]}\times \frac{[\text{HA}][\text{OH}^{-}]}{[\text{A}^{-}]}\ \text{ }\ \text{ }&=[\text{H}_{\text{3}}\text{O}^{\text{+}}][\text{ OH}^{-}]\end{align}
$K_{a}\text{(HA)}\times K_{b}\text{(A}^{-}\text{)}=K_{w}$ (1)
If we divide both sides of this equation by the units and take negative logarithms of both sides, we obtain
\begin{align} \text{p}K_{a}&=-\text{log}\frac{K_{a}\text{(HA)}}{\text{mol dm}^{-\text{3}}}-\text{log}\frac{K_{b}\text{(A}^{-}\text{)}}{\text{mol dm}^{-\text{3}}}\ \text{ }\ \text{ }&=-\text{log}\frac{\text{10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ dm}^{-\text{6}}}{\text{mol}^{\text{2}}\text{ dm}^{-6}}\end{align} $\text{p}K_{a}\left(\text{HA}\right) + \text{ }\text{p}K_{b}\text{(A}^{-}\text{)}=\text{p}K_{w}$ (2) Thus the product of the acid constant for a weak acid and the base constant for the conjugate base must be Kw, and the sum of pKa and pKb for a conjugate acid-base pair is 14.
Equation (1) or (2) enables us to calculate the base constant of a conjugate base from the acid constant of the acid, and vice versa. Given the acid constant for a weak acid like HOCl, for instance, we are able to calculate not only the pH of HOCl solutions but also the pH of solutions of salts like NaOCl or KOCl which are, in effect, solutions of the conjugate base of HOCl, namely, the hypochlorite ion, OCl.
EXAMPLE 1 Find the pH of (a) 35% (m/v%) HOCl (hypochlorous acid) and (b) 35% (m/v%) commercial solution of NaOCl (sodium hypochlorite) from the value for Ka given in the table of Ka values.
Solution
a) For 35% (m/v%) HOCl, the solution contains 35 g of hypochlorous acid in 100 mL of solution and
\begin{align} \text{n}_{\text{HOCl}}&=\frac{\text{35.0 g}\text{HOCl}}{\text{52.46 g}\text{ mol}^{-1}}\ \text{ }&=\text{6.67 }\times\text{10}^{-1}\text{mol }\text{HOCl}\ \end{align} and its concentration is then \begin{align} \text{ }[\text{HOCl}]\text{ }&=\frac{\text{n}_{\text{HOCl}}}{V_{\text{solution}}}=\frac{\text{6.67}\times \text{10}^{-1}\text{ mol }}{\text{1.0}\times \text{10}^{-1}\text{ dm}^{3}}\ \text{ }&=\text{6.67 }\text{mol dm}^{-3}\ \end{align} The concentration of hydronium-ions generated from the acid can be calculated using the following equation, discussed in the pH of Solutions of Weak Acids, \begin{align} \left[\text{H}_{3}\text{O}^{+}\right]&\approx\sqrt{K_{a}c_{a}}\ \text{ }&\approx\sqrt{\text{3.9}\times\text{10}^{-8}\text{ mol dm}^{-\text{3}}\times \text{6.67} \text{ mol dm}^{-3}}\ \text{ }&\approx\text{5.10}\times \text{ 10}^{-4}\text{ mol dm}^{-\text{3}} \end{align} Checking the accuracy of the approximation we find
\begin{align} \frac{[\text{H}_{3}\text{O}^{+}]\text{ }}{c_{a}}&=\frac{\text{5.10}\times\text{10}^{-4}}{\text{6.67}}\ \text{ }&=\text{7.6}\times\text{10}^{-5}, \,that \,is, \, \text{0.0076 percent}\ \end{align}
so that, the approximation is valid and the pH of the solution is \begin{align} \text{pH}&=-\text{log}\left(\text{5.10}\times\text{10}^{-4}\right)\ \text{ }&=\text{3.29} \end{align} The relatively high concentration of the acid accounts for a weakly acidic solution in spite that the Ka for hypochlorous acid has an order of magnitude of 10-8.
b) To calculate de pH for a 35% (m/v%) commercial solution of NaOCl, we need to calculate the number of moles that correspond to 35 g NaOCl in 100 mL of solution
\begin{align} \text{n}_{\text{NaOCl}}&=\frac{\text{35.0 g}\text{ NaOCl}}{\text{74.44 g}\text{ mol}^{-1}}\ \text{ }&=\text{4.7}\times\text{10}^{-1}\text{mol }\text{NaOCl}\ \end{align} and its concentration is then \begin{align} \text{ }[\text{NaOCl}]\text{ }&=\frac{\text{n}_{\text{NaOCl}}}{V_{\text{solution}}}=\frac{\text{4.7}\times \text{10}^{-1}\text{ mol }}{\text{1.0}\times \text{10}^{-1}\text{ dm}^{3}}\ \text{ }&=\text{4.7 }\text{mol dm}^{-3}\ \end{align} Since 1 mol of NaOCL dissociates in 1 mol of Na+ and 1 mol OCl-, this solution contains 4.7 mol dm-3 OCl- ions. Now, we must calculate Kb: \begin{align}K_{b}\text{(OCl}^{-}\text{)}&=\frac{K_{w}}{K_{a}\text{(HOCl)}}\ \text{ }&=\frac{\text{1.00}\times \text{10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{3.9}\times \text{ 10}^{-8}\text{ mol dm}^{-3}}\ \text{ }&=\text{2.56}\times \text{10}^{-7}\text{ mol dm}^{-3} \end{align} Thus \begin{align}\left[\text{OH}^{-}\right]&\approx\sqrt{K_{b}c_{b}}\ \text{ }&\approx\sqrt{\text{2.56}\times \text{10}^{-7}\text{ mol}\text{ dm}^{-3}\times \text{4.7}\text{ mol}\text{ dm}^{-3}}\ \text{ }&\approx\text{1.1}\times \text{ 10}^{-3}\text{ mol dm}^{-3}\end{align} Checking the accuracy of the approximation we find \begin{align} \frac{[\text{OH}^{-}]\text{ }}{c_{b}}&=\frac{\text{1.1}\times\text{10}^{-3}}{\text{4.7}}\ \text{ }&=\text{2.3}\times\text{10}^{-4}, \,that \,is, \, \text{0.023 percent}\ \end{align} The pOH of this solution is then \begin{align} \text{pOH}&=-\text{log}\left(\text{1.1}\times\text{10}^{-3}\right)\ \text{ }&=\text{2.96} \end{align} and the pH \begin{align}\text{pH}&=\text{14}-\text{pOH}\ \text{ }&=\text{14}-\text{2.96}\ \text{ }&=\text{11.04} \end{align} Again, the concentration of the base , OCl, accounts for a pH around 11.0 in this solution. What makes HClO/ClO- disinfecting agents?
size=175</chemeddl-jmol2>
Hypochlorous acid
HOCl and ClO- are both strong oxidizing agents and react with a wide variety of biological molecules including proteins, amino acids, lipids, and DNA. Such reactivity...
How does ClO- clean a surface?
A detergent functions by minimizing the magnitude of attractive forces between soil and the solid surface by adsorption of detergent components both on soil and on the solid surface. Breaking the organic soil...
Factors affecting the effectiveness of NaClO solutions
The most relevant factors affecting the effectiveness of cleaning and disinfecting products...
Environmental and health risks
Chlorine based disinfectants can react with organic matter in water and form by-products like...
Task: Look for actual numbers of volumes of NaClO used in the industrySverazo 00:49, 3 December 2009 (UTC) ( Sofia Erazo )
Conjugate acid-base pairs and microbial nutrition
Ammonium salts such as chloride and sulfate are employed in media growth for microorganisms as a source of nitrogen. Ammonium chloride is called "yeast food" in diverse fermentation processes including bread baking (S. cerevisiae) and production of citric acid (Y. lipolytica).
Sacharomyces cerevisiae cells under differential interference microscopy
Wet yeast
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Yeast leavened bread
EXAMPLE 2 Find the pH of 0.05 M NH4Cl (ammonium chloride), using the value Kb(NH3) = 1.8 × 10–5 mol dm–3.
Solution We regard this solution as a solution of the weak acid NH4+ and start by finding Ka for this species:
\begin{align}K_{a}\text{(NH}_{\text{4}}^{\text{+}}\text{)}=\frac{K_{w}}{K_{b}\text{(NH}_{\text{3}}\text{)}}&=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ dm}^{-\text{6}}}{\text{1}\text{.8 }\times \text{ 10}^{-\text{5}}\text{ mol dm}^{-\text{3}}}\ \text{ }&=\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol dm}^{-\text{3}}\end{align} We can now evaluate the hydronium-ion concentration with the usual approximation: \begin{align}\left[\text{ H}_{3}\text{O}^{+}\right]&=\sqrt{K_{a}c_{a}}\ \text{ }&=\sqrt{\text{5}\text{.56 }\times \text{ 10}^{-\text{10}}\text{ mol dm}^{-\text{3}}\times \text{ 0}\text{.05 mol dm}^{-\text{3}}}\ \text{ }&=\text{5}\text{.27 }\times \text{ 10}^{-6}\text{ mol dm}^{-\text{3}}\end{align} hence, $\text{pH} = \text{log}\left( \text{5.27} \times \text{10}^{-6}\right) = \text{5.28}$ Note: The ammonium ion is a very weak acid (as seen in the Tables of Ka and and Kb values). A solution of NH4+ ions will thus not produce a very acidic solution. A pH of 5 is about the same pH as that of black coffee, not very acidic. Before the Brönsted-Lowry definition of acids and bases and the idea of conjugate acid-base pairs became generally accepted, the interpretation of acid-base behavior revolved very much around the equation $\text{Acid} + \text{Base} \rightarrow \text{Salt} + \text{Water}$ In consequence the idea prevailed that when an acid reacted with a base, the resultant salt should be neither acidic or basic, but neutral. In order to explain why a solution of sodium acetate was basic or a solution of ammonium chloride was acidic, a special term called hydrolysis had to be invoked. Thus, for instance, sodium acetate was said to be hydrolyzed because the acetate ion reacted with water according to the reaction $\text{CH}_{3}\text{COO}^{-} + \text{H}_{2}\text{O} \rightleftharpoons \text{CH}_{3} \text{COOH} + \text{OH}^{-}$ From the Brönsted-Lowry point of view there is, of course, nothing special about such a hydrolysis. It is a regular proton transfer. Nevertheless you should be aware of the existence of the term hydrolysis since it is still often used in this context.
Because the Brönsted-Lowry definition is so successful at explaining why some salt solutions are acidic and some basic, one must beware of making the mistake of assuming that no salt solutions are neutral. Many are. A good example is 0.10 M NaNO3. This solution is neutral because neither the Na+ ion nor the NO3 ion shows any appreciable acidic or basic properties. Since NO3 is the conjugate base of HNO3 we might expect it to produce a basic solution, but NO3 is such a weak base that it is almost impossible to detect such an effect. Just how weak a base NO3 is can be demonstrated using the value of Ka (HNO3) = 20 mol dm–3 obtained from the Tables of Ka and and Kb values.
\begin{align}K_{b}\text{(NO}_{\text{3}}^{-}\text{)}&=\frac{K_{w}}{K_{a}\text{(HNO}_{\text{3}}\text{)}}\ \text{ }&=\frac{\text{1}\text{.00 }\times \text{ 10}^{-\text{14}}\text{ mol}^{\text{2}}\text{ dm}^{-\text{6}}}{\text{20 mol dm}^{-\text{3}}}\ \text{ }&=\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\text{ mol dm}^{-\text{3}}\end{align} If we now apply the conventional formula from equation 4 from the section on the pH of weak base solutions to calculate [OH] in 0.10 M NaNO3, we obtain \begin{align}\left[\text{OH}^{-}\right]&=\sqrt{K_{b}c_{b}}\ \text{ }&=\sqrt{\text{5}\text{.0 }\times \text{ 10}^{-\text{16}}\times \text{ 1}\text{.0 }\times \text{ 10}^{-\text{1}}\text{ mol dm}^{-\text{3}}}\ \text{ }&=\text{7}\text{.1 }\times \text{ 10}^{-\text{9}}\text{ mol dm}^{-\text{3}}\end{align} But this is less than one-tenth the concentration of OH ion which would have been present in pure H2O, with no added NaNO3. Essentially all the OH ions are produced by H2O, and the pH turns out to be only slightly above 7.00. (Note also that the derivation of equation 4 from the pH of weak base solutions section assumed that the [OH] produced by H2O was negligible. To get an accurate result in this case requires a completely different equation.)
In general, all salts that combine group I and group II cations with anions (conjugate bases) derived from of strong acids yield neutral solutions when dissolved in water. Examples are CaI2, LiNO3, KCl, and Mg(ClO4)2.
There is only one exception to this rule. The hydrated beryllium ion, Be(H2O)42+, is a weak acid (Ka = 3.2 × 10–7 mol dm–3) so that solutions of beryllium salts are acidic.
EXAMPLE 3 The following are salts used in food processing, classify them as acidic, basic, or neutral: (a) 1 M Potassium bromate; (b) 1 M trisodium citrate; (c)1 M trisodium phosphate; (d) 1 M Sodium aluminum sulfate; (e) 1 M potassium hydrogen sulfate; (f) 1 M ammonium chloride.
Solution
Salt Cation Anion Overall Used in[1] [2]
KBrO3 Neutral Neutral Neutral Flour conditioning: Oxidation of wheat glutathione into its disulfide.
Na3C6H5O7 Neutral Basic Basic Condensed milk for pH adjustment and binding of calcium to avoid aggregation of casein.
Na3PO4 Neutral Basic Basic Emulsifier, protein modifier, buffering agent in confectionery,cereals, and processed cheese.
NaAl(SO4)2 Neutral, Acidic Neutral Acidic Because of its slow reaction rate, it is used in combination with other leavening acids to provide tunneling or blistering effects in baked products.
KH2PO4 Neutral Acidic Acidic Buffering agent, mineral supplement.
NH4Cl Acidic Neutral Acidic Yeast fermentation processes as a source of nitrogen for yeast metabolism.
The table lists the acid-base properties of some of the more frequently encountered ions and provides a quick reference for deciding whether a given salt will be acidic, basic, or neutral in solution. Note that the table tells us nothing about the strength of any acid or base. If we need to know more about the pH, other than whether it is above, below, or equal to 7, we need information about the actual value of the acid or base constant. The table also lists the SO42–ion as neutral, though classifying it as very feebly basic would be more accurate. The Acid-Base Properties of Some Common Ions
Cations Anions Anion
Acidic
Cr3+, Fe3+, Al3+
Hg2+, Be2+ NH4+, H3O+
HSO4
Neutral
Mg2+, Ca2+, Sr2+, Ba2+
Li+, Na+, K+ Ag+
NO3, ClO4
Cl, Br, I SO42– (very weakly basic)
Basic None
PO43, CO32, SO32
F, CN, OH, S2– CH3COO, HCO3
EXAMPLE 4 Without actually doing any calculations, match the following solutions and pH values, using the Tables of Ka and and Kb values, and the table on this page.
Aqueous Solution
1 M
pH
?
NH4NO3 8.0
Na4P2O7 11.7
NaNO3 9.4
MgSO4 7.0
(CH3COO)2Ca
(calcium acetate)
1.0
KHSO4 4.6
Solution The pH of 7.0 is easiest to pick. Only one of the salt solutions given has both a neutral anion and a neutral cation. This is NaNO3. In the case of MgSO4 the Mg2+ ion is neutral but the SO42 ion is very feebly basic; this would agree with a pH of 8.0, only slightly basic. The SO42 ion is such a feeble base because its conjugate acid, HSO4, is quite a strong acid, certainly the most acidic of all the ions featured. Accordingly we expect 1 M KHSO4 to correspond to the lowest pH, namely, 1.0. The only other acidic solution is 4.6, and this must correspond to 1 M NH4NO3 since NH4+ is the only other acidic ion present. Among basic ions the pyrophosphate ion,P2O7, is the strongest. The most basic pH, 11.7, thus corresponds to 1 M Na4P2O7. Only one solution is left: 1 M (CH3COO)2Ca. This should be feebly basic and so matches the remaining pH of 9.4 rather well.
Aqueous Solution
1 M
pH Application in food[3] [4]
NH4NO3 4.6 Fertilizer and fermentation processes.
Na4P2O7 11.7 Used in milk based beverages, sea food, processed meat, pet food, and confectionery products as chelating agent, protein modifier, dispersing agent, coagulant, pH adjustment.[3]
NaNO3 7.0 Preserve red color of meat and antimicrobial activity against C. botulinum (its activiy is pH dependent). Also found in fruits and vegetables (i.e. lettuce, radish, rhubarb and strawberries).
MgSO4 8.0 Supplementation for magnesium in soil and livestock diet.[3] Coagulation of soy proteins in production of tofu.
(CH3COO)2Ca (calcium acetate) 9.4 Chelating agent for stabilization of color, aroma, and texture.
KHSO4 1.0 Antimicrobial and acidulant.
References
1. Food Chemistry 3rd Ed. 2004 Belitz, et al.
2. Food Additives, 2nd ed. 2002, Branen, A., Davidson, M.P., Salminen, S. and Thorngate III, J.H.
3. Humphreys, J.L., Carlson, M.S., Lorenzen, C.L. 2009. Dietary supplementation of magnesium sulfate and sodium bicarbonate and its effect on pork quality during environmental stress. Livestock Sci. 125:1:15-21 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.07%3A_Conjugate_Acid-Base_Pairs_and_pH/14.7.01%3A_Foods-_From_Cleaning_and_Disinfection_to_Microbial_Nutrition_and_Protein_Modificat.txt |
So far in discussing pH we have dealt only with solutions obtained by adding a single acid, such as acetic acid, or a single base, such as the acetate ion, to water. We must now turn to a consideration of solutions to which both an acid and a base have been added. The simplest case of such a solution occurs when the acid and base are conjugate to each other and also present in comparable amounts. Solutions of this special kind are called buffer solutions because, as we shall shortly see in the video below, it is difficult to change their pH even when an appreciable amount of strong acid or strong base is added.
As a typical example of a buffer solution, let us consider the solution obtained when 3.00 mol acetic acid (HC2H3O2) and 2.00 mol sodium acetate (Na C2H3O2) are added to sufficient water to produce a solution of total volume 1 L. The stoichiometric concentration of acetic acid, namely, ca, is then 3.00 mol L–1, while the stoichiometric concentration of sodium acetate, cb, is 2.00 mol L–1. As a result of mixing the two components, some of the acetic acid, say x mol L–1, is converted to acetate ion and hydronium ion. We can now draw up a table in order to find the equilibrium concentrations in the usual way.
Species
Initial Concentration (mol L-1)
Change in Concentration (mol L-1)
Equilibrium Concentration (mol L-1)
H3O+ 10-7 (negligible) x x
CH3COO- 2.00 x 2.00 + x
CH2COOH 3.00 (-x) 3.00 - x
We can now substitute concentrations in the equilibrium expression
$K_{a}=\frac{[\text{CH}_{3}\text{COO}^{-}][\text{H}_{3}\text{O}^{+}]}{[\text{CH}_{3}\text{COOH}]} \nonumber$
from which we obtain
$\text{1.8}\times \text{10}^{-5}\text{ mol L}^{-1}=\frac{\text{(2}\text{.00 + }x\text{)}x}{\text{3.00}-x}\text{ mol L}^{-1}\label{2}$
In order to solve this equation, we make the approximation that x is negligibly small compared with both 2.00 and 3.00, that is, that only a minute fraction of acetic acid has converted to acetate ion. We then have
$\frac{\text{2.00}x}{\text{3.00}}=\text{1.8}\times \text{10}^{-5} \nonumber$
or
\begin{align}x=\frac{\text{3.00}}{\text{2.00}}\times \text{ 1.8 }\times \text{ 10}^{-5}\\text{ }=\text{2.7}\times \text{10}^{-5}\end{align}
Obviously, our approximation is a very good one. Since x is only 0.001 percent of 2.00 or 3.00, there really is no point in obtaining a second approximation by feeding x back into Eq. $\ref{2}$. We can thus safely conclude that
$[\text{H}_{3}\text{O}^{+}] = \text{2.7} \times \text{10}^{-5}\text{mol L}^{-1}$
and
$\text{pH} = {4.57}$
The example we have just considered demonstrates two obvious features:
1. When the acid and its conjugate base are mixed, very little of the acid is converted to base, or vice versa. (x was small compared with 2.00 and 3.00.)
2. In a buffer mixture, the hydronium-ion concentration and the hydroxide-ion concentration are small compared with the concentrations of acid and conjugate base. ([H3O+] = 2.7 × 10–5 mol L–1; [HO] = 3.7 × 10–10 mol L–1 as compared with [CH3COO] = 2.00 mol L–1 and [CH3COOH] = 3.00 mol L–1)
By assuming that these features are common to all buffer solutions, we make it very easy to handle them from a mathematical standpoint.
Let us now consider the general problem of finding the pH of a buffer solution which is a mixture of a weak acid HA, of stoichiometric concentration ca, and its conjugate base A, of stoichiom
$[\text{H}_{3}\text{O}^{+}]=K_{a}\times \frac{[\text{HA}]}{[\text{A}^{-}]}\label{6}$
Taking negative logarithms of both sides, we obtain
$-\text{log }[\text{H}_{3}\text{O}^{+}]=-\text{log }K_{a}-\text{log}\frac{[\text{HA}]}{[\text{A}^{-}]} \nonumber$
$\text{pH}=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\label{8}$
Equation $\ref{8}$ is called the Henderson-Hasselbalch equation and is often used by chemists and biologists to calculate the pH of a buffer.
As we saw in the case of the acetic acid―sodium acetate buffer described earlier, the equilibrium concentrations of HA and A are usually almost identical to the stoichiometric concentrations. That is,
$[\text{HA}] \approx \text{c}_{a}$ and $[\text{A}^{-}]\approx\text{c}_{b}$
We can substitute these values into Eqs. $\ref{6}$ and $\ref{8}$ to obtain the very useful approximations
$[\text{H}_{3}\text{O}^{+}]\approx K_{a}\times \frac{c_{a}}{c_{b}}\label{9}$
and
$\text{pH}\approx \text{ p}K_{a}\text{ + log}\frac{c_{b}}{c_{a}} \nonumber$
Example $1$: pH of Solution
Find the pH of the solution obtained when 1.00 mol NH3 and 0.40 mol NH4Cl are mixed to give 1 L of solution. Kb(NH3) = 1.8 × 10–5 mol L–1.
Solution
In order to use Eq. $\ref{9}$,we need first to have the value of
\begin{align}K_{a}\left(\text{NH}_{4}^{+}\right)=\frac{K_{w}}{K_{b}\left(\text{NH}_{3}\right)}\\text{ }=\frac{\text{1.00}\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{2}}{\text{1.8 }\times \text{ 10}^{-5}\text{ mol L}^{-1}}\\text{ }=\text{5.56}\times \text{ 10}^{-10}\text{ mol L}^{-1}\end{align}
We also have ca = 0.40 mol L–1 and cb = 1.00 mol L–1. Thus
\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]=K_{a}\times \frac{c_{a}}{c_{b}}\\text{ }=\text{5.56}\times \text{ 10}^{-10}\text{ mol L}^{-1}\times \frac{\text{ 0.4 mol L}^{-1}}{\text{1.0 mol L}^{-1}}\\text{ }=\text{2.22 }\times \text{ 10}^{-10}\text{ mol L}^{-1}\end{align} \nonumber
from which
and $\text{pH} = {9.65}$
To see why a mixture of an acid and its conjugate base is resistant to a change in pH, let us go back to our first example: a mixture of acetic acid (3 mol L–1)and sodium acetate (2 mol L–1). What would happen if we now added 0.50 mol sodium hydroxide to 1 L of this mixture? The added hydroxide ion will attack both the acids present, namely, the hydronium ion and acetic acid. Since the hydronium-ion concentration is so small, very little hydroxide ion will be consumed by reaction with the hydronium ion. Most will be consumed by reaction with acetic acid. Further, since the hydroxide ion is such a strong base, the reaction
$\text{CH}_{3}\text{COOH}+ \text{OH}^{-} \rightarrow \text{CH}_{3}\text{COO}^{-} + \text{H}_{2}\text{O} \nonumber$
will go virtually to completion, and 0.50 mol acetic acid will be consumed. The same amount of acetate ion will be produced. In tabular form:
Species
Initial Concentration
mol L-1
Change in Concentration
mol L-1
Equilibrium Concentration
mol L-1
H3O+ 2.7 x 10-5 Small approx. 2.7 x 10-5
CH3COO- 2.00 0.50 2.50 + 2.7 x 10-5 = 2.50
CH2COOH nbsp; 3.00 (-0.50) 2.50 - 2.7 x 10-5 = 2.50
Substituting the equilibrium concentrations of base (acetate ion) and conjugate acid (acetic acid) into the Henderson-Hasselbalch equation, Eq. $\ref{8}$, we have
\begin{align}\text{pH}=\text{p}K_{a}\text{ + log}\frac{[\text{A}^{-}]}{[\text{HA}]}\\text{ }=-\text{log(1.8} \times \text{10}^{-5}\text{) + log}\frac{\text{(2.50 mol L}^{-1}\text{)}}{\text{(2.50 mol L}^{-1}\text{)}}\\text{ }=-\left(\text{0.25}-\text{5} \right)+ \text{log}\left(\text{1}\right)\\text{ }=\text{4.74 + 0}=\text{4.74}\end{align} \nonumber
The addition of 0.5 mol sodium hydroxide to buffer mixture has thus succeeded in raising its pH from 4.57 to only 4.74. If the same 0.5 mol had been added to a cubic decimeter of pure water, the pH would have jumped all the way from 7.00 up to 13.7! The buffer is extremely effective at resisting a change in pH because the added hydroxide ion attacks the weak acid (in very high concentration) rather than the hydronium ion (in very low concentration). The major effect of the addition of the hydroxide ion is thus to change the ratio of acid to conjugate base, i.e., to change the value of
$\frac{[\text{CH}_{3}\text{COOH}]}{[\text{CH}_{3}\text{COO}^{-}]} \nonumber$
As long as the amount of weak acid is much larger than the amount of base added, this ratio is not altered by very much. Since the hydronium-ion concentration is governed by
$[\text{H}_{3}\text{O}^{+}]=K_{a}\frac{[\text{CH}_{3}\text{COOH}]}{[\text{CH}_{3}\text{COO}^{-}]} \nonumber$
the hydronium-ion concentration and pH are also altered to only a small extent.
The ability of a buffer solution to resist large changes in pH has a great many chemical applications, but perhaps the most obvious examples of buffer action are to be found in living matter. If the pH of human blood, for instance, gets outside the range 7.2 to 7.6, the results are usually fatal. The pH of blood is controlled by the buffering action of several conjugate acid-base pairs. The most important of these is undoubtedly the H2CO3/HCO3 pair, but side chains of the amino acid histidine in the hemoglobin molecule also play a part. (Hemoglobin, a protein, is the red substance in the blood. It is responsible for carrying oxygen away from the lungs.) Most enzymes (biological catalysts) can only function inside a rather limited pH range and must therefore operate in a buffered environment. The enzymes which start the process of digestion in the mouth at a pH of around 7 become inoperative in the stomach at a pH of 1.4. The stomach enzymes in turn cannot function in the slightly basic environment of the intestines.
14.08: Buffer Solutions
back to Buffer Solutions
Aspartame
Soft drinks and buffers
Citric acid
So far you have learned about pH in solutions where either a single acid, such as citric acid, or a single base, such as the citrate ion have been added to water. Now, let us consider solutions prepared with both an acid and a base. The simplest case of such a solution occurs when the acid and base are conjugate to each other and also are present in comparable amounts. Solutions of this special kind are called buffer solutions because it is difficult to change their pH even when an appreciable amount of strong acid or strong base is added. Why are buffer solutions important in foods?
Buffering solutions in foods play an important role in maintaining specific pH values for optimum activity of enzymes, protein solubility, and functionality. As discussed in previous exemplars, pH may also modify the color and flavor of foods and it is a critical factor in the preservation of many processed foods. Buffering solutions are also used as reaction media in the production of food ingredients and additives. Overall pH control is a major factor in maintaining the physical, chemical, and microbiological stability of foods.
Foods contain numerous compounds able to form buffering systems. Molecules with acid-base properties naturally found in foods include amino acids, organic acids, proteins, and charged polysaccharides. Other buffering systems are intentionally added to processed foods, examples of these are the weak acids discussed in the pH of weak acids in foods and their corresponding conjugate bases.
How do we calculate the pH of buffer solutions?
As an example of a buffer solution, let us consider the solution obtained when 3.00 mol citric acid (H3C6H5O7) and 2.00 mol monosodium citrate (NaH2C6H5O7) are added to sufficient water to produce a solution of total volume 1 dm³. The stoichiometric concentration of citric acid, namely, ca, is then 3.00 mol dm–3, while the stoichiometric concentration of sodium citrate, cb, is 2.00 mol dm–3. As a result of mixing the two components, some of the citric acid, say x mol dm–3, is converted to citrate ion and hydronium ion. We can now draw up a table in order to find the equilibrium concentrations in the usual way.
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 10–7 (negligible) x x
H2C6H5O7 2.00 x 2.00 + x
H3C6H5O7 3.00 (–x) 3.00 – x
We can now substitute concentrations in the equilibrium expression $K_{a}=\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}][\text{H}_{3}\text{O}^{+}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}$ from which we obtain $\text{1.4}\times \text{10}^{-3}\text{ mol dm}^{-3}=\frac{\text{(2.00 + }x\text{)}x}{\text{3.00}-x}\text{ mol dm}^{-3}$...(1) In order to solve this equation, we make the approximation that x is negligibly small compared with both 2.00 and 3.00, that is, that only a minute fraction of citric acid has converted to citrate ion. We then have $\frac{\text{2.00}x}{\text{3.00}}=\text{1.4}\times \text{10}^{-3}$
or \begin{align}x&=\frac{\text{3.00}}{\text{2.00}}\times \text{ 1.4 }\times \text{ 10}^{-3}\ \text{ }&=\text{2.1}\times \text{10}^{-3}\end{align}
Since x is only 0.1 percent of 2.00 or 3.00, the approximation is valid and there is no need to obtain a second approximation by feeding x back into Eq. (1). We can thus conclude that $[\text{H}_{3}\text{O}^{+}] = \text{2.1} \times \text{10}^{-3}\text{mol dm}^{-3}$ and $\text{pH} = \text{2.67}\,$ This example demonstrates two obvious features: 1 When the acid and its conjugate base are mixed, very little of the acid is converted to base, or vice versa. (x is small compared with 2.00 and 3.00.)
2 In a buffer mixture, the hydronium-ion concentration and the hydroxide-ion concentration are small compared with the concentrations of acid and conjugate base. ([H3O+] = 2.7 × 10–5 mol dm–3; [HO] = 3.7 × 10–10 mol dm–3 as compared with [H2C6H5O7] = 2.00 mol dm–3 and [H3C6H5O7] = 3.00 mol dm–3)
The Henderson-Hasselbach equation
Assuming that the above features are common to all buffer solutions, we make it very easy to handle them from a mathematical standpoint. Let us now consider the general problem of finding the pH of a buffer solution which is a mixture of a weak acid HA, of stoichiometric concentration ca, and its conjugate base A, of stoichiometric concentration cb. We can rearrange the expression for Ka of the weak acid (Equation 2 on the pH of solutions of weak acids) as follows:
$[\text{H}_{3}\text{O}^{+}]=K_{a}\times \frac{[\text{HA}]}{[\text{A}^{-}]}\label{6}$
(2)
Taking negative logarithms of both sides, we obtain
$-\text{log }[\text{H}_{3}\text{O}^{+}]=-\text{log }K_{a}-\text{log}\frac{[\text{HA}]}{[\text{A}^{-}]} \nonumber$
$\text{pH}=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\label{8}$
(3)
Equation $\ref{8}$ is called the Henderson-Hasselbalch equation and is often used by chemists and biologists to calculate the pH of a buffer.
As we saw in the case of the citric acid―sodium citrate buffer described earlier, the equilibrium concentrations of HA and A are usually almost identical to the stoichiometric concentrations. That is,
$[\text{HA}] \approx \text{c}_{a}$ and $[\text{A}^{-}]\approx\text{c}_{b}$
We can substitute these values into Eqs. (2) and (3) to obtain the very useful approximations $[\text{H}_{3}\text{O}^{+}]\approx K_{a}\times \frac{c_{a}}{c_{b}}$ (4) and $\text{pH}\approx \text{ p}K_{a}\text{ + log}\frac{c_{a}}{c_{b}}$ (5)
Example $1$: pH of a Buffer Solution
Calculate the pH of a buffer containing 3.93 g of NaH2PO4 and 4.31 g of Na2HPO4 per 450 mL of solution
Solution: First, we need to calculate the concentration of both the acid (NaH2PO4) and the base (Na2HPO4) in solution. For NaH2PO4 we have that the number of moles is
\begin{align}\text{n}_{\text{NaH}_{2}\text{PO}_{4}} &= \frac{\text{3.93 g}}{\text{119.98 g mol}^{-1}} \ \text{ } &= \text{3.27}\times \text{10}^{-2} \text{ mol} \end{align} and its concentration \begin{align}\left[\text{NaH}_{2}\text{PO}_{4}\right] &= \frac{\text{3.27}\times \text{10}^{-2} \text{ mol}}{\text{0.45 dm}^{3}}\ \text{ } &= \text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3} \end{align} and for Na2HPO4 \begin{align}\text{n}_{\text{Na}_{2}\text{HPO}_{4}} &= \frac{\text{4.31 g}}{\text{141.96 g mol}^{-1}} \ \text{ } &= \text{3.04}\times \text{10}^{-2} \text{ mol} \end{align}
\begin{align}\text{n}_{\text{Na}_{2}\text{HPO}_{4}} &= \frac{\text{4.31 g}}{\text{141.96 g mol}^{-1}} \ \text{ } &= \text{3.04}\times \text{10}^{-2} \text{ mol} \end{align}
Which means that the concentrations of H2PO4 and HPO42 are respectively 7.27 x 10–2 mol dm–3 and 7.55 x 10–2 mol dm–3. Using these values and a pKa2= 7.21 for the pair H2PO4/HPO42 (calculated from Ka2) in the Henderson Hasselbach equation, the pH of the buffer becomes \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&= \text{ 7.21}\text{+ log}\frac{\text{6.75} \times \text{10}^{-2}\text {mol dm}^{-3}}{\text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3}}\ \text{ }&= \text{7.17} \end{align} Phosphate buffers help to control the pH of physiological fluids and are often used in carbonated soft drinks.
Buffer solutions and the stability of food additives
Figure $1$ Aspartame, N-α-aspartyl-L-phenylalanine methyl ester
Aspartame is a high intensity sweetener composed by two amino acids L-aspartic acid and phenylalanine forming the dipeptide N-α-aspartyl-L-phenylalanine methyl ester. Aspartame is approximately 200 times sweeter than sucrose and presents synergistic interactions with other sweeteners including saccharin, cyclamate, stevioside, acesulfame K, and various sugars.[1] Aspartame is currently used as non-nutritive sweetener in various products including soft drinks, dry foods, ice cream, yogurt, fruit juices, and tabletop sweeteners.[2]
As a dipeptide, the stability of aspartame can be affected by extreme values of pH, temperature, and water activity. The degradation of aspartame generates the di-peptide aspartylphenylalanine and methanol. In some instances, aspartame can first form the cyclic compound diketopiperazine with the release of methanol. Ultimately, aspartylphenylalanine will be hydrolyzed into aspartate and phenylalanine, which posses a mayor risk for people with phenylketonuria. None of the degradation products has a sweet taste resulting in the loss of sweetness.[1]
The soft drink industry is one of the most important users of aspartame for the manufacture of diet beverages. Given that pH is a major factor affecting the stability of aspartame, it is critical to have a way to maintain its value within certain limits to ensure the sweetness expected in the product. Such control can be attained by having a buffering system in the formulation. In order to select the appropriate buffering system, it is necessary to consider not only the target pH, but also the concentration and chemical composition of the buffer as well as the potential interactions with other components in the formulation.
The concentration and the chemical composition of a buffer can affect the stability of aspartame. An study on the stability of this sweetener in solution showed that its degradation rate constants were smaller at pH 3 than at pH 7 and they increased as the concentration of the buffer was increased between 0.01 and 1.0 M. The study also investigated the effect of the composition of the buffer using phosphate and citrate acid-base pairs. At both pH 3 and 7 the degradation rates were higher in phosphate buffer. The smallest rate constant values for this buffer were registered at concentrations between 0.01 and 0.1 M. Even though these results can be explained in part by differences in the dissociation constants of the buffer components, the overall difference in the rate constants suggested that there might be additional causes affecting the degradation of aspartame in these buffers. The authors attributed such difference to the fact that phosphate buffer ions can act as bifunctional catalysts because they can both donate protons and accept groups in close proximity. This activity of the phosphate ion promotes the cyclization of aspartame to form diketopiperazine and methanol. Although citrate can also accept and donate electrons with aspartame, due to its larger size, it is not as efficient as the phosphate ions resulting in a slower degradation rate. On the other hand, the difference on the degradation rates at pH 3 and 7 can be explained considering that a pH 3, the free amino group of aspartame is protonated reducing its nucleophilic activity and preventing its interaction with the carbonyl group to form diketopiperazine. [3]
Example $2$: Buffer Preparation
Make the calculations to prepare a buffer to be used in the manufacture of a soft drink containing aspartame and will preserve its activity as sweetener.
Solution: This buffer should have a pH of 3 and be made with citric acid/citrate with a Ka1= 1.4 x 10–3 in a concentration between concentrations 0.01 to 0.1 M.
With this information we can use the Henderson-Hasselbach equation as follows \begin{align}\text{pH}&=\text{p}K_{a}\text{ + log}\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\ \text{3}&=-\text{log(1.4} \times \text{10}^{-3}\text{) + log}\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\ \text{3}&=\text{2.85}+ \text{log}\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\ \text{0.15}&=\text{log}\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\end{align} Applying the antilogarithm function in both sides of the equation we have \begin{align}\text{10 }^{ 0.15}&= \frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\ \text{1.41} &= \frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}\ \end{align} Now that we have the value for the ratio [H2C6H5O7]/[H3C6H5O7], we need to determine the concentrations for each compound. Since the concentration required for the buffer is 0.1, we need to consider the following equilibrium
$\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7} + \text{H}_{2}\text{O} \rightarrow \text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-} + \text{H}_{3}\text{O}^{+}$
and the concentration at the equilibrium for the acid an the base are
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 10–7 (negligible) x x
H2C6H5O7 0.00 x x
H3C6H5O7 0.10 (–x) 0.10 – x
Expressing the concentration of the acid and the base in terms of x, we have
\begin{align}\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}= \frac{\text{x}}{\text{0.1 - x}}=\text{1.41}\end{align} Solving for x \begin{align}\text{x}&=\text{1.41}\left(\text{0.1} - \text{x}\right)\ \text{x}&=\text{0.141}-\text{1.41x}\ \text{2.41x}&=\text{0.141}\ \text{x}&=\frac{\text{0.141}}{\text{2.41}}=\text{0.0585} \end{align} Therefore,
$[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]=\text{0.0585 mol dm}^{-3}$ and
$[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]=\text{0.1} - \text{0.0585}=\text{0.415 mol dm}^{-3}$
To confirm that this mixture of citric acid and citrate will yield a buffer of pH 3 we can calculate \begin{align}\text{pH}&=-\text{log(1.4} \times \text{10}^{-3}\text{) + log}\frac{(\text{0.0585 mol dm}^{-3})}{( \text{0.0415 mol dm}^{-3})}\ \text{ }&=\text{2.85} + \text{0.149}&\ \text{pH}&= 2.99 \end{align}
How do buffers work?
To better understand why a mixture of an acid and its conjugate base is resistant to a change in pH, let us go back to our first example: a mixture of citric acid (3 mol dm–3) and sodium citrate (2 mol dm–3). What would happen if we now added 0.50 mol sodium hydroxide to 1 dm3 of this mixture? The added hydroxide ion will attack both the acids present, namely, the hydronium ion and citric acid. Since the hydronium-ion concentration is so small, very little hydroxide ion will be consumed by reaction with the hydronium ion. Most of it will be consumed by reaction with citric acid. Further, since the hydroxide ion is such a strong base, the reaction
$\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7} + \text{OH}^{-} \rightarrow \text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-} + \text{H}_{2}\text{O}$ will go virtually to completion, and 0.50 mol citric acid will be consumed. The same amount of citrate ion will be produced. In tabular form:
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 2.1 x 10–4 (negligible) Small approx. 2.1 x 10–4
H2C6H5O7 2.00 0.5 2.00 + 0.5 = 2.5
H3C6H5O7 3.00 (– 0.5) 3.00 – 0.5 = 2.5
Substituting the equilibrium concentrations of base (citrate ion) and conjugate acid (citric acid) into the Henderson-Hasselbalch equation, Eq. (3), we have
\begin{align}\text{pH}&=\text{p}K_{a}\text{ + log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&=-\text{log(1.4} \times \text{10}^{-3}\text{) + log}\frac{\text{(2.50 mol dm}^{-3}\text{)}}{\text{(2.50 mol dm}^{-3}\text{)}}\ \text{ }&=-\left(\text{0.15}-\text{3} \right)+ \text{log}\left(\text{1}\right)\ \text{ }&=\text{2.85 + 0}=\text{2.85}\end{align}
The addition of 0.5 mol sodium hydroxide to the buffer mixture raised its pH from 2.5 to only 2.85.
Now, what would be the pH of the solution if the 0.5 mol of sodium hydroxide would have been added to a cubic decimeter of pure water in absence of citric acid?
This amount of sodium hydroxide would produce a solution containing 0.5 M of hydroxide ions with a pOH equal to
$\text{pOH}=-\text{log}(\text{0.5}) = \text{0.3}\,$ and $\text{pH}= \text{14} - \text{ 0.3} = \text{13.7}\,$ This shows you that the buffer is extremely effective at resisting a change in pH because the added hydroxide ion attacks the weak acid (in very high concentration) rather than the hydronium ion (in very low concentration). The major effect of the addition of the hydroxide ion is thus to change the ratio of acid to conjugate base, i.e., to change the value of $\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ As long as the amount of weak acid is much larger than the amount of base added, this ratio is not altered by very much. Since the hydronium-ion concentration is governed by $[\text{H}_{3}\text{O}^{+}]=K_{a}\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ Similarly, if instead of a strong base, a strong acid such as hydrochloric was added to the buffer mixture, it would react with the citrate in solution. Although, the pH would decrease, the change would be minor again.
Example $3$ pH of a Solution
Find the pH of the solution obtained when 2.00 mol H2C6H5O7 and 0.80 mol H3C6H5O7 are mixed to give 2.5 dm3 of solution. Kb(H2C6H5O7) = 7.1 × 10–12 mol dm–3.
Solution In order to use Eq. (4),we need first to have the value of
\begin{align}K_{a}\left(\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}\right)&= \frac{K_{w}}{K_{b}\left(\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}\right)}\ \text{ }&=\frac{\text{1.00}\times \text{ 10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{7.1 }\times \text{ 10}^{-12}\text{ mol dm}^{-3}}\ \text{ }&=\text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{-3} \end{align}
Given that the final volume of the solution is 2.5 dm3, the concentration of the acid and base in solution are ca = 0.32 mol dm–3 and cb = 0.80 mol dm–3. Thus
\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]&\approx K_{a}\times \frac{c_{a}}{c_{b}}\ \text{ }&\approx \text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{3}\times \frac{\text{ 0.32 mol dm}^{-3}}{\text{0.80 mol dm}^{-3}}\ \text{ }&\approx \text{5.6}\times \text{ 10}^{-4}\text{ mol dm}^{-3}\end{align}
from which
and $\text{pH} = \text{3.25}\,$
From ChemPRIME: 14.7: Buffer Solutions
References
1. Chemistry and technology of soft drinks and fruit juices. 2nd ed. 2005 Ashurst, P. R.
2. Food Additives, 2nd ed. 2002, Branen, A., Davidson, M.P., Salminen, S. and Thorngate III, J.H.
3. Bell, L.N. and Wetzel, C.R. 1995. Aspartame degradation in solution as impacted by buffer type and concentration. J. Agric. Food Chem. 43:2608-2612. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.08%3A_Buffer_Solutions/14.8.01%3A_Foods-_Food_Additives.txt |
So far you have learned about pH in solutions where either a single acid, such as citric acid, or a single base, such as the citrate ion have been added to water. Now, let us consider solutions prepared with both an acid and a base. The simplest case of such a solution occurs when the acid and base are conjugate to each other and also are present in comparable amounts. Solutions of this special kind are called buffer solutions because it is difficult to change their pH even when an appreciable amount of strong acid or strong base is added.
Why are buffer solutions important in foods?
Buffering solutions in foods play an important role in maintaining specific pH values for optimum activity of enzymes, protein solubility, and functionality. As discussed in previous exemplars, pH may also modify the color and flavor of foods and it is a critical factor in the preservation of many processed foods. Buffering solutions are also used as reaction media in the production of food ingredients and additives. Overall pH control is a major factor in maintaining the physical, chemical, and microbiological stability of foods.
Foods contain numerous compounds able to form buffering systems. Molecules with acid-base properties naturally found in foods include amino acids, organic acids, proteins, and charged polysaccharides. Other buffering systems are intentionally added to processed foods, examples of these are the weak acids discussed in the pH of weak acids in foods and their corresponding conjugate bases.
How do we calculate the pH of buffer solutions?
As an example of a buffer solution, let us consider the solution obtained when 3.00 mol citric acid (H3C6H5O7) and 2.00 mol monosodium citrate (NaH2C6H5O7) are added to sufficient water to produce a solution of total volume 1 dm³. The stoichiometric concentration of citric acid, namely, ca, is then 3.00 mol dm–3, while the stoichiometric concentration of sodium citrate, cb, is 2.00 mol dm–3. As a result of mixing the two components, some of the citric acid, say x mol dm–3, is converted to citrate ion and hydronium ion. We can now draw up a table in order to find the equilibrium concentrations in the usual way.
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 10–7 (negligible) x x
H2C6H5O7 2.00 x 2.00 + x
H3C6H5O7 3.00 (–x) 3.00 – x
We can now substitute concentrations in the equilibrium expression $K_{a}=\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}][\text{H}_{3}\text{O}^{+}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}$ from which we obtain $\text{1.4}\times \text{10}^{-3}\text{ mol dm}^{-3}=\frac{\text{(2.00 + }x\text{)}x}{\text{3.00}-x}\text{ mol dm}^{-3}$...(1) In order to solve this equation, we make the approximation that x is negligibly small compared with both 2.00 and 3.00, that is, that only a minute fraction of citric acid has converted to citrate ion. We then have $\frac{\text{2.00}x}{\text{3.00}}=\text{1.4}\times \text{10}^{-3}$
or \begin{align}x&=\frac{\text{3.00}}{\text{2.00}}\times \text{ 1.4 }\times \text{ 10}^{-3}\ \text{ }&=\text{2.1}\times \text{10}^{-3}\end{align}
Since x is only 0.1 percent of 2.00 or 3.00, the approximation is valid and there is no need to obtain a second approximation by feeding x back into Eq. (1). We can thus conclude that $[\text{H}_{3}\text{O}^{+}] = \text{2.1} \times \text{10}^{-3}\text{mol dm}^{-3}$ and $\text{pH} = \text{2.67}\,$ This example demonstrates two obvious features: 1 When the acid and its conjugate base are mixed, very little of the acid is converted to base, or vice versa. (x is small compared with 2.00 and 3.00.)
2 In a buffer mixture, the hydronium-ion concentration and the hydroxide-ion concentration are small compared with the concentrations of acid and conjugate base. ([H3O+] = 2.7 × 10–5 mol dm–3; [HO] = 3.7 × 10–10 mol dm–3 as compared with [H2C6H5O7] = 2.00 mol dm–3 and [H3C6H5O7] = 3.00 mol dm–3)
The Henderson-Hasselbalch equation
Assuming that the above features are common to all buffer solutions, we make it very easy to handle them from a mathematical standpoint. Let us now consider the general problem of finding the pH of a buffer solution which is a mixture of a weak acid HA, of stoichiometric concentration ca, and its conjugate base A, of stoichiometric concentration cb. We can rearrange the expression for Ka of the weak acid (Equation 2 on the pH of solutions of weak acids) as follows:
$[\text{H}_{3}\text{O}^{+}]=K_{a}\times \frac{[\text{HA}]}{[\text{A}^{-}]}$ (2) Taking negative logarithms of both sides, we obtain $-\text{log }[\text{H}_{3}\text{O}^{+}]=-\text{log }K_{a}-\text{log}\frac{[\text{HA}]}{[\text{A}^{-}]}$ $\text{pH}=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}$ (3) Equation (3) is called the Henderson-Hasselbalch equation and is often used by chemists and biologists to calculate the pH of a buffer.
As we saw in the case of the citric acid―sodium citrate buffer described earlier, the equilibrium concentrations of HA and A are usually almost identical to the stoichiometric concentrations. That is,
$[\text{HA}] \approx \text{c}_{a}$ and $[\text{A}^{-}]\approx\text{c}_{b}$
We can substitute these values into Eqs. (2) and (3) to obtain the very useful approximations $[\text{H}_{3}\text{O}^{+}]\approx K_{a}\times \frac{c_{a}}{c_{b}}$ (4) and $\text{pH}\approx \text{ p}K_{a}\text{ + log}\frac{c_{a}}{c_{b}}$ (5)
Example $1$ pH of Buffer
Calculate the pH of a buffer containing 3.93 g of NaH2PO4 and 4.31 g of Na2HPO4 per 450 mL of solution
Solution: First, we need to calculate the concentration of both the acid (NaH2PO4) and the base (Na2HPO4) in solution. For NaH2PO4 we have that the number of moles is
\begin{align}\text{n}_{\text{NaH}_{2}\text{PO}_{4}} &= \frac{\text{3.93 g}}{\text{119.98 g mol}^{-1}} \ \text{ } &= \text{3.27}\times \text{10}^{-2} \text{ mol} \end{align} and its concentration \begin{align}\left[\text{NaH}_{2}\text{PO}_{4}\right] &= \frac{\text{3.27}\times \text{10}^{-2} \text{ mol}}{\text{0.45 dm}^{3}}\ \text{ } &= \text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3} \end{align} and for Na2HPO4 \begin{align}\text{n}_{\text{Na}_{2}\text{HPO}_{4}} &= \frac{\text{4.31 g}}{\text{141.96 g mol}^{-1}} \ \text{ } &= \text{3.04}\times \text{10}^{-2} \text{ mol} \end{align} \begin{align}\left[\text{Na}_{2}\text{HPO}_{4}\right] &= \frac{\text{3.04}\times \text{10}^{-2} \text{ mol}}{\text{0.45 dm}^{3}}\ \text{ } &= \text{6.75} \times \text{10}^{-2} \text {mol dm}^{-3} \end{align}
Which means that the concentrations of H2PO4 and HPO42 are respectively 7.27 x 10–2 mol dm–3 and 7.55 x 10–2 mol dm–3. Using these values and a pKa2= 7.21 for the pair H2PO4/HPO42 (calculated from Ka2) in the Henderson Hasselbach equation, the pH of the buffer becomes
\begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&= \text{ 7.21}\text{+ log}\frac{\text{6.75} \times \text{10}^{-2}\text {mol dm}^{-3}}{\text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3}}\ \text{ }&= \text{7.17} \end{align} Phosphate buffers help to control the pH of physiological fluids and are often used in carbonated soft drinks.
Buffer solutions and the production of food ingredients
Another example of the importance of buffers in food is the production of food ingredients and bioactive compounds derived from foods. In the last few years, numerous research papers have been published regarding the enzymic modification of food proteins to improve their functionality, sensory, and bioavailability properties. Buffer solutions are employed to conduct enzymic reactions maintaining the pH for optimum enzyme activity throughout the process.
As example of enzymic modification of food proteins we have the oligomerization of milk proteins. The enzymes employed in this process include transglutaminase, peroxidase, laccase, monoamine oxidase, and tyrosine. They act via the oxidation of aromatic and sulphur containing amino acid residues and bonding of oxidized protein groups. Buffer systems for these enzymes include sodium borate decahydrate, potasium phosphate, tris, succinate, and sodium phosphate. [1]
Oligomerization with lactoperoxidase, lipoxygenase, and β-galactosidase of milk proteins is significantly enhanced when conducted in buffer solutions instead of demineralized water. The extent of oligomerization and therefore the size of the resulting molecules depends on the buffer system. Besides the expected effect on the three-dimensional structure and catalytic activity of the enzyme and substrate, the buffers have shown further effects on the oligomerization of milk proteins, it appears that there exists a complex interaction between the buffer, protein, and enzyme accounting for the difference in the molecular weight of the oligomers.[1] In addition, polymerization of hemoglobin and milk proteins has been observed in borate and phosphate buffers without the addition of enzyme.[2] [3]
Boric acid
Example $2$: Buffer Solution
What is the pH of a solution prepared by mixing 250 mL of 0.5 M boric acid (B(OH)3(H2O)) and 750 ml of 0.8 M sodium borate (NaB(OH)4)?
Since both the acid and the base were diluted when the solution was prepared we need to calculate their concentrations with the new volume of 1000 mL = 1 dm3.
Using the relationship
$V_{old} \times c_{old} = V_{new} \times c_{new}$ The new concentration becomes $c_{new}=\frac{V_{old} \times c_{old}}{V_{new}}$ So, for boric acid we have
\begin{align} c_{new}\text{ }_{\text{B}\left(\text{OH}\right)_{3}\text{H}_{2}\text{O}}&=\frac{\text{0.50 mol dm}^{-3} \times \text{ 0.25 dm}^{-3}}{\text{1.0 dm}^{-3}}\ \text{ }&=\text{0.125 mol dm}^{-3} \end{align}
Similarly for borate The Ka for the equilibrium $\text{B}\left(\text{OH}\right)_{3}\left(\text{H}_{2}\text{O}\right) + \text{H}_{2}\text{O}\rightleftharpoons \text{H}_{3}\text{O}^{+} + \text{B}\left(\text{OH}\right)_{4}^{-}$ From our collection of acid-base resources is Ka= 5.8 x 10-10 Thus, using equation (4) \begin{align}\left[\text{H}_{3}\text{O}^{+}\right]&\approx K_{a}\times \frac{c_{a}}{c_{b}}\ \text{ }&\approx \text{5.8}\times \text{ 10}^{-10}\text{ mol dm}^{3}\times \frac{\text{ 0.125 mol dm}^{-3}}{\text{0.60 mol dm}^{-3}}\ \text{ }&\approx \text{1.2}\times \text{ 10}^{-10}\text{ mol dm}^{-3}\end{align} The pH of the solution is then and \begin{align}\text{pH} &= -\text{log}\left(\text{1.2}\times \text{ 10}^{-10}\right)\ \text{ } &= \text{9.92}\end{align}
Boric acid tends to accumulate in adipose tissue, specially in the central nervous system. Since the risks associated with this compound in the human body are still unknown, it is no longer used in food.[4] However, food additives obtained through enzymic processes usually undergo several separation or purification steps which would eliminate or significantly reduce the amount of residual borate buffer in the final product.
How do buffers work?
To better understand why a mixture of an acid and its conjugate base is resistant to a change in pH, let us go back to our first example: a mixture of citric acid (3 mol dm–3) and sodium citrate (2 mol dm–3). What would happen if we now added 0.50 mol sodium hydroxide to 1 dm3 of this mixture? The added hydroxide ion will attack both the acids present, namely, the hydronium ion and citric acid. Since the hydronium-ion concentration is so small, very little hydroxide ion will be consumed by reaction with the hydronium ion. Most of it will be consumed by reaction with citric acid. Further, since the hydroxide ion is such a strong base, the reaction
$\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7} + \text{OH}^{-} \rightarrow \text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-} + \text{H}_{2}\text{O}$ will go virtually to completion, and 0.50 mol citric acid will be consumed. The same amount of citrate ion will be produced. In tabular form:
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 2.1 x 10–4 (negligible) Small approx. 2.1 x 10–4
H2C6H5O7 2.00 0.5 2.00 + 0.5 = 2.5
H3C6H5O7 3.00 (– 0.5) 3.00 – 0.5 = 2.5
Substituting the equilibrium concentrations of base (citrate ion) and conjugate acid (citric acid) into the Henderson-Hasselbalch equation, Eq. (3), we have
\begin{align}\text{pH}&=\text{p}K_{a}\text{ + log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&=-\text{log(1.4} \times \text{10}^{-3}\text{) + log}\frac{\text{(2.50 mol dm}^{-3}\text{)}}{\text{(2.50 mol dm}^{-3}\text{)}}\ \text{ }&=-\left(\text{0.15}-\text{3} \right)+ \text{log}\left(\text{1}\right)\ \text{ }&=\text{2.85 + 0}=\text{2.85}\end{align}
The addition of 0.5 mol sodium hydroxide to the buffer mixture raised its pH from 2.5 to only 2.85.
Now, what would be the pH of the solution if the 0.5 mol of sodium hydroxide would have been added to a cubic decimeter of pure water in absence of citric acid?
This amount of sodium hydroxide would produce a solution containing 0.5 M of hydroxide ions with a pOH equal to
$\text{pOH}=-\text{log}(\text{0.5}) = \text{0.3}\,$ and $\text{pH}= \text{14} - \text{ 0.3} = \text{13.7}\,$ This shows you that the buffer is extremely effective at resisting a change in pH because the added hydroxide ion attacks the weak acid (in very high concentration) rather than the hydronium ion (in very low concentration). The major effect of the addition of the hydroxide ion is thus to change the ratio of acid to conjugate base, i.e., to change the value of $\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ As long as the amount of weak acid is much larger than the amount of base added, this ratio is not altered by very much. Since the hydronium-ion concentration is governed by $[\text{H}_{3}\text{O}^{+}]=K_{a}\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ Similarly, if instead of a strong base, a strong acid such as hydrochloric was added to the buffer mixture, it would react with the citrate in solution. Although, the pH would decrease, the change would be minor again.
Example $3$ pH of Solution
Find the pH of the solution obtained when 2.00 mol H2C6H5O7 and 0.80 mol H3C6H5O7 are mixed to give 2.5 dm3 of solution. Kb(H2C6H5O7) = 7.1 × 10–12 mol dm–3.
Solution In order to use Eq. (4),we need first to have the value of
\begin{align}K_{a}\left(\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}\right)&= \frac{K_{w}}{K_{b}\left(\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}\right)}\ \text{ }&=\frac{\text{1.00}\times \text{ 10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{7.1 }\times \text{ 10}^{-12}\text{ mol dm}^{-3}}\ \text{ }&=\text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{-3} \end{align}
Given that the final volume of the solution is 2.5 dm3, the concentration of the acid and base in solution are ca = 0.32 mol dm–3 and cb = 0.80 mol dm–3. Thus
\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]&\approx K_{a}\times \frac{c_{a}}{c_{b}}\ \text{ }&\approx \text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{3}\times \frac{\text{ 0.32 mol dm}^{-3}}{\text{0.80 mol dm}^{-3}}\ \text{ }&\approx \text{5.6}\times \text{ 10}^{-4}\text{ mol dm}^{-3}\end{align}
from which
and $\text{pH} = \text{3.25}\,$
From ChemPRIME: 14.7: Buffer Solutions
References
1. Hiller, B. and Lorenzen P.-C. 2008 Effect of buffer systems on the oligomerization of milk proteins. LWT-Food Sci. Technol. 41:1140-1144.
2. Chen, K., Ballas, S.K., Hantgan, R.R., and Kim-Shapiro, D.B. 2004 Aggregation of normal and sickle hemoglobin in high concentration phosphate buffer. Biophys. J. 87:4113-4121.
3. Parker, W.C., and Bearn, A.G. 1963. Boric acid-induced heterohenity of conalbumin by starch-gel electrophoresis. Nature, 199, 1184-1186.
4. Food Chemistry 3rd Ed. 2004 Belitz, et al. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.08%3A_Buffer_Solutions/14.8.02%3A_Foods-_Production_of_Food_Ingredients.txt |
So far you have learned about pH in solutions where either a single acid, such as citric acid, or a single base, such as the citrate ion have been added to water. Now, let us consider solutions prepared with both an acid and a base. The simplest case of such a solution occurs when the acid and base are conjugate to each other and also are present in comparable amounts. Solutions of this special kind are called buffer solutions because it is difficult to change their pH even when an appreciable amount of strong acid or strong base is added.
Why are buffer solutions important in foods?
Buffer solutions in foods play an important role in maintaining specific pH values for optimum activity of enzymes, protein solubility, and functionality. As discussed in previous exemplars, pH may also modify the color and flavor of foods and it is a critical factor in the preservation of many processed foods. Buffering solutions are also used as reaction media in the production of food ingredients and additives. Overall pH control is a major factor in maintaining the physical, chemical, and microbiological stability of foods.
Foods contain numerous compounds able to form buffering systems. Molecules with acid-base properties naturally found in foods include amino acids, organic acids, proteins, and charged polysaccharides. Other buffering systems are intentionally added to processed foods, examples of these are the weak acids discussed in the pH of weak acids in foods and their corresponding conjugate bases.
How do we calculate the pH of buffer solutions?
As an example of a buffer solution, let us consider the solution obtained when 3.00 mol citric acid (H3C6H5O7) and 2.00 mol monosodium citrate (NaH2C6H5O7) are added to sufficient water to produce a solution of total volume 1 dm³. The stoichiometric concentration of citric acid, namely, ca, is then 3.00 mol dm–3, while the stoichiometric concentration of sodium citrate, cb, is 2.00 mol dm–3. As a result of mixing the two components, some of the citric acid, say x mol dm–3, is converted to citrate ion and hydronium ion. We can now draw up a table in order to find the equilibrium concentrations in the usual way.
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 10–7 (negligible) x x
H2C6H5O7 2.00 x 2.00 + x
H3C6H5O7 3.00 (–x) 3.00 – x
We can now substitute concentrations in the equilibrium expression $K_{a}=\frac{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}][\text{H}_{3}\text{O}^{+}]}{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}$ from which we obtain $\text{1.4}\times \text{10}^{-3}\text{ mol dm}^{-3}=\frac{\text{(2.00 + }x\text{)}x}{\text{3.00}-x}\text{ mol dm}^{-3}$...(1) In order to solve this equation, we make the approximation that x is negligibly small compared with both 2.00 and 3.00, that is, that only a minute fraction of citric acid has converted to citrate ion. We then have $\frac{\text{2.00}x}{\text{3.00}}=\text{1.4}\times \text{10}^{-3}$
or \begin{align}x&=\frac{\text{3.00}}{\text{2.00}}\times \text{ 1.4 }\times \text{ 10}^{-3}\ \text{ }&=\text{2.1}\times \text{10}^{-3}\end{align}
Since x is only 0.1 percent of 2.00 or 3.00, the approximation is valid and there is no need to obtain a second approximation by feeding x back into Eq. (1). We can thus conclude that $[\text{H}_{3}\text{O}^{+}] = \text{2.1} \times \text{10}^{-3}\text{mol dm}^{-3}$ and $\text{pH} = \text{2.67}\,$ This example demonstrates two obvious features: 1 When the acid and its conjugate base are mixed, very little of the acid is converted to base, or vice versa. (x is small compared with 2.00 and 3.00.)
2 In a buffer mixture, the hydronium-ion concentration and the hydroxide-ion concentration are small compared with the concentrations of acid and conjugate base. ([H3O+] = 2.7 × 10–5 mol dm–3; [HO] = 3.7 × 10–10 mol dm–3 as compared with [H2C6H5O7] = 2.00 mol dm–3 and [H3C6H5O7] = 3.00 mol dm–3)
The Henderson-Hasselbach equation
Assuming that the above features are common to all buffer solutions, we make it very easy to handle them from a mathematical standpoint. Let us now consider the general problem of finding the pH of a buffer solution which is a mixture of a weak acid HA, of stoichiometric concentration ca, and its conjugate base A, of stoichiometric concentration cb. We can rearrange the expression for Ka of the weak acid (Equation 2 on the pH of solutions of weak acids) as follows:
$[\text{H}_{3}\text{O}^{+}]=K_{a}\times \frac{[\text{HA}]}{[\text{A}^{-}]}$ (2) Taking negative logarithms of both sides, we obtain $-\text{log }[\text{H}_{3}\text{O}^{+}]=-\text{log }K_{a}-\text{log}\frac{[\text{HA}]}{[\text{A}^{-}]}$ $\text{pH}=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}$ (3) Equation (3) is called the Henderson-Hasselbalch equation and is often used by chemists and biologists to calculate the pH of a buffer.
As we saw in the case of the citric acid―sodium citrate buffer described earlier, the equilibrium concentrations of HA and A are usually almost identical to the stoichiometric concentrations. That is,
$[\text{HA}] \approx \text{c}_{a}$ and $[\text{A}^{-}]\approx\text{c}_{b}$
We can substitute these values into Eqs. (2) and (3) to obtain the very useful approximations $[\text{H}_{3}\text{O}^{+}]\approx K_{a}\times \frac{c_{a}}{c_{b}}$ (4) and $\text{pH}\approx \text{ p}K_{a}\text{ + log}\frac{c_{a}}{c_{b}}$ (5)
EXAMPLE 1 Calculate the pH of a buffer containing 3.93 g of NaH2PO4 and 4.31 g of Na2HPO4 per 450 mL of solution.
Solution First, we need to calculate the concentration of both the acid (NaH2PO4) and the base (Na2HPO4) in solution. For NaH2PO4 we have that the number of moles is
\begin{align}\text{n}_{\text{NaH}_{2}\text{PO}_{4}} &= \frac{\text{3.93 g}}{\text{119.98 g mol}^{-1}} \ \text{ } &= \text{3.27}\times \text{10}^{-2} \text{ mol} \end{align} and its concentration \begin{align}\left[\text{NaH}_{2}\text{PO}_{4}\right] &= \frac{\text{3.27}\times \text{10}^{-2} \text{ mol}}{\text{0.45 dm}^{3}}\ \text{ } &= \text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3} \end{align} and for Na2HPO4 \begin{align}\text{n}_{\text{Na}_{2}\text{HPO}_{4}} &= \frac{\text{4.31 g}}{\text{141.96 g mol}^{-1}} \ \text{ } &= \text{3.04}\times \text{10}^{-2} \text{ mol} \end{align} \begin{align}\left[\text{Na}_{2}\text{HPO}_{4}\right] &= \frac{\text{3.04}\times \text{10}^{-2} \text{ mol}}{\text{0.45 dm}^{3}}\ \text{ } &= \text{6.75} \times \text{10}^{-2} \text {mol dm}^{-3} \end{align} Which means that the concentrations of H2PO4 and HPO42 are respectively 7.27 x 10–2 mol dm–3 and 7.55 x 10–2 mol dm–3. Using these values and a pKa2= 7.21 for the pair H2PO4/HPO42 (calculated from Ka2) in the Henderson Hasselbach equation, the pH of the buffer becomes \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&= \text{ 7.21}\text{+ log}\frac{\text{6.75} \times \text{10}^{-2}\text {mol dm}^{-3}}{\text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3}}\ \text{ }&= \text{7.17} \end{align} Phosphate buffers help to control the pH of physiological fluids and are often used in carbonated soft drinks.
The pH of buffer solutions and the effect of polyols in foods
The pH of foods is influenced not only by the compounds with obvious acid-base properties. There exist both ionic (salts, NaCl) and non-ionic (sucrose and glycerol) compounds present in foods with ability to alter their pH even in the presence of a buffering system. Since the pH of foods affects both their quality and stability, it is very important to know how the pH of a food product will be affected by its composition and how it will evolve over time.
Polyols in foods
Xylitol Sorbitol
Mannitol Lactitol
Maltitiol Isomaltitiol
Due to their sweetening properties, short chain polyols (compounds containing more than 2 alcohol groups) are often used to substitute sugars. Besides their sweetening properties, polyols can perform a variety of functions in food. They are added as bulking agents, antimicrobials, prebiotics, humectants, and modifiers of freezing point, crystallization, and mouth-feel.[1] Polyols are naturally found in food or obtained by hydrogenation of the corresponding sugar. [2] The compounds shown in the table on the side are used in the formulation of sugar-free dietary foods including products for diabetics, however their use is limited by their laxative and hygroscopic properties.
Polyols have been reported to lower the pH of hydrochloric acid solutions and phosphate buffers. The ability of lowering the pH of a buffer solution appeared to be directly proportional to the molecular weight of the polyol with phosphate buffers being more sensitive to the effect of polyols than citrate buffers. [3] [4] The lower pH value of phosphate buffer solutions when sucrose was added has been attributed to a reduction in the apparent pKa of the phosphate buffer.[5] The apparent pKa values for the second dissociation of phosphoric acid decreased slightly from 6.79 to 6.74 when 2 molal glycerol was added into solution, while a significant decrease to 6.56 was observed with 2 molal sucrose. The authors of this investigation explained that such reduction in the apparent pKa value could be the result of a reduction in the hydration sphere around the buffer ions which would favor proton dissociation. The level of hydration of the ions in turn depends on polyol-water interactions and the charge density of the buffer ions.
size=150</chemeddl-jmol2>
Sucrose
EXAMPLE 2 Calculate the pH of the phosphate buffer solution from EXAMPLE 1 if in addition to NaH2PO4 and Na2HPO4, it contains 2 molal sucrose.
Solution From EXAMPLE 1 we know that the concentrations in solution for NaH2PO4 and Na2HPO4 are 7.27 x 10 –2 and 6.75 x 10 –2 mol dm–2 respectively. A phosphate buffer solution like this containing 2 molal sucrose would register a shift in apparent pKa from 6.79 to 6.56. Under these conditions, the Henderson-Hasselbach equation becomes \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&= \text{ 6.56}\text{ + log}\frac{\text{6.75} \times \text{10}^{-2}\text {mol dm}^{-3}}{\text{7.27} \times \text{10}^{-2} \text {mol dm}^{-3}}\ \text{ }&= \text{6.59} \end{align} this pH value is lower compared to 6.82, the pH of this solutions in the absence of sucrose. How do buffers work?
To better understand why a mixture of an acid and its conjugate base is resistant to a change in pH, let us go back to our first example: a mixture of citric acid (3 mol dm–3) and sodium citrate (2 mol dm–3). What would happen if we now added 0.50 mol sodium hydroxide to 1 dm3 of this mixture? The added hydroxide ion will attack both the acids present, namely, the hydronium ion and citric acid. Since the hydronium-ion concentration is so small, very little hydroxide ion will be consumed by reaction with the hydronium ion. Most of it will be consumed by reaction with citric acid. Further, since the hydroxide ion is such a strong base, the reaction
$\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7} + \text{OH}^{-} \rightarrow \text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-} + \text{H}_{2}\text{O}$ will go virtually to completion, and 0.50 mol citric acid will be consumed. The same amount of citrate ion will be produced. In tabular form:
Species
Initial Concentration
mol dm–3
Change in Concentration
mol dm–3
Equilibrium Concentration
mol dm–3
H3O+ 2.1 x 10–4 (negligible) Small approx. 2.1 x 10–4
H2C6H5O7 2.00 0.5 2.00 + 0.5 = 2.5
H3C6H5O7 3.00 (– 0.5) 3.00 – 0.5 = 2.5
Substituting the equilibrium concentrations of base (citrate ion) and conjugate acid (citric acid) into the Henderson-Hasselbalch equation, Eq. (3), we have
\begin{align}\text{pH}&=\text{p}K_{a}\text{ + log}\frac{[\text{A}^{-}]}{[\text{HA}]}\ \text{ }&=-\text{log(1.4} \times \text{10}^{-3}\text{) + log}\frac{\text{(2.50 mol dm}^{-3}\text{)}}{\text{(2.50 mol dm}^{-3}\text{)}}\ \text{ }&=-\left(\text{0.15}-\text{3} \right)+ \text{log}\left(\text{1}\right)\ \text{ }&=\text{2.85 + 0}=\text{2.85}\end{align}
The addition of 0.5 mol sodium hydroxide to the buffer mixture raised its pH from 2.5 to only 2.85.
Now, what would be the pH of the solution if the 0.5 mol of sodium hydroxide would have been added to a cubic decimeter of pure water in absence of citric acid?
This amount of sodium hydroxide would produce a solution containing 0.5 M of hydroxide ions with a pOH equal to
$\text{pOH}=-\text{log}(\text{0.5}) = \text{0.3}\,$ and $\text{pH}= \text{14} - \text{ 0.3} = \text{13.7}\,$ This shows you that the buffer is extremely effective at resisting a change in pH because the added hydroxide ion attacks the weak acid (in very high concentration) rather than the hydronium ion (in very low concentration). The major effect of the addition of the hydroxide ion is thus to change the ratio of acid to conjugate base, i.e., to change the value of $\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ As long as the amount of weak acid is much larger than the amount of base added, this ratio is not altered by very much. Since the hydronium-ion concentration is governed by $[\text{H}_{3}\text{O}^{+}]=K_{a}\frac{[\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}]}{[\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}]}$ Similarly, if instead of a strong base, a strong acid such as hydrochloric was added to the buffer mixture, it would react with the citrate in solution. Although, the pH would decrease, the change would be minor again.
EXAMPLE 3 Find the pH of the solution obtained when 2.00 mol H2C6H5O7 and 0.80 mol H3C6H5O7 are mixed to give 2.5 dm3 of solution. Kb(H2C6H5O7) = 7.1 × 10–12 mol dm–3.
Solution In order to use Eq. (4),we need first to have the value of
\begin{align}K_{a}\left(\text{H}_{3}\text{C}_{6}\text{H}_{5}\text{O}_{7}\right)&= \frac{K_{w}}{K_{b}\left(\text{H}_{2}\text{C}_{6}\text{H}_{5}\text{O}_{7}^{-}\right)}\ \text{ }&=\frac{\text{1.00}\times \text{ 10}^{-14}\text{ mol}^{2}\text{ dm}^{-6}}{\text{7.1 }\times \text{ 10}^{-12}\text{ mol dm}^{-3}}\ \text{ }&=\text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{-3} \end{align}
Given that the final volume of the solution is 2.5 dm3, the concentration of the acid and base in solution are ca = 0.32 mol dm–3 and cb = 0.80 mol dm–3. Thus
\begin{align}\left[\text{H}_{3}\text{O}^{+}\right]&\approx K_{a}\times \frac{c_{a}}{c_{b}}\ \text{ }&\approx \text{1.4}\times \text{ 10}^{-3}\text{ mol dm}^{3}\times \frac{\text{ 0.32 mol dm}^{-3}}{\text{0.80 mol dm}^{-3}}\ \text{ }&\approx \text{5.6}\times \text{ 10}^{-4}\text{ mol dm}^{-3}\end{align}
from which
and $\text{pH} = \text{3.25}\,$
References
1. Food Additives, 2nd ed. 2002, Branen, A., Davidson, M.P., Salminen, S. and Thorngate III, J.H.
2. Food Chemistry, 3rd Ed. 2004 Belitz, et al.
3. Boyer, J.P.H., Corriu, R.J., Perz, R.J.M. and Reye, C.G. 1975. Acidity in aqueous mixed solvent systems - III. Influence of the organic solvent molecular structure on the acidity of aqueo-organic mixtures (acidity function, water structure, hydrophilic and hydrophobic effects). Tetrahedron, 31:2075-2079.
4. Bell, L.N. and Labuza, T.P. 1992. Compositional influence of the pH of reduced-moisture solutions. J. Food Sci. 57:732-734.
5. Chuy, S. and Bell, L.N. 2006. Buffer pH and pKa values as affected by added glycerol and sucrose. Food. Res. Intl. 39:342-348. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.08%3A_Buffer_Solutions/14.8.03%3A_Foods-_The_Effect_of_Polyols.txt |
We mentioned that in most titrations it is necessary to add an indicator which produces a sudden color change at the equivalence point. A typical indicator for acid-base titrations is phenolphthalein, HC20H13O4. Phenolphthalein, whose structure is shown below, is a colorless weak acid (Ka = 3 × 10–10 mol/L). Its conjugate base, C20H13O4 has a strong pinkish-red color. In order to simplify, we will write the phenolphthalein molecule as HIn (protonated indicator) and its pink conjugate base as In. In aqueous solution, phenolphthalein will present the following equilibrium
$\ce{HIn + H_{2}O \rightleftharpoons In^{-} + H_{3}O^{+}}\label{1}$
According to Le Chatelier’s principle, the equilibrium expressed in equation $\ref{1}$ will be shifted to the left if H3O+ is added. Thus in a strongly acidic solution we expect nearly all the pink In to be consumed, and only colorless HIn will remain. On the other hand, if the solution is made strongly basic, the equilibrium will shift to the right because OH ions will react with HIn molecules, converting them to In. Thus the phenolphthalein solution will become pink.
Clearly there must be some intermediate situation where half the phenolphthalein is in the acid form and half in the colored conjugate-base form. That is, at some pH
$\ce{[HIn] = [In^{-}]} \nonumber$ This intermediate pH can be calculated by applying the Henderson-Hasselbalch equation to the indicator equilibrium:
$\text{pH}=\text{p}K_{a}\text{ + log}\frac{[\text{ In}^{-}]}{[\text{ HIn }]} \nonumber$
Thus at the point where half the indicator is conjugate acid and half conjugate base,
$\text{pH}=\text{p}K_{a}\text{+log1}=\text{p}K_{a} \nonumber$
For phenolphthalein, we have
$\text{pH}=\text{p}K_{a}=-\text{log(3 }\times \text{ 10}^{-10}\text{)}=\text{9.5} \nonumber$
so we expect phenolphthalein to change color in the vicinity of pH = 9.5.
The way in which both the color of phenolphthalein and the fraction present as the conjugate base varies with the pH is shown in detail in Figure $1$. The change of color occurs over quite a limited range of pH―roughly pKa ± 1. In other words the color of phenolphthalein changes perceptibly between about pH 8.3 and 10.5. Observe the actual color change for this indicator in Figure $2$.
Other indicators behave in essentially the same way, but for many of them both the acid and the conjugate base are colored. Their pKa’s also differ from phenolphthalein, as shown in the following table. The indicators listed have been selected so that their pKa values are approximately two units apart. Consequently, they offer a series of color changes spanning the whole pH range.
Properties of Selected Indicators
Color
Name pKa Effective pH range Acid form Basic form
Thymol blue 1.6 1.2 - 2.8 Red Yellow
Methyl orange 4.2 3.1 - 4.4 Red Orange
Methyl red 5.0 4.2 - 6.2 Red Yellow
Bromothymol blue 7.1 6.0 - 7.8 Yellow Blue
Phenophthalein 9.5 8.3 - 10.0 Colorless Red
Alizarin yellow 11.0 10.1 - 12.4 Yellow Red
Indicators are often used to make measurements of pH which are precise to about 0.2 or 0.3 units. Suppose, for example, we add two drops of bromothymol blue to a sample of tap water and obtain a green-blue solution. Since bromothymol blue is green at a pH of 6 and blue at a pH of 8, we conclude that the pH is between these two limits. A more precise result could be obtained by comparing the color in the tap water with that obtained when two drops of indicator solution are added to buffer solutions of pH 6.5 and 7.5.
If a careful choice of both colors and pKa is made, it is possible to mix several indicators and obtain a universal indicator which changes color continuously over a very wide pH range. With such a mixture it is possible to find the approximate pH of any solution within this range. So-called pH paper, as seen below, is impregnated with one or several indicators. When a strip of this paper is immersed in a solution, its pH can be judged from the resulting color.
Example $1$: Indicators
What indicator, from those listed in the table, would you use to determine the approximate pH of the following solutions:
a) 0.1 M CH3COONa (sodium acetate) b) 0.1 M CH3COOH (acetic acid) c) A buffer mixture of sodium acetate and acetic acid d) 0.1 M NH4Cl (ammonium chloride)
Solution
a) A solution of sodium acetate will be mildly basic with a pH of 9 or 10. Phenolphthalein would probably be best. b) A solution of acetic acid, unless very dilute, has a pH in the vicinity of 3. Both thymol blue and methyl orange should be tried. c) Since Ka for acetic acid is 1.8 × 10–5 mol/L, we can expect this buffer to have a pH not far from 5. Methyl red would be a good indicator to try. d) Since NH4+is a very weak acid, this solution will be only faintly acidic with a pH of 5 or 6. Again methyl red would be a good indicator to try, though bromothymol blue is also a possibility. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.09%3A_Indicators.txt |
When an acid is titrated with a base, there is typically a sudden change in the pH of the solution at the equivalence point (where the amount of titrant added equals the amount of acid originally present). If a few drops of indicator solution have been added, this sharp increase in pH causes an abrupt change in color, which is called the endpoint of the indicator. The actual magnitude of the jump in pH, and the pH range which it covers depend on the strength of both the acid and the base involved, and so the choice of indicator can vary from one titration to another. To learn how to choose an appropriate indicator, we need to study in some detail the variation of pH during a titration.
For reference as you read this section, cm3 is equivalent to mL.
First we shall consider titration of a strong acid such as HCl with a strong base such as NaOH. Suppose we place 25.00 cm³ (mL) of 0.10 M HCl solution in a flask and add 0.10 M NaOH from a buret. The pH of the solution in the flask varies with added NaOH, as shown in Figure 1a. The pH changes quite slowly at the start of the titration, and almost all the increase in pH takes place in the immediate vicinity of the endpoint.
The pH change during this titration is caused by the proton-transfer reaction
$\text{H}_{3}\text{O}^{+} + \text{ OH}^{-} \rightarrow \text{ H}_{2}\text{O } + \text{ H}_{2}\text{O} \nonumber$
which occurs as hydroxide ions are added from the buret. Though hydronium ions are being consumed by hydroxide ions in the early stages of the titration, the hydronium-ion concentration remains in the vicinity of 10–1 or 10–2 mol L–1. As a result, the pH remains in the range 1 to 2. As an example of this behavior let us consider the situation halfway to the endpoint, i.e., when exactly 12.50 cm³ of 0.10 M NaOH have been added to 25.00 cm3 (mL) of 0.10 M HCl in the flask. The amount of hydronium ion has been reduced at this point from an original 2.5 mmol to half this value, 1.25 mmol. At the same time the volume of solution has increased from 25 cm3 to (25 + 12.50) cm3 = 37.50 cm3. Therefore, the hydronium-ion concentration is 1.25 mmol/37.50 cm3 = 0.0333 mol L–1, and the resultant pH is 1.48. Though the titration is half completed, this is not very different from the initial pH of 1.00.
The pH of the solution in the flask will only change drastically when we reach that point in the titration when only a minute fraction of the hydronium ions remain unconsumed, i.e., as we approach the endpoint, seen on the graph. Only then will we have reduced the hydronium-ion concentration by several powers of 10, and consequently increased the pH by several units. When 24.95 cm3 of base have been added, we are only 0.05 cm3 (approximately one drop) short of the endpoint. At this point 24.95 cm3 × 0.10 mmol cm–3 = 2.495 mmol hydroxide ions have been added. These will have consumed 2.495 mmol hydronium ions, leaving (2.5 – 2.495) mmol = 0.005 mmol hydronium ions in a volume of 49.95 cm3. The hydronium-ion concentration will now be:
$[\text{ H}_{3}\text{O}^{+}]=\frac{\text{0.005 mmol}}{\text{49.95 cm}^{3}}=\text{1.00 }\times \text{ 10}^{-4}\text{ mol L}^{-1} \nonumber$
giving a pH of 4.00. Because almost all the hydronium ions have been consumed, only a small fraction (one five-hundredth) remains and the volume of solution has nearly doubled. This reduces the hydronium-ion concentration by a factor of 10–3, and the pH increases by three units from its original value of 1.00.
When exactly 25.00 cm3 of base have been added, we have reached the theoretical equivalence point, and the flask will contain 2.5 mmol of both sodium and chloride ions in 50 cm3 of solution; i.e., the solution is 0.05 M NaCl. Furthermore its pH will be exactly 7.00, as seen on the graph, since neither the sodium ion nor the chloride ion exhibits any appreciable acid-base properties.
Immediately after this equivalence point the addition of further NaOH to the flask results in a sudden increase in the concentration of hydroxide ions, since there are now virtually no hydronium ions left to consume them. Thus even one drop (0.05 cm3) of base added to the equivalence point solution adds 0.005 mmol hydroxide ions and produces a hydroxide-ion concentration of 0.005 mmol/50.05 cm3 = 1.00 × 10–4 mol L–1. The resultant pOH = 4.00, and the pH = 10.00. The addition of just two drops of base results in a pH jump from 4.00 to 7.00 to 10.00. This rapid rise causes the indicator to change color, so the endpoint matches the equivalence point if the indicator is chosen properly.
Titration of a strong base with a strong acid can be handled in essentially the same way as the strong acid-strong base situation we have just described.
Example $1$ : pH During Titration
Suppose that 25.00 cm3 of 0.010 M KOH is titrated with 0.01 M HNO3. Find the pH of the solution in the flask (a) before any HNO3 is added; (b) halfway to the endpoint; (c) one drop (0.05 cm3) before the endpoint; (d) one drop after the endpoint.
Solution:
a) Since KOH is a strong base, [OH] = 0.010 = 10–2. Thus $\text{pOH}={2}$ and $\text{pH}={14} - {2} = {12}$ b) Halfway to the endpoint means that half the OH has been consumed. The original amount of OH was
\begin{align}n_{\text{OH}^{-}}=V_{\text{NaOH}}\times c_{\text{NaOH}}\\text{ }=\text{25.00 cm}^{3}\times \frac{\text{0.010 mmol}}{\text{1 cm}^{3}}\={0.25 mmol}\end{align}
so the amount remaining is 0.125 mmol OH. It must have required 0.125 mmol HNO3 to consume the other 0.125 mmol OH, and so the volume of HNO3 added is
\begin{align}V_{\text{HNO}_{3}}=\frac{n_{\text{HNO}_{3}}}{c_{\text{HNO}_{3}}}\\text{ }=\frac{\text{0.125 mmol}}{\text{0.010 mmol cm}^{-3}}\={12.50 cm}^{3}\end{align}
The total volume of solution is thus (12.50 + 25.00) cm3 = 37.50 cm3, and
\begin{align}\text{ }[\text{ OH}^{-}]= \frac{\text{0.125 mmol}}{\text{37.50 cm}^{3}}\\text{ }=\text{3.33 }\times \text{ 10}^{-3}\text{ mmol cm}^{-3}\\text{ }=\text{3.33 }\times \text{ 10}^{-3}\text{ mol L}^{-1}\end{align}
$\text{pOH} = -\text{log }(\text{3.33} \times \text{10}^{-3})= {2.48}$ and $\text{pH}={14}-{2.48}={11.52}$ c) When 24.95cm3 of HNO3 solution has been added, the amount of H3O+ added is
\begin{align}n_{\text{H}_{3}\text{O}^{+}}=\text{24.95 cm}^{3}\times \frac{\text{0.010 mmol}}{\text{1 cm}^{3}}\\text{ }=\text{0.2495 mmol}\end{align}
This would consume an equal amount of OH, and so the amount of OH remaining is (0.2500 – 0.2495) mmol = 0.0005 mmol. Thus
\begin{align}\text{ }[\text{ OH}^{-}]=\frac{\text{0.0005 mmol}}{\text{(25.00 + 24.95) cm}^{3}}\\text{ }=\text{1.0 }\times \text{ 10}^{-5}\text{ mmol cm}^{-3}\\text{ }=\text{1.0 }\times \text{ 10}^{-5}\text{ mol L}^{-1}\end{align}
$\text{pOH} = {5.0}$ and $\text {pH}= {9.0}$ d) An excess of 0.05 cm3 of acid will add
\begin{align}n_{\text{H}_{3}\text{O}^{+}}=\text{0.05 cm}^{3}\times \frac{\text{0.010 mmol}}{\text{1 cm}^{3}}\\text{ }=\text{5 }\times \text{ 10}^{-4}\text{ mmol}\end{align}
of H3O+ to a neutral solution whose volume is (25.00 + 25.05)cm3 = 50.05 cm3.
Thus
$[\text{ H}_{3}\text{O}^{+}]=\frac{\text{5 }\times \text{ 10}^{-4}\text{ mmol}}{\text{50.05 cm}^{3}}=\text{1 }\times \text{ 10}^{-5}\text{ mol L}^{-1}$
$\text {pH}= {5}$
Note:
In this case, because the solutions were one-tenth as concentrated as in the titration of HCl with NaOH worked out in the text, the jump in pH (from 9 to 5) at the endpoint is smaller.
A wide choice of indicators like this is not possible for titrations involving weak acids or bases. When 25.00 cm3 of 0.10M CH3COOH is titrated with 0.10M NaOH, for instance, there is a very much smaller change in pH at the equivalence point, as shown in Figure 1b, and the choice of indicators is correspondingly narrowed. The behavior of the pH in this case is very different from that of the titration of HCl with NaOH, because the acid-base reaction is different.
When CH3COOH is titrated with NaOH, the OH ions consume CH3COOH molecules according to the equation:
$\text{ CH}_{3}\text{COOH } + \text{OH}^{-} \rightarrow \text{ CH}_{3}\text{COO}^{-} + \text{ H}_{2}\text{O } \nonumber$
As a result, the solution in the titration flask soon becomes a buffer mixture with appreciable concentrations of the CH3COO ion as well as its conjugate acid. The [H3O+] and the pH are then controlled by the ratio of acid to conjugate base (equations 2 and 3 in the section on buffer solutions). When we are halfway to the endpoint, for example, [CH3COOH] will be essentially the same as [CH3COO], and
\begin{align}\text{ }[\text{ H}_{3}\text{O}^{\text{+}}]=K_{a}\times \frac{[\text{ CH}_{3}\text{COOH }]}{[\text{ CH}_{3}\text{COO}^{-}]}\\text{ }\approx \text{ }K_{a}=\text{1.8 }\times \text{ 10}^{-5}\text{ mol L}^{-1}\end{align} \nonumber
while the pH will be given by the Henderson-Hasselbalch equation as
\begin{align}\text{pH}=\text{p}K_{a}\text{ + log }\frac{[\text{ CH}_{3}\text{COOH }]}{[\text{ CH}_{3}\text{COO}^{-}]}\\text{ }\approx \text{ p}K_{a}=\text{4.74}\end{align} \nonumber
Comparing this to the pH of 1.78 calculated above for the halfway stage in the titration of HCl, we find a difference of roughly three pH units. The effect of the buffering action of the CH3COOH/ CH3COO conjugate pair is thus to keep the pH some three units higher than before and hence to cut the jump in pH at the endpoint by approximately this amount.
Exactly at the equivalence point we no longer have a buffer mixture but a 0.05-M solution of sodium acetate. This solution is slightly basic, and its pH of 8.72 can be calculated from equation 4 on the section covering the pH of weak base solutions. Beyond this equivalence point, the story is much the same as in the strong-acid case. Addition of even a drop (0.05cm3) of excess base raises the OH concentration to 10–4 mol L–1 and the pH to 10. Of the three indicators which could be used in the titration of HCl, only one is useful for acetic acid. This is phenolphthalein, which changes color to the pinkish hue, as seen below, when in the pH range 8.3 to 10.0.
The titration of a weak base with a strong acid also involves a buffer solution and consequently requires a more careful choice of indicator.
Example $2$ : Titration
For the titration of 25.00 cm3 of 0.010 M NH3 with 0.010 M HCl, calculate the pH (a) before any acid is added; (b) after 12.50 cm3 of HCl has been added; (c) at the equivalence point; (d) after 30.00 cm3 of HCl has been added.
Solution:
a) Before any acid is added, we have a solution of a weak base, and equation 4 from the section on weak bases applies:
\begin{align}\text{ }[\text{ OH}^{-}]=\sqrt{K_{b}c_{b}}\\text{ }=\sqrt{\text{1.8 }\times \text{ 10}^{-5}\text{ mol L}^{-1}\times \text{ 1.0 }\times \text{ 10}^{-2}\text{ mol L}^{-1}}\\text{ }=\text{4.24 }\times \text{ 10}^{-4}\text{ mol L}^{-1}\end{align}
$\text{pOH} = -\text{log }(\text{4.24} \times \text{10}^{-4})= \text{3.37}$
and $\text{pH}={14}-{3.37}={10.63}$
b) Since the concentrations of base and acid are equal, 12.50 cm3 of HCl is enough to consume half the NH3, converting it to the conjugate acid NH4+.
The Henderson-Hasselbalch equation may be used to obtain the pH of this buffer system, provided we use pKa(NH4+). Rearranging equation 2 from the section on acid-base pairs and pH,
\begin{align}\text{p}K_{a}\text{(NH}_{4}^{+}\text{)}=\text{14}-\text{p}K_{b}\text{(NH}_{3}\text{)}\\text{ }=\text{14 + log(1.8 }\times \text{ 10}^{-5}\text{)}\\text{ }=\text{14}-\text{4.74}=\text{9.26}\end{align}
Since $[\text{ NH}_{4}^{+}]\approx [\text{ NH}_{3}]$
\begin{align}\text{pH}=\text{p}K_{a}\text{ + log}\frac{[\text{ A}^{-}]}{[\text{ HA }]}\\text{ }=\text{p}K_{a}\text{ + log}\frac{[\text{ NH}_{3}]}{[\text{ NH}_{4}^{+}]}\\text{ }\approx \text{p}K_{a}=\text{9.25}\end{align}
c) At the equivalence point all the NH3 has been converted to NH4+, and so we have a solution of NH4Cl. The volume of solution has doubled because 25.00 cm3 of HCl was added, and so cNH4+ must be half the original cNH3; that is, cNH4+ = 0.0005 mol L–1. Ka(NH4+) is obtained from Kb(NH3) using equation 1 from the section on acid-base pairs and pH:
\begin{align}K_{a}\text{(NH}_{4}^{+}\text{)}=\frac{K_{w}}{K_{b}\text{(NH}_{3}\text{)}}\\text{ }=\frac{\text{1.00 }\times \text{ 10}^{-14}\text{ mol}^{2}\text{ L}^{-2}}{\text{1.8 }\times \text{ 10}^{-5}\text{ mol L}^{-1}}\\text{ }=\text{5.56 }\times \text{ 10}^{-10}\text{ mol L}^{-1}\end{align}
Thus
\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=\sqrt{K_{a}c_{a}}\\text{ }=\sqrt{\text{5.56 }\times \text{ 10}^{-10}\text{ mol L}^{-1}\times \text{ 5.0 }\times \text{ 10}^{-3}\text{ mol L}^{-1}}\text{ }=\text{1.67 }\times \text{ 10}^{-6}\text{ mol L}^{-1}\\text{ }\\text{pH}= -\text{log }(\text{1.67} \times \text{10}^{-6})= \text{5.78}\end{align}
d) By adding an excess 5.0 cm3 of HCl we have added a strong acid to a solution which previously contained the very weak acid NH4+. To a first approximation, then, we consider only H3O+ produced by HCl.
\begin{align}\text{ }[\text{ H}_{3}\text{O}^{+}]=K_{a}\times \frac{\text{5.0 cm}^{3}\times \text{ 0.010 mol L}^{-1}}{\text{55.00 cm}^{3}}\\text{ }=\text{9.1 }\times \text{ 10}^{-4}\text{ mol L}^{-1}\\text{ }\\text{pH}= -\text{log }({1.69.17} \times{10}^{-4})= \text{3.04}\end{align}
The pH variation during titrations of strong and weak bases with strong acid are shown in Figure $2$. In the case of the titration of 0.010 M NH3 with 0.010 M HCl, methyl red, but not phenolphthalein, would be a suitable indicator. In general the best indicator for a given titration is the one whose pKa most nearly matches the pH calculated at the theoretical endpoint.
14.10: Titration Curves
The acid value (AV) is a common parameter in the specification of fats and oils. It is defined as the weight of KOH in mg needed to neutralize the organic acids present in 1g of fat and it is a measure of the free fatty acids (FFA) present in the fat or oil. An increment in the amount of FFA in a sample of oil or fat indicates hydrolysis of triglycerides (Structure on the left). Such reaction occurs by the action of lipase enzyme and it is and indicator of inadequate processing and storage conditions (i.e., high temperature and relative humidity, tissue damage). The source of the enzyme can be the tissue from which the oil or fat was extracted or it can be a contaminant from other cells including microorganisms. Besides FFA, hydrolysis of triglycerides produces glycerol. The table below shows the acid value of some common oils and bee's wax.
Oil Acid Value (AV)[1]
Canola 0.071
Maize 0.223
Soya 0.60
Virgin olive oil 6.6[2]
Used frying oil 31[3]
Bee's wax 17-36[4]
FFA are a source of flavors and aromas. On one side, we have short chain FFA which tend to be water soluble and volatile with characteristic smell. On the other side, we have long chain saturated and unsaturated fatty acids. The later are more prone to oxidation in their free form and their breakdown products (aldehydes, ketones, alcohols, and organic acids) provide characteristic flavors and aromas. In most cases these flavors and aromas are considered a defect in oils, fats, and foods that contain them. However, there are instances where hydrolysis of triglycerides and oxidation of FFA are key in the development of desirable flavor and aroma in foods. This is the case of aged cheeses and some processed meats.
Sunflower seed oil
The AOAC method to determine AV in fats and oils is based on a titration in ethanol using phenolphthalein as indicator. Disadvantages of this and similar methods are the use of organic solvents (volume and toxicity), the need for heating the reaction media, incomplete solubility of the oil/fat, the need to pre-neutralize the solvents, use of large amounts of sample, and the possibility of error to detect the color change of the indicator when analyzing colored samples. There are some non-titration methods designed to overcome these disadvantages.[1] However, in spite its drawbacks, the titration method is still the most used due to the fact that it does not require expensive equipment.
Acid-Base Titrations
A titration is a volumetric technique in which a solution of one reactant (the titrant) is added to a solution of a second reactant (the analyte) until the equivalence point is reached. The equivalence point is the point at which the titrant has been added in exactly the right quantity to react stoichiometrically with the analyte.
When an acid is titrated with a base, there is typically a sudden change in the pH of the solution at the equivalence point. If a few drops of indicator solution have been added, this sharp increase in pH causes an abrupt change in color, which is called the endpoint of the indicator (See the animation on the left). The actual magnitude of the jump in pH, and the pH range which it covers depend on the strength of both the acid and the base involved, and so the choice of indicator can vary from one titration to another. To learn how to choose an appropriate indicator, we need to study in some detail the variation of pH during a titration.
To simplify calculations we want to work with acid-base reactions that go to completion. For such reason, the most common cases we will encounter are:
• strong base titrated with strong acid
• strong acid titrated with strong base
• weak base titrated with strong acid
• weak acid titrated with strong base
However, we rarely titrate a weak base with a strong acid and vice versa.
Titration of a strong acid with a strong base
Suppose we place 25.00 cm3 of 0.10 M HCl solution in a flask and add 0.10 M NaOH from a buret. The pH of the solution in the flask varies with added NaOH, as shown in Figure 1a. The pH changes quite slowly at the start of the titration, and almost all the increase in pH takes place in the immediate vicinity of the endpoint.
The pH change during this titration is caused by the proton-transfer reaction
$\text{H}_{3}\text{O}^{+} + \text{ OH}^{-} \rightarrow \text{ H}_{2}\text{O } + \text{ H}_{2}\text{O}$ (1) which occurs as hydroxide ions are added from the buret. Though hydronium ions are being consumed by hydroxide ions in the early stages of the titration, the hydronium-ion concentration remains in the vicinity of 10–1 or 10–2 mol dm–3. As a result, the pH remains in the range 1 to 2. As an example of this behavior let us consider the situation halfway to the endpoint, i.e., when exactly 12.50 cm3 of 0.10 M NaOH have been added to 25.00 cm3 of 0.10 M HCl in the flask. The amount of hydronium ion has been reduced at this point from an original 2.5 mmol to half this value, 1.25 mmol. At the same time the volume of solution has increased from 25 cm3 to (25 + 12.50) cm3 = 37.50 cm3. Therefore, the hydronium-ion concentration is 1.25 mmol/37.50 cm3 = 0.0333 mol dm–3, and the resultant pH is 1.48. Though the titration is half completed, this is not very different from the initial pH of 1.00.
The pH of the solution in the flask will only change drastically when we reach that point in the titration when only a minute fraction of the hydronium ions remain unconsumed, i.e., as we approach the endpoint. Only then we will have reduced the hydronium-ion concentration by several powers of 10, and consequently increased the pH by several units. When 24.95 cm3 of base have been added, we are only 0.05 cm3 (approximately one drop) short of the endpoint. At this point 24.95 cm3 × 0.10 mmol cm–3 = 2.495 mmol hydroxide ions have been added. These will have consumed 2.495 mmol hydronium ions, leaving (2.5 – 2.495) mmol = 0.005 mmol hydronium ions in a volume of 49.95 cm3. The hydronium-ion concentration will now be
$[\text{ H}_{3}\text{O}^{+}]=\frac{\text{0.005 mmol}}{\text{49.95 cm}^{3}}=\text{1.00 }\times \text{ 10}^{-4}\text{ mol dm}^{-3}$ giving a pH of 4.00. Because almost all the hydronium ions have been consumed, only a small fraction (one five-hundredth) remains and the volume of solution has nearly doubled. This reduces the hydronium-ion concentration by a factor of 10–3, and the pH increases by three units from its original value of 1.00.
When exactly 25.00 cm3 of base have been added, we have reached the theoretical equivalence point, and the flask will contain 2.5 mmol of both sodium and chloride ions in 50 cm3 of solution; i.e., the solution is 0.05 M NaCl. Furthermore its pH will be exactly 7.00, since neither the sodium ion nor the chloride ion exhibits any appreciable acid-base properties.
Immediately after this equivalence point the addition of further NaOH to the flask results in a sudden increase in the concentration of hydroxide ions, since there are now virtually no hydronium ions left to consume them. Thus even one drop (0.05 cm3) of base added to the equivalence point solution adds 0.005 mmol hydroxide ions and produces a hydroxide-ion concentration of 0.005 mmol/50.05 cm3 = 1.00 × 10–4 mol dm–3. The resultant pOH = 4.00, and the pH = 10.00. The addition of just two drops of base results in a pH jump from 4.00 to 7.00 to 10.00. This rapid rise causes the indicator to change color, so the endpoint matches the equivalence point if the indicator is chosen properly.
A Titration of a strong base with a strong acid can be handled in essentially the same way as the strong acid-strong base situation we have just described. Strong acids and bases are not naturally found in food, however, they are often used as food processing aids and analysis methods.
Example $1$:
Suppose that 40.00 cm3 of 0.020 M KOH is titrated with 0.030 M HNO3. Find the pH of the solution in the flask (a) before any HNO3 is added; (b) halfway to the endpoint; (c) one drop (0.05 cm3) before the endpoint; (d) one drop after the endpoint.
Solution
a) Since KOH is a strong base, [OH] = 0.020 = 2 x 10–2. Thus
$\text{pOH}=\text{1.7}\,$ and $\text{pH}=\text{14} - \text{2} = \text{12.3}\,$
b) Halfway to the endpoint means that half the OH has been consumed. The original amount of OH was
\begin{align} n_{\text{OH}^{-}}&=V_{\text{NaOH}}\times c_{\text{NaOH}}\ \text{ }&=\text{40.00 cm}^{3}\times \frac{\text{0.020 mmol}}{\text{1 cm}^{3}}=\text{0.8 mmol} \end{align} so the amount remaining is 0.4 mmol OH. It must have required 0.4 mmol HNO3 to consume the other 0.4 mmol OH, and so the volume of HNO3 added is
\begin{align} V_{\text{HNO}_{3}}&=\frac{n_{\text{HNO}_{3}}}{c_{\text{HNO}_{3}}}\ \text{ }&=\frac{\text{0.4 mmol}}{\text{0.030 mmol cm}^{-3}}=\text{13.33 cm}^{3} \end{align} The total volume of solution at this point is thus (13.33 + 40.00) cm3 = 53.33 cm3, and
\begin{align}\text{ } [\text{ OH}^{-}]& = \frac{\text{0.4 mmol}}{\text{53.33 cm}^{3}}\ \text{ }&=\text{7.50 }\times \text{ 10}^{-3}\text{ mmol cm}^{-3}=\text{7.50 }\times \text{ 10}^{-3}\text{ mol dm}^{-3} \end{align}
$\text{pOH} = -\text{log }(\text{7.50} \times \text{10}^{-3})= \text{2.12}$ and
$\text{pH}=\text{14}-\text{2.12}=\text{11.88}\,$ c) If the total volume of of HNO3 required to react with all the OH- in solution is 26.66 cm3, when 26.61 cm3 of HNO3 solution have been added, the amount of H3O+ added is
\begin{align} n_{\text{H}_{3}\text{O}^{+}}=\text{26.61 cm}^{3}\times \frac{\text{0.030 mmol}}{\text{1 cm}^{3}}=\text{0.798 mmol} \end{align} This would consume an equal amount of OH, and so the amount of OH remaining is (0.8 – 0.798) mmol = 0.002 mmol. Thus
\begin{align} \text{ }[\text{ OH}^{-}]&=\frac{\text{0.002 mmol}}{\text{(40 + 26.61) cm}^{3}}\ \text{ }&=\text{3.00 }\times \text{ 10}^{-5}\text{ mmol cm}^{-3}=\text{3.00 }\times \text{ 10}^{-5}\text{ mol dm}^{-3} \end{align}
$\text{pOH} = \text{4.52}\,$ and $\text {pH}= \text{9.48}\,$ d) An excess of 0.05 cm3 of acid will add
\begin{align} n_{\text{H}_{3}\text{O}^{+}}=\text{0.05 cm}^{3}\times \frac{\text{0.030 mmol}}{\text{1 cm}^{3}}=\text{1.5 }\times \text{ 10}^{-3}\text{ mmol} \end{align}
of H3O+ to a neutral solution whose volume is (40.00 + 26.71)cm3 = 66.71 cm3.
Thus
$[\text{ H}_{3}\text{O}^{+}]=\frac{\text{1.5 }\times \text{ 10}^{-3}\text{ mmol}}{\text{66.71 cm}^{3}}=\text{2.25 }\times \text{ 10}^{-5}\text{ mol dm}^{-3}$
$\text {pH}= \text{4.64}\,$
Titration of a weak acid with a strong base
When 25.00 cm3 of 0.10M CH3COOH is titrated with 0.10M NaOH, for instance, there is a very much smaller change in pH at the equivalence point, as shown in Figure 1b, and the choice of indicators is correspondingly narrowed. The behavior of the pH in this case is very different from that of the titration of HCl with NaOH, because the acid-base reaction is different. When CH3COOH is titrated with NaOH, the OH ions consume CH3COOH molecules according to the equation
$\text{OH}^{-} + \text{ CH}_{3}\text{COOH } \rightarrow \text{ H}_{2}\text{O } + \text{ CH}_{3}\text{COO}^{-}$ (2)
As a result, the solution in the titration flask soon becomes a buffer mixture with appreciable concentrations of the CH3COO ion as well as its conjugate acid. The [H3O+] and the pH are then controlled by the ratio of acid to conjugate base (equations 2 and 3 in the section on buffer solutions). When we are halfway to the endpoint, for example, [CH3COOH] will be essentially the same as [CH3COO], and
\begin{align} \text{ }[\text{ H}_{3}\text{O}^{\text{+}}]&=K_{a}\times \frac{[\text{ CH}_{3}\text{COOH }]}{[\text{ CH}_{3}\text{COO}^{-}]}\ \text{ }&\approx \text{ }K_{a}=\text{1.8 }\times \text{ 10}^{-5}\text{ mol dm}^{-3}\end{align}
while the pH will be given by the Henderson-Hasselbalch equation as
\begin{align}\text{pH}&=\text{p}K_{a}\text{ + log }\frac{[\text{ CH}_{3}\text{COOH }]}{[\text{ CH}_{3}\text{COO}^{-}]}\ \text{ }&\approx \text{ p}K_{a}=\text{4.74}\end{align}
Comparing this to the pH of 1.78 calculated above for the halfway stage in the titration of HCl, we find a difference of roughly three pH units. The effect of the buffering action of the CH3COOH/ CH3COO conjugate pair is thus to keep the pH some three units higher than before and hence to cut the jump in pH at the endpoint by approximately this amount.
Exactly at the equivalence point we no longer have a buffer mixture but a 0.05-M solution of sodium acetate. This solution is slightly basic, and its pH of 8.72 can be calculated from equation 4 on the section covering the pH of weak base solutions. Beyond this equivalence point, the story is much the same as in the strong-acid case. Addition of even a drop (0.05cm3) of excess base raises the OH concentration to 10–4 mol dm–3 and the pH to 10. Of the three indicators which could be used in the titration of HCl, only one is useful for acetic acid. This is phenolphthalein, which changes color in the pH range 8.3 to 10.0. This endpoint lies within the pH jump from 7 to 10 which occurs ± 0.05 cm3 (one drop) from the endpoint. Both the other two indicators would give a color change before the true equivalence point. As shown in Figure 1b, the color of methyl red would start changing after only about 4 cm3 of base had been added!
The titration of a weak base with a strong acid also involves a buffer solution and consequently requires a more careful choice of indicator.
The pH variation during titrations of strong and weak bases with a strong acid are shown in Figure 2. In the case of the titration of 0.010 M NH3 with 0.010 M HCl, methyl red, but not phenolphthalein, would be a suitable indicator. In general the best indicator for a given titration is the one whose pKa most nearly matches the pH calculated at the theoretical endpoint.
Example $2$:
Ten grams of olive oil are dispersed in 50 cm3 ethanol for a final volume of 60.86 cm3. The mixture is titrated with 50 cm3 KOH 0.1 M. a) What is the AV of this sample of olive oil? Assuming that the only free fatty acid present in the sample is oleic acid and that we are working with an aqueous solution at all times, what is the pH of the mixture b) before any KOH is added?, when c) 10 cm3, d) 25 cm3, e) 35 cm3 e) an excess of 1.5cm3 KOH 0.1 M have been added?
Note: Oleic acid (c-9-octadecenoic) pKa=9.85[5]
Solution
a) If 50 cm3 of 0.1 M KOH were used to neutralize the sample, the number of moles of KOH is
\begin{align} n_{\text{KOH}}&=V_{\text{KOH}}\times c_{\text{KOH}}\ \text{ }&=\text{50.00 cm}^{3}\times \frac{\text{0.1 mmol}}{\text{1 cm}^{3}}=\text{5.0 mmol} \end{align} The molar mass of KOH is 56.10 g/mol and the mass of KOH employed in mg is \begin{align} m_{\text{KOH}}&=n_{\text{KOH}}\times M_{\text{KOH}}\ \text{ }&=\text{5.0 mmol}\times \frac{\text{56.10 mg}}{\text{mmol}}=\text{280.5 mg} \end{align}
this amount of KOH was used to neutralize 10 g of olive oil. The acid value is the then
\begin{align} \text{AV}&=\frac {\text{mg}_{\text{ KOH}}}{\text{g}_{\text{ sample}}}\ \text{ }&=\frac {\text{280.5 mg}_{\text{ KOH}}}{\text{10 g}_{\text{ sample}}}=\text{28.05} \end{align} This AV is considerable higher than the one shown in the table above. What does this tell you about the oil?
This oil has probably been heated (cooking/frying), it is old, or it has been stored under conditions favoring lipase activity. In the case of refined oils, a high AV indicates inappropriate or incomplete refining procedures.
b) From the number of moles of KOH employed to neutralize the oil we can conclude that there were 5.0 mmol of oleic acid in the sample and the ethanolic dispersion. Given that the volume of the dispersion was 60.83 cm3 the concentration of oleic acid is then
\begin{align} c_{\text{oleic}}&=\frac {n_{\text{oleic}}} UndefinedNameError: reference to undefined name 'V' (click for details) Callstack: at (Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/14:_Ionic_Equilibria_in_Aqueous_Solutions/14.10:_Titration_Curves/14.10.01:_Foods-_Acid_Value_and_the_Quality_of_Fats_and_Oils), /content/body/div[3]/section/div/div/span[1], line 1, column 1 =\frac{\text{5.0 mmol}}{\text{60.83 cm}^{3}}\ \text{ }&=\text{0.082 mmol cm}^{-3}=\text{0.082 mol dm}^{-3} \end{align} Using equation (5) from Weak acids in foods - pH and beyond
\begin{align}\text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\sqrt{K_{a}c_{a}} \ \text{ }&=\sqrt{(\text{1.41}\times \text{ 10}^{-10}\text{ mol dm}^{-3})(\text{ 0.082}\text{ mol dm}^{-3})} \ \text{ }&=\sqrt{\text{1.16}\times \text{ 10}^{-11}\text{ mol}^{\text{2}}\text{ dm}^{-6}}\ \text{ }[\text{ H}_{\text{3}}\text{O}^{\text{+}}\text{ }]\text{ }&=\text{3.4}\times \text{ 10}^{-\text{6}}\text{ mol dm}^{-\text{3}} \ \text{pH}&=-\text{log (3.4 }\times \text{ 10}^{-6}\text{)}=\text{5.47} \ \end{align}
c) When 10 cm3 of KOH solution had been added, the volume of the dispersion is 60.83 + 10 = 70.83 cm3 and the corresponding number of moles of OH- is \begin{align} n_{\text{OH}^{-}}&=V_{\text{OH}^{-}}\times c_{\text{OH}^{-}}\ \text{ }&=\text{10.00 cm}^{3}\times \frac{\text{0.1 mmol}}{\text{1 cm}^{3}}=\text{1.0 mmol} \end{align} This means that 1.0 mmol of oleic acid have reacted and produced 1.0 mmol of oleate, its conjugate base. At this point, the number of moles of oleic acid and oleate are \begin{align}n_{\text{oleic}}&= \text{5.0 mmol} - \text{1.0 mmol} = \text{4.0 } \text{mmol oleic acid}\ n_{\text{oleate}}&= \text{1.0 mmol oleate}\,\end{align} and their concentrations \begin{align} c_{\text{oleic}}&=\frac {n_{\text{oleic}}} UndefinedNameError: reference to undefined name 'V' (click for details) Callstack: at (Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/14:_Ionic_Equilibria_in_Aqueous_Solutions/14.10:_Titration_Curves/14.10.01:_Foods-_Acid_Value_and_the_Quality_of_Fats_and_Oils), /content/body/div[3]/section/div/div/span[2], line 1, column 1 =\frac{\text{4.0 mmol}}{\text{70.83 cm}^{3}}\ \text{ }&=\text{0.056 mmol cm}^{-3}=\text{0.056 mol dm}^{-3}\ \text{ }\ c_{\text{oleate}}&=\frac {n_{\text{oleate}}} UndefinedNameError: reference to undefined name 'V' (click for details) Callstack: at (Bookshelves/General_Chemistry/ChemPRIME_(Moore_et_al.)/14:_Ionic_Equilibria_in_Aqueous_Solutions/14.10:_Titration_Curves/14.10.01:_Foods-_Acid_Value_and_the_Quality_of_Fats_and_Oils), /content/body/div[3]/section/div/div/span[3], line 1, column 1 =\frac{\text{1.0 mmol}}{\text{70.83 cm}^{3}}\ \text{ }&=\text{0.014 mmol cm}^{-3}=\text{0.014 mol dm}^{-3} \end{align} Since we have both the acid and its conjugate base in the dispersion we can use the Henderson-Hasselbach equation to calculate the pH \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{oleate}]}{[\text{oleic}]}\ \text{ }&= \text{ 9.85 }\text{+ log }\frac{\text{0.014}\text {mol dm}^{-3}}{\text{0.056} \text {mol dm}^{-3}}\ \text{ }&= \text{9.25} \end{align} Note that all the OH- has reacted with oleic acid and it is not present in the dispersion. d) When 25 cm3 have been added, we are halfway to the equivalence point. Half of the oleic acid has been converted into oleate and their concentrations are equal. $n_{\text{oleic}}=n_{\text{oleate}} = \text{2.5 mmol}\,$ $c_{\text{oleic}}=c_{\text{oleate}} = \frac {\text{2.5 mmol}} {\text {85.83} \text{ cm}^{3}}= \text {0.029 mmol cm}^{3}= \text {0.029 mol dm}^{3}$ Again, we have both the acid and its conjugate base in dispersion and we use the Henderson-Hasselbach equation \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{oleate}]}{[\text{oleic}]}\ \text{ }&= \text{ 9.85 }\text{+ log }\frac{\text{0.029}\text {mol dm}^{-3}}{\text{0.029} \text {mol dm}^{-3}}\ \text{ }&= \text{9.85} + \text{0}=\text{9.85} \end{align} So halfway the equivalence point $\text{pH} = \text{p}{K}_{a}\,$ e) If 35 cm3 0.1 KOH have been added \begin{align} n_{\text{OH}^{-}}&=V_{\text{OH}^{-}}\times c_{\text{OH}^{-}}\ \text{ }&=\text{35.00 cm}^{3}\times \frac{\text{0.1 mmol}}{\text{1 cm}^{3}}\ \text{ }&=\text{3.5 mmol} \end{align} So 3.5 mmol of oleic acid have been converted into oleate and 1.5 remain in the dispersion. With a new volume of 95.83 cm3, their corresponding concentrations are \begin{align} c_{\text{oleic}}&=\text{0.016 mol dm}^{-3}\ c_{\text{oleate}}&=\text{0.036 mol dm}^{-3} \end{align} and \begin{align}\text{pH}&=\text{p}K_{a}\text{+ log}\frac{[\text{oleate}]}{[\text{oleic}]}\ \text{ }&= \text{ 9.85 }\text{+ log }\frac{\text{0.036}\text {mol dm}^{-3}}{\text{0.016} \text {mol dm}^{-3}}\ \text{ }&= \text{10.20} \end{align}
f) If an excess of 1.5 cm3 are added to the dispersion, its final volume is 112.33 cm3. Since all the oleic acid has been converted into oleate, there are no more protons to react with the OH- ions and it will remain in solution.
The number of moles of OH- ions is
\begin{align} n_{\text{OH}^{-}}&=V_{\text{OH}^{-}}\times c_{\text{OH}^{-}}\ \text{ }&=\text{1.5 cm}^{3}\times \frac{\text{0.1 mmol}}{\text{1 cm}^{3}}=\text{0.15 mmol} \end{align} for a concentration of Since the OH- is the strongest base in solution, it will govern the pH of the reaction mixture. The concentration of OH- generated by oleate is negligible, so we calculate the pH of the dispersion as
\begin{align}\text{pOH}&= -\text{log }[\text{OH}^{-}]\ \text{ }&=-\text{log }(\text{1.33} \times \text{10}^{-3})= \text{2.87}\end{align}
and
$\text{pH}=\text{14}-\text{2.82}=\text{11.18}\,$ Notice the assumptions we made to calculate pH at different stages of this titration. Initially, the oil is dispersed in ethanol, the dissociation constants of acids change depending on the medium and strictly speaking it is probably incorrect to talk about pH in this conditions. The method is also based on the assumption that protons are in the interface ethanol-water and readily react with aqueous NaOH. These are some of the limitations of the method and the reason for the need of a titration instead of direct measurement of pH on the sample.
References
1. Kardash, E. and Tur’yan, Y. I. 2005. Acid Value Determination in Vegetable Oils by Indirect Titration in Aqueous-alcohol Media. Croat. Chem. Acta 78:1:99-103.
2. www.unctad.org/infocomm/angla...c/Cxs_033e.pdf
3. Cao, F., Chen, Y., Zhai, F., Li, J., Wang, J., Wang, X., Wang, S., and Zhu, W. 2008. Biodiesel Production From High Acid Value Waste Frying Oil Catalyzed by Superacid Heteropolyacid. Biotech. Bioeng. 101:1:93-100.
4. Food Lipids. Akoh, C. C., and Min, D.B. 3rd ed.
5. Kanicky, J.R. and Shah, D. O. 2002.Effect of degree, type, and position of unsaturation on the pKa of long-chain fatty acids. J. Colloid Interface Sci. 256:1:201-207 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.10%3A_Titration_Curves/14.10.01%3A_Foods-_Acid_Value_and_the_Quality_of_Fats_and_Oils.txt |
In the section on precipitation reactions, we saw that there are some salts which dissolve in water to only a very limited extent. For example, if BaSO4 crystals are shaken with water, so little dissolves that it is impossible to see that anything has happened, as you will see in the video below. Nevertheless, the few Ba2+(aq) and SO42(aq) ions that do go into solution increase the conductivity of the water, allowing us to measure their concentration. The video below shows the creation of Barium Sulfate in a precipitation reaction between barium chloride and sodium sulfate. Notice the white precipitate that forms, which is barium sulfate.
We find that at 25°C
$[\text{Ba}^{2+}]=\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1} = [\text{SO}_{4}^{2-}]\label{1}$
that we would describe the solubility of BaSO4 as 0.97 × 10–5 mol L–1 at this temperature. The solid salt and its ions are in dynamic equilibrium, and so we can write the equation
$\text{BaSO}_{4} ({s}) \rightleftharpoons \text{Ba}^{2+} ({aq}) + \text{SO}_{4}^{2-} ({aq})\label{2}$
As in other dynamic equilibria we have discussed, a particular Ba2+ ion will sometimes find itself part of a crystal and at other times find itself hydrated and in solution.
Since the concentration of BaSO4 has a constant value, it can be incorporated into Kc for Equation $\ref{2}$. This gives a special equilibrium constant called the solubility product Ksp:
$K_{sp}= K_{c}[\text{BaSO}_{4}] = [\text{Ba}^{2+}][\text{SO}_{4}^{2-}]\label{3}$
For BaSO4, Ksp is easily calculated from the solubility by substituting Equation $\ref{1}$ into $\ref{3}$:
\begin{align}K_{sp} & = (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1}) (\text{0.97} \times \text{10}^{-5}\text{ mol L}^{-1})\ &= \text{0.94} \times \text{10}^{-10}\text{ mol}^{2} \text{ L}^{-2}\end{align} \nonumber
In the general case of an ionic compound whose formula is $A_xB_y$, the equilibrium can be written
$\text{A}_{x}\text{B}_{y} ({s}) \rightleftharpoons {x}\text{A}^{m+} ({aq}) + {y}\text{A}^{n+} ({aq}) \nonumber$
The solubility product is then
$\text{K}_{sp} = [\text{A}^{m+}]^{x}[\text{B}^{n+}]^{y} \nonumber$
Solubility products for some of the more common sparingly soluble compounds are given in the table below.
Table $1$ Solubility Product Constants for Some Inorganic Compounds at 25 °C1
Substance Ks Substance Ksp
Aluminum Compounds Barium Compounds
AlAsO4 1.6 × 10-16 Ba3(AsO4)2 8.0 × 10-15
Al(OH)3 amorphous 1.3 × 10-33 BaCO3 5.1 × 10-9
AlPO4 6.3 × 10-19 BaC2O4 1.6 × 10-7
Bismuth Compounds BaCrO4 1.2 × 10-10
BiAsO4 4.4 ×10-10 BaF2 1.0 × 10-6
BiOCl2 7.0 × 10-9 Ba(OH)2 5 × 10-3
BiO(OH) 4 × 10-10 Ba3(PO4)2 3.4 × 10-23
Bi(OH)3 4 ×10-31 BaSeO4 3.5 × 10-8
Bil3 8.1 ×10-19 BaSO4 1.1 × 10-10
BiPO4 1.3 ×10-23 BaSO3 8 × 10-7
Cadmium Compounds BaS2O3 1.6 × 10-5
Cd3(AsO4)2 2.2 ×10-33 Calcium Compounds
CdCO3 5.2 ×10-12 Ca3(AsO4)2 6.8 ×10-19
Cd(CN)2 1.0 ×10-8 CaCO3 2.8 ×10-9
Cd2[Fe(CN)6] 3.2 ×10-17 CaCrO4 7.1 ×10-4
Cd(OH)2 fresh 2.5 ×10-14 CaC2O4 • H2O3 4 × 10-9
Chromium Compounds CaF2 5.3 ×10-9
CrAsO4 7.7 × 10-21 Ca(OH)2 5.5 ×10-6
Cr(OH)2 2 × 10-16 CaHPO4 1 × 10-7
Cr(OH)3 6.3 × 10-31 Ca3(PO4)2 2.0 × 10-29
CrPO4 • 4H2O green 2.4 × 10-23 CaSeO4 8.1 × 10-4
CrPO4 • 4H2O violet 1.0 × 10-17 CaSO4 9.1 × 10-6
Cobalt Compounds CaSO3 6.8 × 10-8
Co3(AsO4)2 7.6 × 10-29 Copper Compounds
CoCO3 1.4 × 10-13 CuBr 5.3 × 10-9
Co(OH)2 fresh 1.6 × 10-15 CuCl 1.2 × 10-6
Co(OH)3 1.6 × 10-44 CuCN 3.2 × 10-20
CoHPO4 2 × 10-7 CuI 1.1 × 10-12
CO3(PO4)2 2 × 10-35 CuOH 1 × 10-14
Gold Compounds CuSCN 4.8 × 10-15
AuCl 2.0 × 10-13 Cu3(AsO4)2 7.6 × 10-36
AuI 1.6 × 10-23 CuCO3 1.4 × 10-10
AuCl3 3.2 × 10-25 Cu2[Fe(CN)6] 1.3 × 10-16
Au(OH)3 5.5 × 10-46 Cu(OH)2 2.2 × 10-20
AuI3 1 × 10-46 Cu3(PO4)2 1.3 × 10-37
Iron Compounds Lead Compounds
FeCO3 3.2 × 10-11 Pb3(AsO4)2 4.0 × 10-36
Fe(OH)2 8.0 × 10-16 PbBr2 4.0 × 10-5
FeC2O4 • 2H2O3 3.2 × 10-7 PbCO3 7.4 × 10-14
FeAsO4 5.7 × 10-21 PbCl2 1.6 × 10-5
Fe4[Fe(CN)6]3 3.3 × 10-41 PbCrO4 2.8 × 10-13
Fe(OH)3 4 × 10-38 PbF2 2.7 × 10-8
FePO4 1.3 × 10-22 Pb(OH)2 1.2 × 10-15
Magnesium Compounds PbI2 7.1 × 10-9
Mg3(AsO4)2 2.1 × 10-20 PbC2O4 4.8 × 10-10
MgCO3 3.5 × 10-8 PbHPO4 1.3 × 10-10
MgCO3 • 3H2O3 2.1 × 10-5 Pb3(PO4)2 8.0 × 10-43
MgC2O4 • 2H2O3 1 × 10-8 PbSeO4 1.4 × 10-7
MgF2 6.5 × 10-9 PbSO4 1.6 × 10-8
Mg(OH)2 1.8 × 10-11 Pb(SCN)2 2.0 × 10-5
Mg3(PO4)2 10-23 to 10-27 Manganese Compounds
MgSeO3 1.3 × 10-5 Mn3(AsO4)2 1.9 × 10-29
MgSO3 3.2 × 10-3 MnCO3 1.8 × 10-11
MgNH4PO4 2.5 × 10-13 Mn2[Fe(CN)6] 8.0 × 10-13
Mercury Compounds Mn(OH)2 1.9 × 10-13
Hg2Br2 5.6 × 10-23 MnC2O4 • 2H2O3 1.1 × 10-15
Hg2CO3 8.9 × 10-17 Nickel Compounds
Hg2(CN)2 5 × 10-40 Ni3(AsO4)2 3.1 × 10-26
Hg2Cl2 1.3 × 10-18 NiCO3 6.6 × 10-9
Hg2CrO4 2.0 × 10-9 2 Ni(CN)2 → Ni2+ + Ni(CN)42 1.7 × 10-9
Hg2(OH)2 2.0 × 10-24 Ni2[Fe(CN)6] 1.3 × 10-15
Hg2l2 4.5 × 10-29 Ni(OH)2 fresh 2.0 × 10-15
Hg2SO4 7.4 × 10-7 NiC2O4 4 × 10-10
Hg2SO3 1.0 × 10-27 Ni3(PO4)2 5 × 10-31
Hg(OH)2 3.0 × 10-26 Silver Compounds
Strontium Compounds Ag3AsO4 1.0 × 10-22
Sr3(AsO4)2 8.1 × 10-19 AgBr 5.0 × 10-13
SrCO3 1.1 × 10-10 Ag2CO3 8.1 × 10-12
SrCrO4 2.2 × 10-5 AgCl 1.8 × 10-10
SrC2O4 • H2O3 1.6 × 10-7 Ag2CrO4 1.1 × 10-12
Sr3(PO4)2 4.0 × 10-28 AgCN 1.2 × 10-16
SrSO3 4 × 10-8 Ag2Cr2O7 2.0 × 10-7
SrSO4 3.2 × 10-7 Ag4[Fe(CN)6] 1.6 × 10-41
Tin Compounds AgOH 2.0 × 10-8
Sn(OH)2 1.4 × 10-28 AgI 8.3 × 10-17
Sn(OH)4 1 × 10-56 Ag3PO4 1.4 × 10-16
Zinc Compounds Ag2SO4 1.4 × 10-5
Zn3(AsO4)2 1.3 × 10-28 Ag2SO3 1.5 × 10-14
ZnCO3 1.4 × 10-11 AgSCN 1.0 × 10-12
Zn2[Fe(CN)6] 4.0 × 10-16
Zn(OH)2 1.2 × 10-17
ZnC2O4 2.7 × 10-8
Zn3(PO4)2 9.0 × 10-33
1. Taken from Patnaik, Pradyot, Dean’s Analytical Chemistry Handbook, 2nd ed., New York: McGraw-Hill, 2004, Table 4.2 (published on the Web by Knovel, http://www.knovel.com).
2. Taken from Meites, L. ed., Handbook of Analytical Chemistry, 1st ed., New York: McGraw-Hill, 1963.
3. Because [H2O] does not appear in equilibrium constants for equilibria in aqueous solution in general, it does not appear in the Ksp expressions for hydrated solids.
No metal sulfides are listed in this table because sulfide ion is such a strong base that the usual solubility product equilibrium equation does not apply. See Myers, R. J. Journal of Chemical Education, Vol. 63, 1986; pp. 687-690.
Example $1$: Equilibrium
When crystals of PbCl2 are shaken with water at 25°C, it is found that 1.62 × 10–2 mol PbCl2 dissolves per cubic decimeter of solution. Find the value of Ksp at this temperature.
Solution
We first write out the equation for the equilibrium:
$\text{PbCl}_{2}({s})\rightleftharpoons \text{Pb}^{2+}({aq}) = \text{2Cl}^{-}({aq}) \nonumber$
so that
$\text{K}_{sp}\text{PbCl}_{2} = [\text{Pb}^{2+}][\text{Cl}^{-}]^{2} \nonumber$
Since 1.62 × 10–2 mol PbCl2 dissolves per cubic decimeter, we have
$[\text{Pb}^{2+}]=\text{1.62} \times \text{10}^{-2} \text{mol L}^{-1} \nonumber$
while
$[\text{Cl}^{-}]=\text{2} \times \text{1.62} \times \text{10}^{-2} \text{mol L}^{-1} \nonumber$
since 2 mol Cl ions are produced for each mol PbCl2 which dissolves. Thus
\begin{align}{K}_{sp}= (\text{1.62}\times \text{10}^{-2}\text{mol L}^{-1})(\text{2 } \times \text{ 1.62 } \times \text{ 10}^{-2} \text{mol L}^{-1})\text{ }^{2}\\text{ } = \text{1.70 } \times \text{ 10}^{-5} \text{ mol}^{3} \text{L} ^{-3}\end{align} \nonumber
Example $2$: Solubility
The solubility product of silver chromate, Ag2CrO4, is 1.0 × 10–12 mol3 L–3. Find the solubility of this salt.
Solution
Again we start by writing the equation
$\text{Ag}_{2}\text{CrO } ({s}) \rightleftharpoons \text {2Ag}^{2+} ({aq}) + \text{CrO}_{4}^{2-} ({aq}) \nonumber$
from which
${K}_{sp}(\text{Ag}_{2}\text{CrO}_{4})= [\text{Ag}^{+}]^{2} [\text{CrO}_{4}^{2-}]= \text{1.0} \times \text{10}^{-12} \text{mol}^{3} \text{L}^{-3} \nonumber$
Let the solubility be x mol L–1. Then
$[\text{CrO}_{4}^{2-}]= {x } \text{ mol } \text{L}^{-1} \nonumber$
and
$[\text{Ag}^{+}] = 2x \text{mol L}^{-1} \nonumber$
Thus
\begin{align}{K}_{sp}= (\text{2}{x} \text{ mol } \text{L}^{-1})^{2} {x} \text{ mol L}^{-1}\\text{ }= (\text{2}{x})^{2} { x}\text{ mol}^{3} \text{ L}^{-3} = \text{1.0} \times \text{10}^{-12}\text{ mol}^{3} \text{ L}^{-3} \, \end{align} \nonumber
or
$4x^{3} = \text{1.0 x 10}^{-12} \nonumber$
and
$x^{\text{3}}=\frac{\text{1.0}}{\text{4}}\text{ }\times \text{ 10}^{12}=\text{2.5 }\times \text{ 10}^{-13}=\text{250 }\times \text{ 10}^{-15} \nonumber$
so that
$x=\sqrt[\text{3}]{\text{250}}\text{ }\times \text{ }\sqrt[\text{3}]{\text{10}^{-\text{15}}}=\text{6}\text{.30 }\times \text{ 10}^{-\text{5}} \nonumber$
Thus the solubility is 6.30 × 10–5 mol L–1 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.11%3A_The_Solubility_Product.txt |
Suppose we have a saturated solution of lead chloride in equilibrium with the solid salt:
$\text{PbCl}_{2}({s})\rightleftharpoons \text{Pb}^{2+}({aq}) + \text{2Cl}^{-}({aq}) \nonumber$
If we increase the chloride-ion concentration, Le Chatelier’s principle predicts that the equilibrium will shift to the left. More lead chloride will precipitate, and the concentration of lead ions will decrease. A decrease in concentration obtained in this way is often referred to as the common-ion effect.
The solubility product can be used to calculate how much the lead-ion concentration is decreased by the common-ion effect. Suppose we mix 10 mL of a saturated solution of lead chloride with 10 mL of concentrated hydrochloric acid (12 M HCl). Because of the twofold dilution, the chloride-ion concentration in the mixture will be 6 mol L–1. Feeding this value into equation 7 from the solubility product section, we then have the result
${K}_{sp} = [\text{Pb}^{2+}][\text{Cl}^{-}]^{2} \nonumber$
or
$\text{1.70} \times \text{10}^{-5} \text{mol}^{3} \text{L}^{-3}=[\text{Pb}^{2+}](\text{6 mol L}^{-1})^{2} \nonumber$
so that
$[\text{ Pb}^{2+}]=\frac{\text{1.70 }\times \text{ 10}^{-5}\text{ mol}^{3}\text{ L}^{-3}}{\text{36 mol}^{2}\text{ L}^{-2}}=\text{4.72 }\times \text{ 10}^{-7}\text{ L} \nonumber$
We have thus lowered the lead-ion concentration from an initial value of 1.62 × 10–2 mol L–1 (see Example 1 from the section on the solubility product) to a final value of 4.72 × 10–7 mol L–1, a decrease of about a factor of 30 000! As a result, we have at our disposal a very sensitive test for lead ions. If we mix equal volumes of 12 M HCl and a test solution, and no precipitate occurs, we can be certain that the lead-ion concentration in the test solution is below 2 × 4.72 × 10–7 mol L–1.
Because it tells us about the conditions under which equilibrium is attained, the solubility product can also tell us about those cases in which equilibrium is not attained. If extremely dilute solutions of Pb(NO3)2 and KCl are mixed, for instance, it may be that the concentrations of lead ions and chloride ions in the resultant mixture are both too low for a precipitate to form. In such a case we would find that the product Q, called the ion product and defined by
$Q=\text{(}c_{\text{Pb}^{2+}}\text{)(}c_{\text{Cl}^{-}}\text{)}^{2} \label{1}$
has a value which is less than the solubility product 1.70 × 10–5 mol3 L–3. In order for equilibrium between the ions and a precipitate to be established, either the lead-ion concentration or the chloride-ion concentration or both must be increased until the value of Q is exactly equal to the value of the solubility product. The opposite situation, in which Q is larger than Ksp, corresponds to concentrations which are too large for the solution to be at equilibrium. When this is the case, precipitation occurs, lowering the concentration of both the lead and chloride ions, until Q is exactly equal to the solubility product.
To determine in the general case whether a precipitate will form, we set up an ion-product expression Q which has the same form as the solubility product, except that the stoichiometric concentrations rather than the equilibrium concentrations are used. Then if $Q > K_{sq}$, precipitation occurs, while if $Q < K_{sq}$, no precipitation occurs.
The video below provides a nice summary to the information above as well as giving you an idea of what this process looks like on a large scale. Notice that at the end of the video, excess chloride ions are added to the solution, causing an equilibrium shift to the side of lead chloride. This addition of chloride ions demonstrates the common ion effect.
Example $1$: Precipitation
Decide whether CaSO4 will precipitate or not when
1. 100 mL of 0.02 M CaCl2 and 100 mL of 0.02 M Na2SO4 are mixed, and also when
2. 100 mL of 0.002 M CaCl2 and 100 mL of 0.002 M Na2SO4 are mixed.
Ksp = 2.4 × 10–5 mol2 L–2.
Solution
a) After mixing, the concentration of each species is halved. We thus have
${c}_{\text{Ca}^{2+}}=\text{0.01 mol L}^{-1}={c}_{\text{SO}_{4}^{2-}}$
so that the ion-product Q is given by
${Q}={c}_{\text{Ca}^{2+}} \times {c}_{\text{SO}_{4}^{2-}} = \text{0.01 mol L}^{-1} \times \text{ 0.01 mol L}^{-1}$
or
${Q}= \text{10}^{-4} \text{ mol}^{2}\text{L}^{-2}$
Since Q is larger than Ksp(2.4 × 10–5 mol2 L–2), precipitation will occur.
b) In the second case
${c}_{\text{Ca}^{2+}}= \text{0.001 mol L}^{-1} = {c}_{\text{SO}_{4}^{2-}}$
and
${Q}={c}_{\text{Ca}^{2+}}\times {c}_{\text{SO}_{4}^{2-}}= \text{1} \times \text{10}^{-6} \text{mol}^{2} \text{L}^{-2}$
Since Q is now less than Ksp, no precipitation will occur.
Example $2$: Precipitate Mass
Calculate the mass of CaSO4 precipitated when 100 mL of 0.0200 M CaCl2 and 100 mL of 0.0200 M Na2SO4 are mixed together.
Solution
We have already seen in part a of the previous example that precipitation does actually occur. In order to find how much is precipitated, we must concentrate on the amount of each species. Since 100 mL of 0.02 M CaCl2 was used, we have
$n_{\text{Ca}^{2+}}=\text{0.0200 }\frac{\text{mmol}}{\text{mL}}\times \text{ 100 ml}=\text{2.00 mmol}$
similarly
$n_{\text{SO}_{4}^{2-}}=\text{0.0200 }\frac{\text{mmol}}{\text{mL}} \times \text{ 100 ml}=\text{2.00 mmol}$
If we now indicate the amount of CaSO4 precipitated as x mmoles, we can set up a table in the usual way:
Species Ca2+ (aq) SO42 (aq)
Initial amount (mmol) ${2.00}$ ${2.00}$
Amount reacted (mmol) $-{x}$ $-{x}$
Equilibrium amount (mmol) $(\text{2}-{x})$ $(\text{2}-{x})$
Equilibrium concentration (mmol mL) $\frac{\text{2}-{x}}{\text{200}}$ $\frac{\text{2}-{x}}{\text{200}}$
Thus
${K}_{sp} = [\text{Ca}^{2+}][\text{SO}_{4}^{2-}]$
or
$\text{2.4 }\times \text{ 10}^{-5}\text{ mol}^{2}\text{ L}^{-2}=\left( \frac{\text{2}-x}{\text{200}}\text{ mol L}^{-1} \right)\left( \frac{\text{2}-x}{\text{200}}\text{ mol L}^{-1} \right)$
Rearranging,
$\text{200}^{2}\times\text{2.4}\times\text{10}^{-5}=\text{0.96}=(\text{2}-{x})^{2}$
or
${2}-{x}=\sqrt{\text{0}\text{.96}}={0.980}$
so that
${x}={2}-{0.980}={1.020}$
Since 1.020 mmol CaSO4 is precipitated, the mass precipitated is given by
$m_{\text{CaSO}_{4}}=\text{1.020 mmol }\times \text{ 136.12 }\frac{\text{mg}}{\text{mmol}}\\text{ }=\text{138.9 mg}=\text{0.139 g}$
Because the solutions are so dilute and because CaSO4 has a fairly large solubility product, only about half (1.02 mmol out of a total of 2.00 mmol) the Ca2+ ions are precipitated. If we wished to determine the concentration of Ca2+ ions in tap water or river water, where it is quite low, it would be foolish to try to precipitate the Ca as CaSO4. Another method would have to be found.
14.12: The Common-Ion Effect
Tartaric acid (TA) is a byproduct of wine production. This organic acid and its salts are used in foods such as fruit jellies, preserves, jams, baked goods, and confections. TA is hardly metabolized and degraded by yeast and spoilage bacteria providing microbiological stability to foods that contain it. In addition to its use as antimicrobial and acidulant, TA and its salts are used as emulsifiers, leavening, and anticacking agents.[1] TA is also used in the beverage industry and has non-food uses in textile coloring, galvanizing, and mirror production.
The increasing popularity in wine consumption in recent years has resulted in the increase of waste from wine making practices.[2] One liter of white wine generates the same amount of water pollution as 3 people in one day.[3] Waste-waters from wine production contain biodegradable compounds and fruit suspended solids; [4] their treatment is of great importance because their high pollutant activity. Moreover, waste treatment is of economic interest because the organic compounds present in waste from the wine making process can have value as additives, ingredients, and substrates in the food and pharmaceutical industries.
Type of Waste Name Derived from Treatment[3]
Waste derived from wine making
Liquid/Sludge Vinasse
Vintage
process
Anaerobic digestion, ozonation, thermophilic anaerobic digestion, aerobic biodegradation, sequencing batch reactor, electrodialysis, and wet oxidation
Solid waste
Grape
marc
Skin, stalks, and seeds Combustion, solid-state fermentation, incineration, composting, and pyrolysis.
Vine
shoots
Vine
pruning
Lees
Grapes
and yeast
Tartaric Acid separated from grape juice
Tartaric Acid crystals
The table above shows sources of waste in wine making as well as methods to treat them. A method devised by Rivas, et al. (2006) to treat distilled lees involves the reaction of tartaric acid with calcium-ions to form calcium tartrate. This salt of limited solubility is then redissolved with HCL to obtain TA. Upon removal of TA, the distilled lees can be used as nutrients for lactic acid bacteria (Lactobacilllus pentosus) for production of lactic acid.
In the method described above, addition of calcium ions prompts the precipitation of calcium tartrate. Once tartaric acid is removed from the solution as tartrate, it will partially dissociate into calcium and tartrate ions establishing the following equilibrium with the solid salt:
$\text{CaC}_{4}\text{H}_{4}\text{O}_{6}({s})\rightleftharpoons \text{Ca}^{2+}({aq}) + \text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}({aq})$ with the corresponding equilibrium constant ${K}_{sp} = [\text{Ca}^{2+}][\text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}]=\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}\,$ If we designate as x the concentration of each of the ions, the concentration of calcium-ions at equilibrium is \begin{align} x^{2}=\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}\end{align} From which $x=\sqrt{\text{7.7}\times \text{10}^{-7}\text{ mol}^{2} \text{dm}^{-6} }= \text{8.77} \times \text{10}^{-4} \text{mol dm}^{-3}$ The concentration of calcium-ions at equilibrium is 8.77 x 10–4 mol dm–3. If now we increase the concentration of tartrate-ions, the equilibrium will shift to the left according to the Le Chatelier’s principle. More calcium tartrate will precipitate, decreasing the concentration of calcium ions. The decrease in concentration obtained in this way is often referred to as the common-ion effect. Similarly, if an excess of calcium-ions is added to the solution, the concentration of tartrate-ion will decrease. Since in this process we are concerned about removing as much tartaric acid in the form of tartrate as possible, addition of calcium ions in excess will minimize dissociation of calcium tartrate increasing its the yield. The solubility product can be used to calculate how much the calcium-ion concentration is decreased by the common-ion effect. Suppose we mix 10 cm3 of a saturated solution of calcium tartrate with 10 cm3 of concentrated sodium tartrate (4 M C4H6O6). Because of the twofold dilution, the concentration of tartrate will be 2 mol dm–3. Feeding this value into equation 7 from the solubility product section, we then have the result or $\text{7.7} \times \text{10}^{-7} \text{mol}^{2} \text{dm}^{-6}=[\text{Ca}^{2+}](\text{2 mol dm}^{-3})$ so that $[\text{ Ca}^{2+}]=\frac{\text{7.7 }\times \text{ 10}^{-7}\text{ mol}^{2}\text{ dm}^{-6}}{\text{2 mol dm}^{-3}}=\text{3.85 }\times \text{ 10}^{-7}\text{ mol}\text{ dm}^{-3}$ We have thus lowered the calcium-ion concentration from an initial value of 8.77 x 10–4 mol dm–3 ) to a final value of 3.85 × 10–7 mol dm–3, a decrease of about a factor of 2000!.
Calcium tartrate
Because it tells us about the conditions under which equilibrium is attained, the solubility product can also tell us about those cases in which equilibrium is not attained. If extremely dilute solutions of Na2C4H6O6 and CaCl2 are mixed, for instance, it may be that the concentrations of calcium ions and chloride ions in the resultant mixture are both too low for a precipitate to form. In such a case we would find that the product Q, called the ion product and defined by
$Q=\text{(}c_{\text{Ca}^{2+}}\text{)(}c_{\text{C}_{4}\text{H}_{4}\text{O}_{6}^{2-}})$ (1)
In this case Q has a value below the solubility product, 7.7 × 10–7 mol2 dm–6. In order for equilibrium between the ions and a precipitate to be established, either the calcium-ion concentration or the tartrate-ion concentration or both must be increased until the value of Q is exactly equal to the value of the solubility product. The opposite situation, in which Q is larger than Ksp, corresponds to concentrations which are too large for the solution to be at equilibrium. When this is the case, precipitation occurs, lowering the concentration of both the lead and chloride ions, until Q is exactly equal to the solubility product. To determine in the general case whether a precipitate will form, we set up an ion-product expression Q which has the same form as the solubility product, except that the stoichiometric concentrations rather than the equilibrium concentrations are used. Then if
$\text{Q}>{K}_{sp}\,$ precipitation occurs
while if $\text{Q}<{K}_{sp}\,$ no precipitation occurs
EXAMPLE 1 Decide whether CaC2O4, calcium oxalate, will precipitate or not when (a)100 cm3 of 0.02 M CaCl2 and 100 cm3 of 0.02 M Na2C2O4 are mixed, and also when (b) 100 cm3 of 0.0001 M CaCl2 and 1000 cm3 of 0.0001 M Na2C2O4 are mixed. Ksp = 2.32 × 10–9 mol2 dm–6.
Solution
a) After mixing, the concentration of each species is halved. We thus have ${c}_{\text{Ca}^{2+}}=\text{0.01 mol dm}^{-3}={c}_{\text{C}_{2}\text{O}_{4}^{2-}}$ so that the ion-product Q is given by ${Q}={c}_{\text{Ca}^{2+}} \times {c}_{\text{C}_{2}\text{O}_{4}^{2-}} = \text{0.01 mol dm}^{-3} \times \text{ 0.01 mol dm}^{-3}$ or ${Q}= \text{10}^{-4} \text{ mol}^{2}\text{dm}^{-6}\,$
Since Q is larger than Ksp (2.32 × 10–9 mol2 dm–6), precipitation will occur.
b) In the second case the concentration of each ion becomes ${c}_{\text{Ca}^{2+}}=\frac{\text{0.0001 mol dm}^{-3} \times \text{100 cm}^{3}}{\text{1100 cm}^{3}} = \text{9.09} \times \text{10}^{-6} \text{mol dm}^{-3}$ and ${c}_{\text{C}_{2}\text{O}_{4}^{2-}}=\frac{\text{0.0001 mol dm}^{-3} \times \text{1000 cm}^{3}}{\text{1100 cm}^{3}} = \text{9.09} \times \text{10}^{-5} \text{mol dm}^{-3}$
thus \begin{align}{Q}&={c}_{\text{Ca}^{2+}}\times {c}_{\text{C}_{2}\text{O}_{4}^{2-}}\ \text{ }&=(\text{9.09} \times \text{10}^{-6}\text{mol dm}^{-3}) (\text{9.09} \times \text{10}^{-5} \text{mol dm}^{-3})\ \text{ }&=\text{8.26} \times \text{10}^{-10}\text{mol}^{2} \text{dm}^{-6}\end{align}
Since Q is now less than Ksp, no precipitation will occur.
EXAMPLE 2 Calculate the mass of CaC2O4 precipitated when 100 cm3 of 0.0200 M CaCl2 and 100 cm3 of 0.0200 M Na2C2O4 are mixed together.
Solution In in part a of the previous example we determined that precipitation does actually occur. In order to find how much calcium oxalate is precipitated, we must concentrate on the amount of each species. Since 100 cm3 of 0.02 M CaCl2 was used, we have
$n_{\text{Ca}^{2+}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}}\times \text{ 100 cm}^{3}=\text{2.00 mmol}$
similarly $n_{\text{C}_{2}\text{O}_{4}^{2-}}=\text{0.0200 }\frac{\text{mmol}}{\text{cm}^{3}} \times \text{ 100 cm}^{3}=\text{2.00 mmol}$ If we designate the amount of CaC2O4 that precipitates as x mmoles, we can set up the following table
Species Ca2+ (aq) C2O42 (aq)
Initial amount (mmol) $\text{2.00}\,$ $\text{2.00}\,$
Amount reacted (mmol) $-{x}\,$ $-{x}\,$
Equilibrium amount (mmol) $(\text{2}-{x})\,$ $(\text{2}-{x})\,$
Equilibrium concentration (mmol cm–3) $\frac{\text{2}-{x}}{\text{200}}$ $\frac{\text{2}-{x}}{\text{200}}$
Thus
${K}_{sp} = [\text{Ca}^{2+}][\text{C}_{2}\text{O}_{4}^{2-}]\,$
or $\text{2.32 }\times \text{ 10}^{-9}\text{ mol}^{2}\text{ dm}^{-6}=\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)\left( \frac{\text{2}-x}{\text{200}}\text{ mol dm}^{-3} \right)$ Rearranging, $\text{200}^{2}\times\text{2.32}\times\text{10}^{-9}=\text{9.28}\times\text{10}^{-3}=(\text{2}-{x})^{2}$ or $\text{2}-{x}=\sqrt{\text{9.28}\times \text{10}^{-3}}=\text{0.096}$ so that ${x}=\text{2}-\text{0.096}=\text{1.904}\,$ Since 1.904 mmol of CaC2O4 precipitated, its mass is \begin{align}m_{\text{CaC}_{2}\text{O}_{4}}&=\text{1.904 mmol }\times \text{ 146.11 }\frac{\text{mg}}{\text{mmol}}\ \text{ }&=\text{278.2 mg}=\text{0.278 g}\end{align} Note: Since the solubility product of CaC2O4 is very small, about 95% of the calcium oxalate originally formed precipitates.
References
1. Food Additives, 2nd ed. 2002, Branen, A., Davidson, M.P., Salminen, S. and Thorngate III, J.H.
2. Pollard, K. 2009. Waste management in the Wine Industry. www.brocku.ca/tren/courses/tr...nt%20Paper.pdf
3. Arvanitoyannis, I. S., Ladas, D., and Mavromatis, A. 2006. Wine waste treatment methodology. Int. J. Food Sci. Tech. 41:1117-1151.
4. Rivas, B., Torrado, A., Moldes, A. B., and Dominguez, J. M. 2006. Tartaric acid recovery from distilled lees and use of the residual solid as an economic nutrient for Lactobacillus. J. Agric. Food Chem. 54:7904-7911.
14.13: The Solubilities of Salts of Weak Acids
In many chemical operations it is an advantage not only to be able to form a precipitate but to be able to redissolve it. Fortunately, there is a wide class of sparingly soluble salts which can almost always be redissolved by adding acid. These are precipitates in which the anion is basic; i.e., they are the salts of weak acids. An example of such a precipitate is calcium carbonate, whose solubility equilibrium is
$\text{CaCO}_{3}\text{ }({s}) \rightleftharpoons \text{Ca}^{2+} ({aq}) + \text{CO}_{3}^{2-} ({aq}) \nonumber$
If acid is now added to this solution, some of the carbonate ions become protonated and transformed into HCO3 ions. As a result, the concentration of the carbonate ion is reduced. In accord with Le Chatelier’s principle, the system will respond to this reduction by trying to produce more carbonate ions. Some solid CaCO3 will dissolve, and the equilibrium will be shifted to the right. If enough acid is added, the carbonate-ion concentration in the solution can be reduced so as to make the ion product (Q = cCa2+ × cCO32) smaller than the solubility product Ksp so that the precipitate dissolves.
A similar behavior is shown by other precipitates involving basic anions. Virtually all the carbonates, sulfides, hydroxides, and phosphates which are sparingly soluble in water can be dissolved in acid. Thus, for instance, we can dissolve precipitates like ZnS, Mg(OH)2, and Ca2(PO4)3 because all the following equilibria
$\text{ZnS}\text{ }({s})\rightleftharpoons \text{Zn}^{2+} ({aq}) + \text{S}^{2-} ({aq}) \nonumber$
$\text{Mg(OH)}_{2}\text{ }({s})\rightleftharpoons \text{Mg}^{2+} ({aq}) + \text{2OH}^{-} ({aq}) \nonumber$
$\text{Ca}_{3}(\text{PO}_{4})_{2}\text{ }({s})\rightleftharpoons \text{3Ca}^{2+} ({aq}) + \text{2PO}_{4}^{3-} ({aq}) \nonumber$
can be shifted to the right by attacking the basic species S2–, OH, and PO43 with hydronium ions. Very occasionally we find an exception to this rule. Mercury(II) sulfide, HgS, is notorious for being insoluble. The solubility product for the equilibrium
$\text{HgS}\text{ }({s})\rightleftharpoons \text{Hg}^{2+} ({aq}) + \text{S}^{2-} ({aq} \nonumber$
is so minute that not even concentrated acid will reduce the sulfide ion sufficiently to make Q smaller than Ksp.
Occasionally the shift in a solubility-product equilibrium caused by a decrease in pH may be undesirable. One example of this was mentioned in the section on chalcogens. Acid rainfall can occur when oxides of sulfur and other acidic air pollutants are removed from the atmosphere. In some parts of the United States pH values as low as 4.0 have been observed. These acid solutions dissolve marble and limestone (CaCO3) causing considerable property damage. This is especially true in Europe, where some statues and other works of art have been almost completely destroyed over the last half century. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/14%3A_Ionic_Equilibria_in_Aqueous_Solutions/14.12%3A_The_Common-Ion_Effect/14.12.01%3A_Foods-_Calcium_Tartrate_and_Treatment_of_Wine_Waster-Waters.txt |
In the sections on Using Chemical Equations in Calculations we indicate that heat is a form of energy and show how the quantity of heat energy absorbed or released by a chemical change can be related to the corresponding chemical equation. We also state the law of conservation of energy, and arguments in other sections have often been based on the idea that energy can neither be created nor destroyed. The law of conservation of energy is the first of three important laws involving energy and matter, which were discovered over a century ago. These laws were originally based on the movement or transfer (dynamics) of heat (thermo), and the law of conservation of energy is therefore referred to as the first law of thermodynamics.
15: Thermodynamics- Atoms Molecules and Energy
In the sections on Using Chemical Equations in Calculations we indicate that heat is a form of energy and show how the quantity of heat energy absorbed or released by a chemical change can be related to the corresponding chemical equation. We also state the law of conservation of energy, and arguments in other sections have often been based on the idea that energy can neither be created nor destroyed. The law of conservation of energy is the first of three important laws involving energy and matter, which were discovered over a century ago. These laws were originally based on the movement or transfer (dynamics) of heat (thermo), and the law of conservation of energy is therefore referred to as the first law of thermodynamics
We assign the symbol ΔH and the name enthalpy change to the quantity of heat absorbed by a chemical or physical change under conditions of constant pressure. You may wonder just how heat energy could be absorbed or given off when atoms and molecules change position and structure during a chemical reaction, but we have not yet developed theories of chemical bonding, molecular structure, intermolecular forces, and molecular motion to the point where a satisfactory explanation can be given. We are in a position to investigate what can happen to molecules when matter absorbs or releases heat. One result of this study will be a clearer understanding of enthalpy. At the same time we will begin to appreciate what molecular factors contribute to making a reaction exothermic or endothermic. This gives us a solid basis for discussing several aspects of what is probably the most important problem facing our technological society today―the energy crisis
The first law of thermodynamics (the law of conservation of energy) states that when heat energy is supplied to a substance, that energy cannot disappear-it must still be present in the atoms or molecules of the substance. Some of the added energy makes the atoms or molecules move faster. This is called translational energy. In the case of molecules, which can rotate and vibrate, some of the added energy increases the rotational and vibrational energies of the molecules. You can investigate vibrations of the ethane molecule above in the Jmol. Finally, any atom or molecule will have a certain electronic energy which depends on how close its electron clouds are to positively charged nuclei.
The total of translational, rotational, vibrational, and electronic energies is the internal energy of an atom or molecule. When chemical reactions occur, the internal energy of the products is usually different from that of the reactants, and the difference appears as heat energy in the surroundings. If the reaction is carried out in a closed container (bomb calorimeter, for example), the increase in internal energy of the atoms and molecules is exactly equal to the heat energy absorbed from the surroundings. If the internal energy decreases, the energy of the surroundings must increase; i.e., heat energy is given off.
When a chemical reaction occurs at constant pressure, as in a coffee-cup calorimeter, there is a change in potential energy of the atmosphere (given by P ΔV) as well as a change in heat energy of the surroundings. Because the heat energy absorbed can be measured more easily than P ΔV, it is convenient to define the enthalpy as the internal energy plus the increased potential energy of the atmosphere. Thus the enthalpy increase equals the heat absorbed at constant pressure.
Enthalpy changes for a variety of reactions may be calculated from standard enthalpies of formation. They may also be estimated by summing the bond enthalpies of all bonds broken and subtracting the bond enthalpies of all bonds formed. Because the dissociation enthalpy for the same type of bond varies from one molecule to another, the second method is not as accurate as the first. However, it has the advantage that enthalpy changes for reactions of a particular compound can be estimated even if the compound has not yet been synthesized.
The enthalpy change for a reaction depends on the relative strengths of the bonds broken and formed and on the relative number of bonds broken and formed. A good fuel is a substance which can combine with oxygen from the air, forming more bonds and/or stronger bonds than were originally present. The fossil fuels, coal, petroleum, and natural gas consist mainly of carbon and hydrogen. When they burn in air, strong O—H and C=O bonds are formed in the resulting H2O and CO2 molecules. The supply of fossil fuels is limited, and they constitute a nonrenewable resource. Coal supplies ought to last another century or two, but petroleum and natural-gas supplies will be essentially depleted in half a century or less. During the next few decades it will be possible to gasify or liquefy coal to extend our supply of gaseous and liquid fuels. Conservation of these fuels can also make a major contribution toward continuing their use. Eventually, however, it will be necessary to develop nuclear or solar energy or some unknown source of energy if we are to continue our current energy-intensive way of life. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.01%3A_Prelude_to_Thermodynamics.txt |
When we supply heat energy from a bunsen burner or an electrical heating coil to an object, a rise in temperature usually occurs. Provided that no chemical changes or phase changes take place, the rise in temperature is proportional to the quantity of heat energy supplied. If q is the quantity of heat supplied and the temperature rises from T1 to T2 then
$q = C * (T_{2} – T_{1}) \nonumber$
OR
$q = C * (\Delta T) \nonumber$
where the constant of proportionality C is called the heat capacity of the sample. The sign of q in this case is + because the sample has absorbed heat (the change was endothermic), and (ΔT) is defined in the conventional way.
If we add heat to any homogenous sample of matter of variable mass, such as a pure substance or a solution, the quantity of heat needed to raise its temperature is proportional to the mass as well as to the rise in temperature. That is,
$q = C * m * (T_2 – T_1) \nonumber$
OR
$q = C * m * (\Delta T) \nonumber$
The new proportionality constant C is the heat capacity per unit mass. It is called the specific heat capacity (or sometimes the specific heat), where the word specific means “per unit mass.”
Specific heat capacities provide a convenient way of determining the heat added to, or removed from, material by measuring its mass and temperature change. As mentioned [|previously], James Joule established the connection between heat energy and the intensive property temperature, by measuring the temperature change in water caused by the energy released by a falling mass. In an ideal experiment, a 1.00 kg mass falling 10.0 m would release 98.0 J of energy. If the mass drove a propeller immersed in 0.100 liter (100 g) of water in an insulated container, its temperature would rise by 0.234oC. This allows us to calculate the specific heat capacity of water:
98 J = C × 100 g × 0.234 oC
C = 4.184 J/goC
At 15°C, the precise value for the specific heat of water is 4.184 J K–1 g–1, and at other temperatures it varies from 4.178 to 4.219 J K–1 g–1. Note that the specific heat has units of g (not the base unit kg), and that since the Centigrade and kelvin scales have identical graduations, either oC or K may be used.
Example $1$: Specific Heat of Water
How much heat is required to raise the temperature of 500 mL of water (D = 1.0) from 25.0 oC to 75.0 oC, given that the specific heat capacity of water is 4.184 J K–1 g–1?
Solution:
q = 4.18 J/goC × 500 g × (75.0 - 25.0)
q = 104 500 J or 104 kJ.
Table $1$ Specific heat capacities (25 °C unless otherwise noted)
Substance phase Cp(see below)
J/(g·K)
air, (Sea level, dry, 0 °C) gas 1.0035
argon gas 0.5203
carbon dioxide gas 0.839
helium gas 5.19
hydrogen gas 14.30
methane gas 2.191
neon gas 1.0301
oxygen gas 0.918
water at 100 °C (steam) gas 2.080
water at T=[1] liquid 0.01°C 4.210
15°C 4.184
25°C 4.181
35°C 4.178
45°C 4.181
55°C 4.183
65°C 4.188
75°C 4.194
85°C 4.283
100°C 4.219
water (ice) at T= [2] solid 0°C 2.050
-10°C 2.0
-20°C 1.943
-40°C 1.818
ethanol liquid 2.44
copper solid 0.385
gold solid 0.129
iron solid 0.450
lead solid 0.127
Electrical Energy Conversion
The most convenient way to supply a known quantity of heat energy to a sample is to use an electrical coil. The heat supplied is the product of the applied potential V, the current I flowing through the coil, and the time t during which the current flows:
$q = V * I * t \nonumber$
If the SI units volt for applied potential, ampere for current, and second time are used, the energy is obtained in joules. This is because the volt is defined as one joule per ampere per second:
1 volt × 1 ampere × 1 second = 1$\begin{matrix}\frac{\text{J}}{\text{A s}}\end{matrix}$ × 1 A × 1 s = 1 J
Example $2$: Heat Capacity
An electrical heating coil, 230 cm3 of water, and a thermometer are all placed in a polystyrene coffee cup. A potential difference of 6.23 V is applied to the coil, producing a current of 0.482 A which is allowed to pass for 483 s. If the temperature rises by 1.53 K, find the heat capacity of the contents of the coffee cup. Assume that the polystyrene cup is such a good insulator that no heat energy is lost from it.
Solution The heat energy supplied by the heating coil is given by
$q = V * I * t = 6.23 \text{V} * 0.482 \text{A} * 483 \text{s} = 1450 \text{V A s} = 1450 \text{J}$
However,
$q = C * (T_{2} – T_{1})$
Since the temperatue rises, T2 > T1 and the temperature change ΔT is positive:
1450 J = C × 1.53 K
so that
\begin{matrix}C=\frac{\text{1450 J}}{\text{1}\text{.53 K}}=\text{948 J K}^{-\text{1}}\end{matrix}
Note
Note: The heat capacity found applies to the complete contents of the cup-water, coil, and thermometer taken together, not just the water.
As discussed in other sections, an older, non-SI energy unit, the calorie, was defined as the heat energy required to raise the temperature of 1 g H2O from 14.5 to 15.5°C. Thus at 15°C the specific heat capacity of water is 1.00 cal K–1 g–1. This value is accurate to three significant figures between about 4 and 90°C.
If the sample of matter we are heating is a pure substance, then the quantity of heat needed to raise its temperature is proportional to the amount of substance. The heat capacity per unit amount of substance is called the molar heat capacity, symbol Cm. Thus the quantity of heat needed to raise the temperature of an amount of substance n from T1 to T2 is given by
$q = C * n * (T_{2} – T_{1})\label{7}$
The molar heat capacity is usually given a subscript to indicate whether the substance has been heated at constant pressure (Cp)or in a closed container at constant volume (CV).
Example $3$: Molar Heat Capacity
A sample of neon gas (0.854 mol) is heated in a closed container by means of an electrical heating coil. A potential of 5.26 V was applied to the coil causing a current of 0.336 A to pass for 30.0 s. The temperature of the gas was found to rise by 4.98 K. Find the molar heat capacity of the neon gas, assuming no heat losses.
Solution The heat supplied by the heating coil is given by
$q = V * I * t$
$= 5.26 \text{V} * 0.336 \text{A} * 30.0 \text{s}$
$= 53.0 \text{V A s}$
$= 53.0 \text{J}$
Rearranging Eq. $\ref{7}$, we then have
\begin{matrix}C_{m}=\frac{q}{n\text{(T}_{\text{2}}-\text{T}_{\text{1}}\text{)}}=\frac{\text{53}\text{.0 J}}{\text{0}\text{.854 mol }\times \text{ 4}\text{.98 K}}=\text{12}\text{.47 J K}^{-\text{1}}\text{ mol}^{-\text{1}}\end{matrix}
However, since the process occurs at constant volume, we should write
$C_{V} = 12.47 \text{J K^{-1} mol^{-1}}$
15.02: Heat Capacities
Mercury (10 cc) and 10 cc of water in test tubes at room temperature (with thermistors interfaced to computer, if possible). Allow both tubes to equilibrate to room temperature (~17 oC), then immerse them simultaneously in a water bath at 34 oC. Will 136 g of mercury heat up faster or slower than 10 g of water? Measure T every 1 second with interfaced computer. Plot. Paradoxically, mercury increases in T much faster. The amount of heat absorbed by each is roughly equivalent, since the test tubes and volumes are identical. Small differences result from the change in rate of heat transfer across the glass with ΔT.
Figure \(1\) Mercury and water in test tubes
Choose Tf at any time, and calculate ΔT for Hg and water. Assume q absorbed is same for both, and calculate Cp for Hg from q = m x C x ΔT for water.
The q/T Paradox: Which "Contains More Heat", a Cup of Coffee at 95 °C or a Liter of Icewater?[1]
A small mass of water at 0oC is added to a measured mass of liquid nitrogen, and the amount that evaporates is compared to the mass that evaporates when a larger mass of water at 95oC is added to liquid nitrogen. This demonstration requires knowledge of both specific heat and heat capacity. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.02%3A_Heat_Capacities/15.2.01%3A_Lecture_Demonstrations.txt |
Let us turn our attention from the macroscopic to the microscopic level. According to the first law of thermodynamics, the heat energy absorbed as we raise the temperature of a substance cannot be destroyed. But where does it go? In the case of a monatomic gas, like neon, this question is easy to answer. All the energy absorbed is converted into the kinetic energy of the neon molecules (atoms). In other sections, we found that the kinetic energy of the molecules in a sample of gas is given by the expression
$E_{k}=\tfrac{\text{3}}{\text{2}}nRT \nonumber$
Thus if the temperature of a sample of neon gas is raised from T1 to T2, the kinetic energy of the molecules increases from 3/2 nRT1 to 3/2 nRT2, a total change of
$\tfrac{3}{2}nR\left( T_{2}-T_{1}\right) =\left( \tfrac{3}{2}R\right) n\left( T_{2}-T_{1}\right) \nonumber$
Inserting the value of R in appropriate units, we obtain
\begin{matrix}\frac{3}{2}\left(8.314\frac{\text{J}}{\text{K mol}}\right)n\left(T_{2}-T_{1}\right)=\left(12.47\frac{\text{J}}{\text{K mol}}\right)n\left(T_{2}-T_{1}\right)\end{matrix}
This is the same quantity that is obtained by substituting the experimental value of CV for neon (calculated in Example 2 from Heat Capacities) into Eq. (4) from Heat Capacities. In other words the quantity of heat found experimentally exactly matches the increase in kinetic energy of the molecules required by the kinetic theory of gases.
Table $1$ lists the CV values not only for neon but for some other gases as well. We immediately notice that only the noble gases and other mon-atomic gases such as Hg and Na have molar heat capacities equal to 3/2R, or 12.47 J K–1 mol–1. All other gases have higher molar heat capacities than this. Moreover, as the table shows, the more complex the molecule, the higher the molar heat capacity of the gas. There is a simple reason for this behavior.
Table $1$: Molar Heat Capacities at Constant Volume (CV) for Various Gases (Values at 298 K Unless Otherwise Stated).
Gas Cv /J K-1 mol-1 Gas Cv /J K-1 mol-1
Monoatomic Gases Triatomic Gases
Ne 12.47 CO2 28.81
Ar 12.47 N2O 30.50
Hg 12.47 (700K) SO2 31.56
Na 12.47 (1200K)
Diatomic Gases Alkanes
N2 20.81 CH4 27.42
O2 21.06 C2H6 44.32
Cl2 25.62 C3H8 65.20
C4H10 89.94
A molecule which has two or more atoms is not only capable of moving from one place to another (translational motion), it can also rotate about itself, and it can change its shape by vibrating. When we heat a mole of Cl2 molecules, for example, we not only need to supply them with enough energy to make them move around faster (increase their translational kinetic energy), we must also supply an additional quantity of energy to make them rotate and vibrate more strongly than before. For a mole of more complex molecules like n-butane even more energy is required since the molecule is capable of changing its shape in all sorts of ways. In the butane molecule there are three C—C bonds around which segments of the molecule can rotate freely. All the bonds can bend or stretch, and the whole molecule can rotate as well. Such a molecule is constantly flexing and writhing at room temperature. As we raise the temperature, this kind of movement occurs more rapidly and extra energy must be absorbed in order to make this possible.
When we heat solids and liquids, the situation is somewhat different than for gases. The rapid increase of vapor pressure with temperature makes it virtually impossible to heat a solid or liquid in a closed container, and so heat capacities are always measured at constant pressure rather than at constant volume. Some Cp values for selected simple liquids and solids at the melting point are shown in Table $2$. In general the heat capacities of solids and liquids are higher than those of gases. This is because of the intermolecular forces operating in solids and liquids. When we heat solids and liquids, we need to supply them with potential energy as well as kinetic energy. Among the solids, the heat capacities of the metals are easiest to explain since the solid consists of individual atoms. Each atom can only vibrate in three dimensions. According to a theory first suggested by Einstein, this vibrational energy has the value 3RT, while the heat capacity is given by 3R = 24.9 J K–1 mol–1.
Table $2$: Molar Heat Capacities at Constant Pressure Cp for Various Solids and Liquids at the Melting Point.
Substance Cp (solid)/J K-1 mol-1 Cp (liquid)/J K-1 mol-1
Monoatomic Substances
Hg 27.28 27.98
Pb 29.40 30.33
Na 28.20 31.51
Diatomic Substances
Br2 53.8 75.7
I2 54.5 80.7
HCl 50.5 62.2
HI 47.5 68.6
Polyatomic Substances
H2O 37.9 76.0
NH3 49.0 77.0
Benzene 129.0 131.0
n-Heptane 146.0 203.1
As can be seen from the table, most monatomic solids have Cp values slightly larger than this. This is because solids expand slightly on heating. The atoms get farther apart and thus increase in potential as well as vibrational energy.
Solids which contain molecules rather than atoms have much higher heat capacities than 3R. In addition to the vibration of the whole molecule about its site in the crystal lattice, the individual atoms can also vibrate with respect to each other. Occasionally molecules can rotate in the crystal, but usually rotation is only possible when the solid melts. As can be seen from the values for molecular liquids in Table $2$, this sudden ability to rotate causes a sharp increase in the heat capacity. For monatomic substances, where there is no motion corresponding to the rotation of atoms around each other, the heat capacity of the liquid is only very slightly higher than that of the solid. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.03%3A_Heat_Capacity_and_Microscopic_Changes.txt |
When matter absorbs or releases heat energy, we cannot always explain this energy change on the microscopic level in terms of a speeding up or a slowing down of molecular motion as we do for heat capacities. This is particularly true when the heat change accompanies a chemical change. Here we must consider changes in the kinetic and potential energies of electrons in the atoms and molecules involved, that is, changes in the electronic energy.
As a simple example of a chemical change, let us consider an exothermic reaction involving only one kind of atom, the decomposition of ozone, O3:
$\ce{2O_{3}(g) \rightarrow 3O_2(g)} \,\,\, 25°C\label{1}$
O3 is a gas which occurs in very low but very important concentrations in the upper atmosphere. It can be produced in somewhat higher concentrations in the laboratory by using an electric spark discharge and can then be concentrated and purified. The result is a blue gas which is dangerously unstable and liable to explode without warning. Let us now suppose that we have a pure sample of 2 mol O3(g) in a closed container at 25°C and are able to measure the quantity of heat evolved when it subsequently explodes to form O2 gas according to Eq. $\ref{1}$ (see Figure $1$ ). It is found that 287.9 kJ is released. This energy heats up the surroundings.
Where do these 287.9 kJ come from? Certainly not entirely from the translational kinetic energy of the molecules. To begin with we had 2 mol O3 at 25°C and at the end 3 mol O2 at 25°C. Since the translational kinetic energy of any gas is 3/2 nRT, this corresponds to an increase of 3/2 (3 mol – 2 mol)RT = 3/2 × 1 mol × 8.314 J K–1 mol–1 × 298 K = 3.72 kJ. After the reaction the translational energy of the molecules is higher, and so heat should have been absorbed, not given off. In any case 3.73 kJ is only 1.3 percent of the total heat change. The changes in rotational and vibrational energy are even smaller, accounting for a decrease in energy of the substance in the container by only 0.88 kJ.
From the standpoint of energy, the most important thing that happens as 2 mol O3 is converted into 3 mol O2 is rearrangement of the valence electrons so that they are closer to positively charged O nuclei. We see in other sections that the closer electrons are to nuclei, the lower their total energy. Thus three O2 molecules have less electronic energy than the two O3 molecules from which they were formed. The remaining energy first appears as kinetic energy of the O2 molecules. Immediately after the reaction the O2 is at a very high temperature. Eventually this energy finds its way to the surroundings, and the O2 cools to room temperature.
A detailed summary of the various energy changes which occur when O3 reacts to form O2 is given in Table $1$. The important message of this table is that 99 percent of the energy change is attributable to the change in electronic energy. This is a typical figure for gaseous reactions. What makes a gaseous reaction exothermic or endothermic is the change in the bonding. Changes in the energies of molecular motion can usually be neglected by comparison.
TABLE $1$ Detailed Balance Sheet of the Energy Changes Occurring in the Reaction $2O_3(g) \rightarrow 3O_{2}(g)$ 25°C, constant volume
Type of Energy Initial Value/kJ Final Value/kJ Change*in Energy/kJ
Electronic
x
x – 290.70
–290.70
Translational
7.43
11.15
+3.72
Rotational and vibrational
8.32
7.44
– 0.88
Total
x + 15.75
x – 272.11
–287.86
* Final value – initial value.
† There is no experimental means of determining the initial or final value of the electronic energy—only the change in electronic energy can be measured. Highly accurate calculations of electronic energy from wave-mechanical theory require complicated mathematics end a great deal of computer power. Therefore we have represented the initial electronic energy by x.
The sum of all the different kinds of energy which the molecules of a substance can possess is called the internal energy and given the symbol U. (The symbol E also widely used.) In a gas we can regard the internal energy as the sum of the electronic, translational, rotational, and vibrational energies. In the case of liquids and solids the molecules are closer together, and we must include the potential energy due to their interactions with each other. Noncovalent interactions (intermolecular forces) attract one molecule to others. In addition, the motion of one molecule now affects its neighbors, and we can no longer subdivide the energy into neat categories as in the case of a gas.
Measurement of Internal Energy
Equation $\ref{2}$ tells us how to detect and measure changes in the internal energy of a system. If we carry out any process in a closed container the volume remains constant), the quantity of heat absorbed by the system equals the increase in internal energy.
$q_{V} = \triangle U\label{2}$
A convenient device for making such measurements is a bomb calorimeter (Figure $2$ ), which contains a steel-walled vessel (bomb) with a screw-on gas-tight lid. In the bomb can be placed a weighed sample of a combustible substance together with O2(g) at about 3 MPa (30 atm) pressure. When the substance is ignited by momentarily passing electrical current through a heating wire, the heat energy released by its combustion raises the temperature of water surrounding the bomb. Measurement of the change in temperature of the water permits calculation ofqV (and thus ΔU), provided the heat capacity of the calorimeter is known.
The heat capacity can be determined as in Example 1 from Heat Capacities or by igniting a substance for which ΔU is already known.
Example $1$: Energy
When 0.7943g of glucose, C6H12O6, is ignited in a bomb calorimeter, the temperature rise is found to be 1.841 K. The heat capacity of the calorimeter is 6.746 kJ K–1. Find ΔUm for the reaction
$C_{6}H_{12}O_{6}(s) + 6O_{2}(g) \rightarrow 6CO_{2}(g) + 6H_{2}O(l)$
under the prevailing conditions.
Solution
The heat energy absorbed by the calorimeter in increasing its temperature by 1.841 K is given by
$q = C \triangle T = 6.745 \text{kJ K^{-1} }* 1.841 \text{K} = 12.42 \text{kJ}\) \nonumber$
Since this heat energy was released by the reaction system, we must regard it as negative. Accordingly,
$q_{V} = –12.42 \text{kJ} = \triangle U \nonumber$
We need now only to calculate the change in internal energy per mole, that is, ΔUm. Now
$n_{\text{glucose}}=\text{0}\text{.7953 g }\times \text{ }\frac{\text{1 mol}}{\text{180}\text{.16 g}}=\text{4}\text{.409 }\times \text{ 10}^{-\text{3}}\text{ mol} \nonumber$
Thus $\triangle U_{m}= \frac{-12.42 kJ}{4.409} \times 10^{-3} \text{mol}= -2817 \text{kJ mol}^{-1} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.04%3A_Internal_Energy.txt |
We define the internal energy in molecular terms in another section, but it is important to realize that we can deal with this property purely in macroscopic terms as well. Thus we can talk about the internal energy of a substance without even knowing its chemical formula. More importantly, we can usually measure the change in the internal energy which accompanies a physical or chemical change without having the slightest idea of what is happening on the molecular level. In order to show how this is done, we first need to introduce the following conventions and terms:
1 System. The term system is used to describe any sample of matter in which we are particularly interested. Thus, in the conversion of O3 to O2 described in Figure $1$, the system would denote the contents of the flask before, after, or during the chemical change which occurs. Initially the system would thus consist of 2 mol O3 and finally of 3 mol O2.
2 Initial and final states. In thermodynamics our principal concern is with the initial state before any changes begin, and the final state, when no more changes occur. By convention we refer to the initial state with the subscript 1 and to the final state with the subscript 2. Referring again to the reaction illustrated in Figure $1$, if the volume of the flask is 24.47 dm³, the initial pressure P1 will be 202.7 kPa (2.00 atm) and the final pressure P2 will be 304.0 kPa (3.00 atm). Likewise we could refer to the initial value of the internal energy as U1 and the final value as U2. 3 The delta convention. When a chemical or physical change occurs, many of the properties of the system change. We conventionally refer to a change in a property with the symbol Δ (delta). In the O3–O2 reaction, for example, the pressure P increases by 101.3 kPa, and we can indicate this fact by writing
$\triangle P = 101.3 \text{kPa} \nonumber$
More formally, if a property X changes from an initial value X1 to a final value X2, then ΔX is defined as
$\triangle X = X_{2} – X_{1} \nonumber$
Thus, in the above case.
$\triangle P = P_{2} – P_{1} = 304.0 \text{kPa} – 202.7 \text{kPa} = 101.3 \nonumber$
We can also describe changes in internal energy using the delta convention. It is apparent from Table 1 in Internal Energy that the conversion of 2 mol O3 to 3 mol O2 results in a decrease of 287.9 kJ in the internal energy. In the delta convention we express this as a negative quantity:
$\triangle U = U_{2} – U_{1} = –287.9 \text{kJ} \nonumber$
since the initial value of U is larger than the final value. 4 Heat energy. Our final convention refers to the heat energy q absorbed by the system. If the heat change occurs at constant volume, the symbol qV is used, if at constant pressure, qp. By convention we refer to the heat energy absorbed by the system as being positive and the heat energy released by the system as being negative. In the O3 to O2 conversion described in Figure $1$, for example, we would write
$q_{V} = –287.9 \text{kJ} \nonumber$
since the system releases 287.9 kJ of heat energy to the surroundings. Now that we are familiar with these conventions, we can use them to state an important general principle. When a chemical or physical change occurs at constant volume, the heat energy absorbed by the system is equal to the increase in its internal energy, or in mathematical language
$q_{V} = \triangle U \nonumber$
This general behavior of matter is an immediate consequence of the law of conservation of energy. If an endothermic reaction occurs at constant volume and heat energy is absorbed by the reaction system, then this heat energy cannot disappear: It all goes toward increasing U. Alternatively, if the internal energy decreases, as in Figure $1$, no energy is lost: It all appears as heat energy and is transferred to the surroundings. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.05%3A_Thermodynamic_Terms_and_Conventions.txt |
In chemistry we are interested not only in those changes occurring in a closed container at constant volume but also in those occurring in an open container at constant (i.e., atmospheric) pressure. When a change occurs at constant pressure, there is another energy factor we must consider in addition to the heat absorbed and the change in internal energy. This is the expansion work wexp which the system does as its volume expands against the external pressure.
In order to understand the nature and magnitude of this expansion work, let us consider the simple example illustrated in Figure $1$. Here a sample of oxygen gas is heated at a constant pressure P from an initial temperature T1 to a final temperature T2 by means of an electrical heating coil. The gas is confined in a cylinder by a piston, and the pressure is maintained by placing a weight of the correct magnitude on top of the piston. The whole apparatus is maintained in a vacuum so that there is no atmospheric pressure to consider in addition to the effect of the weight and piston.
When the heating coil in this apparatus is switched on, the temperature rises and the gas expands in compliance with Charles’ law, lifting the piston and weight in the process. Energy must he supplied from the heating coil not only to increase the energy of the oxygen molecules but also to lift the weight. In other words,
$q_{P} = \triangle U + w_{exp}\label{1}$
where
• qp = heat absorbed
• ΔU = change in internal energy
• wexp = expansion work, i.e., work done in lifting the weight
To calculate the magnitude of this expansion work, we begin with the definition of pressure:
$P=\frac{\text{force}}{\text{area}}=\frac{F}{A} \nonumber$
or
$F = PA \nonumber$
where F is the total farce exerted by the weight and piston on the oxygen gas and A is the area of the piston. As the gas sample is heated, the volume increases from an initial value V1 to a final value V2, while the piston and its weight move from a height h1 to a height h2. The work done is thus given by the expression
$w_{exp} = \text{force exerted × distance moved}$
$= (P * A) * (h_{2} – h_{1})$
$= P * (A * h_{2} – A * h_{1})$
However, since the volume of a cylinder is the area of the base times the height,
$A * h_1 = V_1$ initial volume of gas
and $A * h_2 = V_2$ final volume
Thus $w_{exp} = P * (V_2 – V_1)$
or $w_{exp} = P \triangle V\label{2}$
Inserting this value for wexp into Eq. $\ref{1}$, we obtain the final result
$q_{P} = \triangle U + P \triangle V\label{3}$
It is important to realize that the expansion work P ΔV does not depend on our sample being in the apparatus of Figure $1$ in which there is an obvious gain in the potential energy of a weight. If instead of a weight we allowed the atmosphere to exert a pressure P on the gas, the result would still be P ΔV. In this second case, instead of lifting a visible weight, the expanding gas would push back the atmosphere and hence be lifting invisible air molecules. The work done, and hence the increase in the potential energy of the atmosphere, would still be P ΔV.
This simple example of an expanding gas helps us to see what is involved in the general case of a chemical or physical change occurring at constant pressure. In any such case the heat energy absorbed by the system, qp, will always exactly account for the increase in internal energy ΔU and the expansion work P ΔV. In other words the relationship $q_{P} = \triangle U + P \triangle V$ is valid in the general case.
Example $1$
When 1 mol liquid H2O is boiled at 100°C and 101.3 kPa (1.000 atm) pressure, its volume expands from 19.8 cm3 in the liquid state to 30.16 dm³ in the gaseous state. The heat energy absorbed by the vaporization process is found experimentally to be 40.67 kJ. Calculate the increase in internal energy ΔU of H2O.
Solution
We must first calculate the expansion work P ΔV.
$w_{exp} = P * (V_2 – V_1)$
$= 101.3 \text{kPa} * (30.16 \text{dm}^3– 0.0198 \text{dm}^3) = 3053 \text{kPa dm}^3$
In the section on Pressure, we see that 1Pa × 1 dm3 = 1 J, and we have
$w_{exp} = 3053 \text{J} = 3.055 \text{kJ}$
Also, since
$q_P = \triangle U + P \triangle V$
we have
$\triangle U = q_P – P \triangle V = 40.67 \text{kJ} – 3.05 \text{kJ} = 37.63 \text{kJ}$
Note: As we see in the sections on Properties of Organic Compounds and Other Covalent Substances and Solids, Liquids, and Solutions, the vaporization of a liquid is always an endothermic process. Since the molecules attract each other, energy must be supplied to separate them as vaporization occurs. However, not all the energy supplied when a liquid boils goes to increasing the potential energy of the molecules. A significant proportion is needed to increase the potential energy of the air as well.
It is far easier to carry out most chemical reactions in a container open to the atmosphere than in a closed system like a bomb calorimeter. However, when a reaction is carried out at constant atmospheric pressure, it is necessary to measure P, V1 and V2 (as in Example $1$ ) as well as qp in order to calculate the change in internal energy from the equation ΔU = qpP ΔV. These extra measurements are a nuisance, and they can be avoided by introducing a new quantity which is related to the internal energy but also includes the potential energy of the atmosphere. This quantity is also used in the sections on Using Chemical Equations in Calculations. It is called the enthalpy, symbol H, and is defined by
$H = U + PV\label{4}$
From Eq. $\ref{4}$ we can see that the enthalpy is always larger than the internal energy by a quantity PV. This extra energy is added to take account of the fact that whenever a body of volume V is introduced into the atmosphere, the potential energy of the atmosphere is increased by PV [by the same argument used to derive Eq. $\ref{2}$]. The enthalpy is thus of the form
Enthalpy = internal energy + potential energy of atmosphere
When a system undergoes a chemical or physical change at constant pressure, the change in enthalpy is given by
Change in enthalpy = change in internal energy + change in potential energy of atmosphere
The change in enthalpy thus includes both the energy changes for which heat energy must be supplied from the surroundings. In more formal language
$\triangle H = \triangle U + \triangle (PV) = \triangle U + (P_2V_2 – P_1V_1) \nonumber$
If we consider conditions of constant pressure, P2 = P1 = P, and
$\triangle H = \triangle U + (PV_2 – PV_1) = \triangle U + P(V_2 – V_1) = \triangle U + P \triangle V \nonumber$
but from Eq. $\ref{3}$
$q_P = \triangle U + P \triangle V \nonumber$
Thus
$q_P= \triangle H \nonumber$
In other words, the heat energy absorbed by a system in any change at constant pressure is equal to the increase in its enthalpy. The change in enthalpy ΔH can be obtained from a single experimental measurement: the heat energy absorbed at constant pressure. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.06%3A_Enthalpy.txt |
In an elementary laboratory, enthalpy changes are often measured in a “coffee-cup calorimeter” such as that shown in Figure $1$. Suppose the reaction to be measured is between two solutions. One of these solutions is introduced into the coffee cup, and the temperature of both solutions is measured. The second solution is now introduced, the mixture stirred, and the rise in temperature recorded. Since the cup is made of polystyrene foam, a very good insulator, very little heat energy escapes. It may be measurable, however, and may be represented by the calorimeter constant.
The Calorimeter Constant
A simple experiment can be used to determine how much energy is lost to the calorimeter, the thermometer, and the surroundings. This energy loss, divided by the temperature change in the calorimeter that caused it, is often called the "calorimeter heat capacity" or "calorimeter constant" as illustrated by the following example:
Example $1$: Calorimeter Constant
A styrofoam cup calorimeter contains 50 mL (50.00 g) of water at 22.00°C, and 40.00 mL (42.00 g) of water at 32.00°C is added. The mixture reaches a maximum temperature of 27.30°C. What is the calorimeter constant?
Solution
The hot water loses a quantity of heat $q_{hot}$;
\begin{align} q_{hot} &= C\, m\, ΔT \nonumber \[4pt] &= 4.18 \, J/g°C \times 50.0\, g \times (27.30°C - 32.00°C) \nonumber \[4pt] &= -982\, J \nonumber \end{align} \nonumber
The cold water gains a quantity of heat $q_{cold}$;
\begin{align} q_{cold} &= C\, m\, ΔT \nonumber \[4pt] &= 4.18 \, J/g°C \times 40.0\, g \times (27.30°C - 22.00°C) \nonumber \[4pt] &= 886\, J \nonumber \end{align} \nonumber
If no heat were lost to the surroundings,qhot would equal - qcold and their sum would be zero. However, it appears that more heat was lost by the hot water than was gained by the cold. The missing heat was absorbed by the calorimeter and surroundings:
\begin{align} q_{cal} &= q_{hot} + q_{cold} \[4pt] &= 96\, J.\nonumber \end{align} \nonumber
The calorimeter heat capacity is this heat divided by the temperature change of the calorimeter:
Ccal = q/ΔT = 96 J / 5.3 °C = 18.1 J/°C
Using the Calorimeter Constant
The heat absorbed by the calorimeter and its contents is equal to the heat released by a chemical reaction, as illustrated by the following example:
Example $2$: ΔH
The calorimeter above contains 250.0 cm3 of 1.000 M HCl (D = 1.02 g/cm3) at 20.38°C and to this is added, with stirring,250.0 cm3 of 1.000 M NaOH (D = 1.04 g/cm3) also at 20.38°C. The temperature of the mixture rises to 27.80°C. Assuming that the specific heat capacity of the NaCl solution that results is 4.20J/g°C, calculate ΔH, for the reaction
$\ce{H_{3}O^{+} (aq) + OH^{-} (aq) \rightarrow 2H_{2}O (l)} \nonumber$
Solution
The heat change for the solution is
qsolution = C m ΔT = 4.20 J/g°C × (250.0 × 1.02 + 250 × 1.04) g × (27.80°C - 20.38°C) = 16 049 J
The heat change for the calorimeter is
qcalorimeter = CcalΔT = 18.1 J/°C × 7.42°C = 134 J
The total heat change of the surroundings is
qtotal = qsolution + qcalorimeter = 16 183 J or 16.2 kJ
The heat change by the system is therefore
qsystem = -qsurroundings = -16 183 J or -16.2 kJ
To calculate ΔH, we need to divide the enthalpy change by the amount of product, in moles. Since 250 mL of 1 M solutions of each reactant are mixed and they are in a 1:1 stoichiometric ratio, we will produce 0.250 mol (n = C × V mol = 0.250 L x 1.00 mol/L) of product, so:
$\Delta H_{m}=\frac{-\text{16}\text{.2 kJ}}{\text{0}\text{.250 mol}}=\text{64}\text{.7 kJ mol}^{-\text{1}} \nonumber$
This value is slightly different from the accepted value (58.7 kJ/mol) because of inaccuracies in the heat capacity of the solution and crudeness of the method.
Another Approach
In order to find the quantity of heat energy evolved, we need to know the rise in temperature but also the heat capacity of the calorimeter contents and calorimeter. These can be combined, and determined in a separate experiment: After the final solution has cooled to room temperature, it can be reheated to the higher temperature by means of an electrical coil. The heat supplied in this second experiment can be calculated in the usual way from the applied potential, the current flowing, and the time.
Example $3$: Coffee-cup Calorimeter
1.000 M HCl (250.0 cm3) at 20.38°C and 250.0 cm3 of 1.000 M NaOH also at 20.38°C are mixed in a coffee-cup calorimeter similar to that shown in the Figure. The temperature of the mixture rises to 27. 80°C. The mixture is then cooled to 20.45°C after which 24.06 V is applied to the heating coil, allowing 2.13 A to flow for 300.0 s. The temperature rises to 28.23°C. Find ΔH, for the reaction
$\ce{H_{3}O^{+} (aq) + OH^{-} (aq) \rightarrow 2H_{2}O (l)} \nonumber$
Solution
We first calculate the heat capacity of the contents of the calorimeter. The heat supplied by the coil qcoil is given by
q2 = 24.06 V × 2.13 A × 300.0 s = 15 374 J
so that
$C=\frac{q_{\text{coil}}}{\Delta T}=\frac{\text{15 374 J}}{\text{(28}\text{.23}-\text{20}\text{.45)K}}=\text{1976 J K}^{-\text{1}} \nonumber$
Knowing the heat capacity, we can now calculate qcontents, the heat change in the contents of the experiment:
qcontents = C ΔT = 1976 J K–1 (27.80 – 20.38)K = 14.66 kJ
where the sign is positive since the contents have gained energy.
The system (the chemical reactants) released the same amount of energy, so it is
qsystem = - qcontents = -14.66 kJ
The reaction was exothermic.
Because the change occurs at constant pressure, we can equate qsystem to the enthalpy change ΔH. However, the question asks for the enthalpy change per mole of acid (that is, ΔHm), while only 250 cm3 × 1 mol dm–3 = 0.250 mol acid was used. Accordingly,
$\Delta H_{m}=\frac{-\text{14}\text{.66 kJ}}{\text{0}\text{.250 mol}}=\text{58}\text{.7 kJ mol}^{-\text{1}} \nonumber$
Constant Volume ("Bomb") Calorimetry
Even though most chemical reactions are more conveniently studied at constant pressure than at constant volume, there are some reactions for which the opposite is true. When one of the reactants is a gas, for instance, the reaction is much easier to carry out in a closed container called a bomb calorimeter. If a reaction is measured at constant volume, of course, we find ΔU rather than ΔH. However, almost all thermochemical data are recorded in terms of the enthalpy rather than the internal energy. It is therefore often necessary to convert a measured ΔU value to a ΔH value. This is done as shown in the following example:
Example $4$: Calculate ΔH
Use the data in Table 1 in the Internal Energy section to find ΔH for the reaction
$\ce{2O_{3}(g) \rightarrow 3O_2(g)} \nonumber$ 25°C, 1 bar
Solution
ΔH for a constant-pressure change is given by
$\Delta H = \Delta U + P \Delta V \nonumber$
Since ΔU is known from Table 1 in the Internal Energy section, we need only calculate P ΔV. The initial volume of the system V1 is that of 2 mol O3 at 25°C.
$V_{\text{1}}=\frac{nRT}{P}\text{ }=\text{2 mol }\frac{RT}{P} \nonumber$
Similarly, the final volume is that of 3 mol O2 at 25°C:
$V_{\text{2}}=\text{ 3 mol }\frac{RT}{P} \nonumber$
Thus
$\Delta V=V_{\text{2}}-V_{\text{1}}=\frac{RT}{P}\text{ (3}-\text{2)mol} \nonumber$
and
$P\text{ }\Delta V=P\left[ \frac{RT}{P}\text{ (3}-\text{2)mol} \right]=RT\text{(3}-\text{2)mol} \nonumber$
$\text{= 8}\text{.3143 }\frac{\text{J}}{\text{K mol}}\text{ }\times \text{ 298}\text{.15 K }\times \text{ 1 mol}=\text{2479 J}=\text{2}\text{.479 kJ} \nonumber$
Thus
$\Delta H = \Delta U + P \Delta V = - 287.9 \text{kJ} + 2.5 \text{kJ} = - 285.5 \text{kJ} \nonumber$
ΔH vs. ΔU
Perhaps because values of ΔH are usually easier to measure than values of ΔU, chemists have chosen to concentrate exclusively on ΔH rather than on ΔU as a way of recording thermochemical data. Though enthalpies of formation are easy to find, equivalent tables of internal energies are nonexistent. In many ways this insistence on ΔH rather than on ΔU is a pity. In particular, it suggests that somehow the enthalpy H has more fundamental significance on the molecular level than the internal energy U. It is important to realize that this is not the case. It is the internal energy which has a simple molecular interpretation, namely, the total energy of all the molecules in the system. By contrast the enthalpy includes not only the total energy of the molecules in the system but the potential energy of the atmosphere outside the system as well. We use the enthalpy so often because of its convenience rather than because of its molecular significance.
A further point worth making about the enthalpy is that the difference between ΔH and ΔU is not often of great chemical importance. This is particularly true of reactions which involve only gases, such as the decomposition of ozone. In a gaseous reaction the main factor determining both ΔU and ΔH is the change in electronic energy. Changes in molecular energy and also the expansion work P ΔV are usually small compared with this change. In the decomposition of ozone, for example, the change in electronic energy is – 290.7 kJ per 2 mol ozone. The value of ΔU is –287.9 kJ, while that of ΔH is –285.4 kJ. The three quantities are all within a few percent of each other. For many purposes, differences of this order of magnitude are immaterial. When this is the case, we can equate both ΔU and ΔH to the change in electronic energy.
15.07: Measuring the Enthalpy Change
Calorimeter Constant
To a styrofoam cup calorimeter containing 250 mL of water at 24.0 oC is added a known amount of heat as 250 mL of water from a second styrofoam cup at 32.0 oC. The final temperature, measured by a computer-interfaced thermistor, is 27.7 oC. The computer-generated plot of T vs. time should be projected[1]. What is the calorimeter constant?
ΔT calorimeter = 27.7 oC – 24.0 oC = 3.7 oC
ΔT cold = 27.5 oC – 24.0 oC = 3.7 oC
ΔThot = 27.7 – 32.0 = -4.3 oC
qhot = m x S.H. x ΔT = 250 g x 4.18 J/goC x -4.3oC = -4494 J
qcold = m x S.H. x ΔT = 250 x 4.18 x 3.7 oC = 3867 J
qcalorimeer + qhot + qcold = 0
qcalorimeter -4494 + 3867 = 0
qcalorimeter = 627 J.
C = Qcalorimeter/T = 627 J / 3.7 oC = 169 J/oC
The Ammonium Nitrate "Cold Pack"[2]
A styrofoam cup calorimeter contains 250 mL of water at 25.0oC. Solid NH4NO3 (5 g ) is added, and the temperature falls to 23.8oC. The computer-generated plot of T vs. time should be projected[3]. The calorimeter is found to absorb 169 J to change its temperature 1 oC, so it is said to have a calorimeter constant of 169 J/oC. What is the enthalpy change for the dissolution reaction?
qwater = m x S.H. x ΔT = 250 g x 4.18 J/goC x -1.2oC
= -1254 J
qcalorim = C x ΔT = 169 x -1.2oC = -203 J
q tot = -1254 + -203 = -1457 J
q rxn = +1457 J.
ΔH rxn = q (kJ) / n (mol)
n = m / M = 5 g NH4NO3 / 80 g/mol = 0.057 mol
ΔH rxn = q/n = 1.457 kJ / 0.057 mol = +25.6 kJ/mol
The reaction is spontaneous even though it is endothermic, because of the large positive entropy change resulting from water association with the separate ions in solution.
References
1. We used Vernier LoggerPro(R) software
2. J. Chem. Educ., 2004, 81 (1), p 64A
3. We used Vernier LoggerPro(R) software | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.07%3A_Measuring_the_Enthalpy_Change/15.7.01%3A_Lecture_Demonstrations.txt |
Both enthalpy and the internal energy are often described as state functions. This means that they depend only on the state of the system, i.e., on its pressure, temperature, composition, and amount of substance, but not on its previous history. Thus any solution of NaCl at 25°C and 1 bar (100 kPa) which contains a mixture of 1 mol NaCland 50 mol H2O has the same internal energy and the same enthalpy as any other solution with the same specifications. It does not matter whether the solution was prepared by simply dissolving NaCl(s) in H2O, by reacting NaOH(aq) with HCl(aq), or by some more exotic method.
The fact that the internal energy and the enthalpy are both state functions has an important corollary. It means that when a system undergoes any change whatever, then the alteration in its enthalpy (or its internal energy) depends only on the initial state of the system and its final state. The initial value of the enthalpy will be H1, and the final value will be H2. No matter what pathway we employ to get to state 2, we will always end up with the value H2 for the enthalpy. The enthalpy change ΔH = H2 H1 will thus be independent of the path used to travel from state 1 to state 2. This corollary is of course the basis of Hess' law. The change in enthalpy for a given chemical process is the same whether we produce that change in one or in several steps.
15.09: Standard Pressure
You will often find enthalpy changes indicated by Δ and called standard enthalpy changes. The superscript is added to indicate that the enthalpy change has occurred at the standard pressure of 1 bar (100 kPa or 0.9869 atm). Unless very high pressures are involved, ΔH changes very little with a change in pressure, and so we have ignored superscripts up to this point. However, two other properties of matter called the entropy, symbol S, and the free energy, symbol G, are discussed in other sections. These are quite sensitive to pressure, and the inclusion of this superscript is important. For reasons of consistency therefore we will indicate standard enthalpy changes as Δ from now on.
15.10: Bond Enthalpies
The heat changes which accompany a chemical reaction are caused largely by changes in the electronic energy of the molecules. If we restrict our attention to gases, and hence to fairly simple molecules, we can go quite a long way toward predicting whether a reaction will be exothermic by considering the bonds which are broken and made in the course of the reaction. In order to do this we must first become familiar with the idea of a bond enthalpy.
In other sections we point out that when a chemical bond forms, negative charges move closer to positive charges than before, and so there is a lowering of the energy of the molecule relative to the atoms from which it was made. This means that energy is required to break a molecule into its constituent atoms. The bond enthalpy DX–Y of a diatomic molecule X—Y is the enthalpy change for the (usually hypothetical) process:
$\ce{XY(g) \rightarrow X(g) + Y(g)} \nonumber$ $\Delta H^{o} (298 K) = D_{x-y} \nonumber$
We have already used the term bond energy to describe this quantity, though strictly speaking the bond energy is a measure of ΔU rather than ΔH. As we have already seen, ΔU and ΔH are nearly equal, and so either term may be used.
As an example, let us consider the bond enthalpy for carbon monoxide. It is possible to establish the thermochemical equation
$\ce{CO(g) \rightarrow C(g) + O(g)} \nonumber$
$\Delta H^{o}(298 K) = 1073 kJ mol^{-1}\label{3}$
Accordingly we can write
$\ce{C_{C\equiv O}= 1073 kJ mol^{-1}} \nonumber$
even though the process to which Eq. $\ref{3}$ corresponds is hypothetical: Neither carbon nor oxygen exists as a monatomic gas at 298 K. For triatomic and polyatomic molecules, the bond enthalpy is usually defined as a mean. In the case of water, for instance, we have
$\ce{H_{2}O(g) \rightarrow 2H(g) + O(g)} \nonumber$
$\Delta H^{o} (298 K) = 927.2 \text{kJ mol}^{-1} \nonumber$
Since it requires 927.2 kJ to break open two O—H bonds, we take half this value as the mean bond enthalpy and write
$D_{O-H} = 463.6 \text{kJ mol}^{-1} \nonumber$
In methanol, CH3OH,however, a value of 427 kJ mol–1 for the O—H bond enthalpy fits the experimental data better. In other words the strength of the O—H varies somewhat from compound to compound. Because of this fact, we must expect to obtain only approximate results, accurate only to about ± 50 kJ mol–1, from the use of bond enthalpies. Bond enthalpies for both single and multiple bonds are given in Table $1$.
TABLE $1$ Average Bond Energies/kJ mol–1.
As an example of how a table of bond enthalpies can he used to predict the ΔH value for a reaction, let us take the simple case
$\ce{H_{2}(g) + F_{2}(g) \rightarrow 2HF(g)}\label{9}$
298 K, 1 atm
We can regard this reaction as occurring in two stages (Figure $1$ ). In the first stage all the reactant molecules are broken up into atoms:
$\ce{H_{2}(g) + F_{2}(g) \rightarrow 2H(g) + F(g)}\label{10}$ 298 K, 1 atm
For this stage
$\Delta H_I = H_{H-H}+ D_{F-F}\label{11}$
since 1 mol H2 and 1 mol F2 have been dissociated.
In the second stage the H and F atoms are reconstituted to form HF molecules:
$\ce{2H(g) + 2F(g) \rightarrow 2HF(g)} \nonumber$ 298 K, 1 atm
For which
$\Delta H_{II} = – 2D_{H-F} \nonumber$
where a negative sign is necessary since this stage corresponds to the reverse of dissociation.
Since Eq. $\ref{9}$ corresponds to the sum of Equations $\ref{10}$ and $\ref{11}$, Hess's law allows us to add ΔH values:
$\Delta H^{o}_{reaction} = \Delta H_{I} + \Delta H_{II} \nonumber$
$= D_{H-H} + D_{F-F} – 2D{H-F}$
$= (436 + 159 – 2 * 566) \text{kJ mol}^{-1}$
$= –539 \text{kJ mol}^{-1}$
We can work this same trick of subdividing a reaction into a bond-breaking stage followed by a bond-making stage for the general case of any gaseous reaction. In the first stage all the bonds joining the atoms in the reactant molecules are broken and a set of gaseous atoms results. For this stage
$\Delta H_{I} = \sum_{\text{bonds broken}} D \nonumber$
The enthalpy change is the sum of the bond enthalpies for all bonds broken. In the second stage these gaseous atoms are reconstituted into the product molecules. For this second stage therefore
$\Delta H_{II} = – \sum_{\text{bonds formed}} D \nonumber$
where the negative sign is necessary because the reverse of bond breaking is occurring in this stage. The total enthalpy change for the reaction at standard pressure is thus
$\Delta H^{o} = \Delta H_{I} + \Delta H_{II} \nonumber$
or
$\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)} \nonumber$
The use of this equation is illustrated in the next example.
Example $1$: Enthalpy Change
Using Table $1$ calculate the value of ΔH°(298 K) for the reaction
$\ce{CH_{4}(g) + 2O_{2}(g) \rightarrow CO_{2}(g) + 2H_{2}O(g)} \nonumber$
Solution
It is best to sketch the molecules and their bonds in order to make sure that none are missed.
Thus $\Delta H^{o} = \sum D \text{(bond broken)} – \sum D \text{(bond formed)}$
$= 4 D_{C\bond{-}H} + 2 D_{D\bond{=}D} - 2 D_{C\bond{=}O} - 4 D_{O\bond{-}H}$
$= (4 * 416 + 2 * 498 – 2 * 803 – 4 * 467) \text{kJ mol}^{-1}$
$= – 814 \text{kJ mol}^{-1}$
The experimental value for this enthalpy change can be calculated from standard enthalpies of formation. It is –802.4 kJ mol–1. The discrepancy is due to the unavoidable use of mean bond enthalpies in the calculation. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.08%3A_State_Functions.txt |
We are in a position to appreciate the general principle which determines whether a gaseous reaction will be exothermic or endothermic. If less energy is needed to break up the reactant molecules into their constituent atoms than is released when these atoms are reconstituted into product molecules, then the reaction will be exothermic. Usually an exothermic reaction corresponds to the breaking of weak bonds (with small bond enthalpies) and the making of strong bonds (with large bond enthalpies). In the hydrogen-fluorine reaction (Figure 1 from Bond Enthalpies) one quite strong bond (DH–H = 436 kJ mol–1) and one very weak bond (DF–F = 159 kJ mol–1) are broken, while two very strong H―F bonds are made. (Note that with a bond enthalpy of 566 kJ mol–1, H―F is the strongest of all single bonds.) The combustion of methane discussed in Example 1 from Bond Enthalpies is another example of the formation of stronger bonds at the expense of weaker ones. The bond enthalpy of the O―H bond is not much different in magnitude from those of the C―H and O=O bonds which it replaces: All lie between 400 and 500 kJ mol–1. The determining factor making this reaction exothermic is the exceedingly large bond enthalpy of the C=O bond which at 803 kJ mol–1 is almost twice as great as for the other bonds involved in the reaction. Not only this reaction but virtually all reactions in which CO2 with its two very strong C=O bonds is produced are exothermic.
In other cases the number of bonds broken or formed can be important. A nice example of this is the highly exothermic (ΔH°(298 K) = –483.7 kJ mol–1) reaction between hydrogen and oxygen to form water:
$\ce{2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(g)} \nonumber$
All three types of bonds involved have comparable bond enthalpies:
$D_{H{-}H} = 436 \text{kJ mol}^{-1}$ $D_{O{=}O} = 498 \text{kJ mol}^{-1}$ $D_{O{-}H} = 467 \text{kJ mol}^{-1}$
but the reason the reaction is exothermic becomes obvious if we rewrite it to make the bonds visible:
While three bonds must be broken (two H―H and one O=O bond), a total of four bonds are made (four O―H bonds). Since all the bonds are similar in strength, making more bonds than are broken means the release of energy. In mathematical terms
$\triangle H^{o} = 2D_{H{-}H} + D_{O{=}O} + 4D_{O{-}O} \nonumber$
$= 2 * 436 \text{kJ mol}^{-1} + 1 * 498 \text{kJ mol}^{-1} – 4 * 467 \text{kJ mol}^{-1} \nonumber$
$= – 498 \text{kJ mol}^{-1} \nonumber$
In summary, there are two factors which determine whether a gaseous reaction will be exothermic or not: (1) the relative strengths of the bonds as measured by the bond enthalpies, and (2) the relative number of bonds broken and formed. An exothermic reaction corresponds to the formation of more bonds, stronger bonds, or both.
Since the strength of chemical bonds is a factor in determining whether a reaction will release energy or not, it is obviously important to know which kinds of bonds will be strong and which weak, and we can make some empirical generalizations about the magnitudes of bond enthalpies. The first and most obvious of these is that triple bonds are stronger than double bonds which in turn are stronger than single bonds. As can be seen from Table 1 in Bond Enthalpies, triple bonds have bond enthalpies in the range of 800 to 1000 kJ mol–1. Double bonds range between 400 and 800 kJ mol–1 and single bonds are in the range of 150 to 500 kJ mol–1.
A second generalization is that the strengths of bonds usually increase with polarity. The bond enthalpies of the hydrogen halides, for instance, increase in the order HI < HBr < HCl < HF, and a similar order can be noted for bonds between carbon and halogens. There are exceptions to this rule, though. One would expect the C―N bond to be intermediate in strength between the C―C bond and the C―O bond. As the table shows, it is actually weaker than either.
A third factor affecting the strength of bonds is the size of the atoms. For the most part smaller atoms form stronger bonds. The smallest atom, hydrogen, forms four of the five strongest single bonds in the table. This is not entirely a matter of size since hydrogen is also the most electropositive element featured. Difference in electronegativity, however, does not explain why the H―H bond enthalpy is so large. If we look at the halogens (VIIA elements), we find that the bond enthalpies of I―I, Br―Br, and Cl―Cl increase as expected with decreasing size. The F―F bond is an exception, though: DF―F (158 kJ mol–1) is significantly smaller than DCl―Cl (242 kJ mol–1). Other notable exceptions to this rule are the N―N and O―O bonds. The occurrence of these weak single bonds is of considerable importance to the chemistry of compounds which contain them.The value of a particular bond enthalpy in a given molecule can sometimes be very informative about the nature of the bonding in the molecule. This is especially true of molecules in which resonance is a possibility, as the following example shows.
Example $1$: Bond Enthalpy
When benzene is burned in oxygen according to the equation
$\ce{C_{6}H_{6}(g) + \frac{15}{2} O_{2}(g) \rightarrow 6CO_{2}(g) + 3H_{2}O(g)} \nonumber$ calorimetric measurements give the value of [ΔH°(298 K) as –3169 kJ mol–1 benzene. Use this information together with Table 1 from the Bond Enthalpies section to calculate the mean bond enthalpy for the carbon-carbon bond in benzene.
Solution Indicating the required bond enthalpy by the symbol DCC and carefully counting the bonds broken and bonds formed, we have
$\triangle H^{o} = 6D_{C{-}H} + 6D_{CC} + \frac{15}{2} D_{O{=}O} – C_{C{=}O} + 6D_{O{-}H} \nonumber$ or
$–3169 \text{kJ mol}^{-1} = 6 (416 \text{kJ mol}^{-1}) + 6D_{CC} + \frac{15}{2} (498 \text{kJ mol}^{-1}) – 12 (803 \text{mol}^{-1}) – 6 (467 \text{kJ mol}^{-1}) = –6207 \text{kJ mol}^{-1} + 6D_{CC} \nonumber$
Thus $6D_{CC} = (–3169 + 6207) \text{kJ mol}^{-1} = 3038 \text{kJ mol}^{-1} \nonumber$
or $D_{CC} = 506 \text{kJ mol}^{-1} \nonumber$
As expected for a resonance structure the bond-enthalpy value is intermediate between that given in the table for a C―C single bond (348 kJ mol–1) and that given for a C=C double bond (614 kJ mol–1). | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.11%3A_Bond_Enthalpies_and_Exothermic_or_Endothermic_Reactions.txt |
A chemical fuel is any substance which will react exothermically with atmospheric oxygen, is available at reasonable cost and quantity, and produces environmentally acceptable reaction products. During the past century the most important sources of heat energy in the United States and other industrialized countries have been the fossil fuels: coal, petroleum, and natural gas. In 1971, for example, the United States obtained 43.5 percent of its energy from oil, 34.7 percent from natural gas, and 19.7 percent from coal. Only 1.5 percent was obtained from hydroelectric power and 0.6 percent from nuclear power. Other industrialized countries also obtain 95 percent or more of their energy from fossil fuels.
Coal, petroleum, and natural gas consist primarily of carbon and hydrogen, and so it is not hard to see why they make excellent fuels. When they burn in air, the principal products are water and carbon dioxide, compounds which contain the strongest double bond (DC=O = 804 kJ mol–1) and the third-strongest single bond (DO―H = 463 kJ mol–1) in Table 1 from Bond Enthalpies. Thus more energy is liberated by bond formation than is needed to break the weaker C―C and C―H bonds in the fuel.
Example $1$: Enthalpy Change
Use the bond enthalpies given in Table 1 from Bond Enthalpies to estimate the enthalpy change when 1 mol heptane, C7H16, is burned completely in oxygen:
$\ce{C_{7}H_{16}(g) + 11O_{2} \rightarrow 7CO_{2} + 8H_{2}O} \nonumber$
Solution
remembering that the projection formula for heptane is
we can make up the following list of bonds broken and formed:
Bonds Broken Bonds Formed
6 C―C
2 088 kJ mol–1
14 C=O
–11 256 kJ mol–1
16 C―H
6 608 kJ mol–1
16 H―O
– 7 408 kJ mol–1
11 O=O
5 478 kJ mol–1
_______________
Total
14 174 kJ mol–1
Total
–18 664 kJ mol–1
Thus
$\triangle H = 14 174 \text{kJ mol}^{-1} – 18 664 \text{kJ mol}^{-1} = –4490 \text{kJ mol}^{-1} \nonumber$
Apart from the hydrocarbon compounds in fossil fuels, there are few substances which fulfill the criteria for a good fuel. One example is hydrogen gas:
$\ce{2H_{2}(g) + O_{2}(g) \rightarrow 2H_{2}O(l)} \nonumber$
with $\triangle H^{o} (298 K) = –571.7 \text{kJ mol}^{-1}$.
Hydrogen does not occur as the element at the surface of the earth, however, so it must be manufactured. Right now much of it is made as a by-product of petroleum refining, and so hydrogen will certainly not be an immediate panacea for our current petroleum shortage. Eventually, though, it may be possible to generate hydrogen economically by electrolysis of water with current provided by nuclear power plants, and so this fuel does merit consideration.
15.13: Photosynthesis
The adjective fossil describes the fossil fuels very aptly, because all of them are derived from the remains of plants or animals which lived on earth millions of years ago. Coal, for example, began as plant matter in prehistoric swamps, where it was able to decompose in the absence of air. Present-day peat bogs are examples of this first stage in coal formation, and in countries such as Ireland dried peat is an important fuel. Over long periods of time, at high temperatures and pressures under the earth’s surface, peat can be transformed into lignite, a brown, soft form of coal. Continued action of geological forces converts lignite into bituminous, or soft coal, and eventually into anthracite, or hard coal. When burned, these latter two types of coal release considerably more heat per unit mass than do lignite or peat.
A crucial point to realize about fossil fuels is that the energy we release by burning them came originally from the sun. The plants from which the fuels were derived grew as a result of photosynthesis, the combination of carbon dioxide and water under the influence of sunlight to form organic compounds whose empirical formula is approximately
$\ce{CO_{2}(g) + H_{2}O(l) \rightarrow [CH_{2}O](s) + O_{2}(g)} \label{1}$
with $\Delta H \approx 469 \text{kJ mol}^{-1}$
Since a number of different substances are formed by photosynthesis, the empirical formula [CH2O] and the ΔH are only approximate.
Photosynthesis is endothermic, and the necessary energy is supplied by the absorption of solar radiant energy. This energy can be released by carrying out the reverse of Equation \ref{1}, a process which is exothermic. When we burn paper, wood, or dried leaves, the heat given off is really a stored form of sunlight. Plants and animals obtain the energy they need to grow or move about from the oxidation of substances produced by photosynthesis. This oxidation process is called respiration.
After millions of years of geological change, the fossil fuels are significantly different in chemical structure from newly photosynthesized plant or animal material. The changes which occur can be approximated by the equation for formation of methane (natural gas):
$\ce{2[CH_{2}O] \rightarrow CH_{4}(g) + CO_{2}(g)} \nonumber$
with $\Delta H \approx –47 \text{kJ mol}^{-1}$.
This reaction is only slightly exothermic, and so very little of the energy captured from sunlight is lost. However, about half the carbon and all the oxygen are lost as carbon dioxide gas, and so a fossil fuel like methane can release more heat per carbon atom (and per gram) than can wood or other organic materials. This is why anthracite and bituminous coals are better fuels than the peat from which they are formed. The enthalpy changes which occur during photosynthesis, respiration, and formation and combustion of fossil fuels are summarized in Figure $1$. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/15%3A_Thermodynamics-_Atoms_Molecules_and_Energy/15.12%3A_Fossil_Fuels_and_the_Energy_Crisis.txt |
The experiences you have had in the chemical laboratory have probably already taught you that there is an uphill character to some reactions and a downhill character to others. We want to understand why some chemical reactions are downhill while others are uphill. In particular we want to know what kinds of atomic and molecular processes proceed downhill, and which processes require a “push” to go uphill. We find that these questions of uphill and downhill nature of a reaction can be answered using concepts such as spontaneity, entropy, and free energy. Chemical reactions which proceed downhill, are said to be spontaneous.
16: Entropy and Spontaneous Reactions
The experiences you have had in the chemical laboratory have probably already taught you that there is an uphill character to some reactions and a downhill character to others. A simple example is the combination of mercury with bromine which we considered in detail in the section on macroscopic and microscopic views of a chemical reaction:
$\ce{Hg(l) +Br2(l)→HgBr2(s)} \nonumber$
This reaction proceeds from Hg(l) and Br2(l) to HgBr2(s), much as a ball rolls downhill. On the other hand, decomposition of HgBr2 to the elements is an uphill process, and we must “push” this reaction to force it to occur. One way to do this is to dissolve HgBr2 in water and pass an electrical current through the solution. Mercury will appear at one electrode and bromine at the other.
We want to understand why some chemical reactions are downhill while others are uphill. In particular we want to know what kinds of atomic and molecular processes proceed downhill, and which processes require a “push” to go uphill. We find that these questions of uphill and downhill nature of a reaction can be answered using concepts such as spontaneity, entropy, and free energy.
Chemical reactions which proceed downhill, are said to be spontaneous. With spontaneity, we are considering whether a process will occur, but can say nothing about the rate at which it will occur. Spontaneous process may also require energy to overcome an activation barrier. This means that some processes may be spontaneous, but do not occur at a noticeable rate. An everyday example is burning a log of wood where the burning reaction is clearly "downhill" and thus defined spontaneous, but without a flame to supply heat to overcome the activation barrier, the reaction will not occur. A spontaneous process corresponds to a rearrangement of atoms and molecules from a less probable to a more probable situation, as measured by the thermodynamic probability W of the reactants and products. W is defined as the number of alternative microscopic arrangements which correspond to the same macroscopic state. Because most macroscopic samples of matter contain 1015 particles or more, very large values of W are encountered. Therefore it is convenient to use the entropy S, which is proportional to the logarithm of W, as a measure of spontaneity. According to the second law of thermodynamics, when a spontaneous process occurs, there must be an increase in total entropy.
The entropy of a perfectly ordered crystal at absolute zero is zero according to the third law of thermodynamics. Increasing the temperature, volume, or amount of substance increases the entropy. For a given amount of substance, the heavier and more complex the molecules, the greater the entropy, while stronger forces between atoms, molecules, or ions result in lower entropy. In general the greater the randomness or disorder of the atoms or molecules the greater the entropy of a substance.
To determine whether a chemical reaction is spontaneous we must calculate the change in total entropy of the chemical system and its surroundings. This can be done directly using ΔS°sys and –ΔH°/T or indirectly using the change in Gibbs free energy ΔG°. For a spontaneous reaction occurring at constant temperature and pressure, ΔG° must be negative. Since ΔG° = ΔH° – T ΔS°, a negative enthalpy change is the most important factor governing spontaneity of a reaction at low temperatures. At high values of T, –T ΔS° becomes large and an increase in entropy of the system is essential for a spontaneous process.
The change in Gibbs free energy at standard pressure can be calculated from tables of ΔGf°. Once obtained it can be used to determine two other useful values, the either the amount of energy available to do work wmax(or the amount of energy needed to drive a process in a non-spontaneous direction), and the standard equilibrium constant K°. For a reaction in the gas phase K° is numerically the same as Kp, provided the latter is expressed in atmospheres. Knowledge of ΔG° thus allows calculation of the equilibrium partial pressures of reactants and products.
16.02: Spontaneous Processes and Molecular Probability
Processes like formation of HgBr2, which have a downhill character, are called spontaneous processes. When we attempt to reverse a spontaneous process, as in decomposition of HgBr2, an uphill battle must invariably be fought and we are dealing with a nonspontaneous process
Below are two videos of a process, one going forward in time, and the other rewinding back through time.
It is clear that the first video of nitrogen triiodide exploding at the touch of a feather is going forward in time. It is also as immediately obvious that the second video, of nitrogen triiodide "unexploding" is going backward. We know instinctively from everyday experience that such a process would be unnatural, and simply would not occur. Exploding is not the only process for which we have this basic intuition on. Below is another video going backward in time to show a processes we know is unnatural and would not occur spontaneously in real life:
In the above video, a dye is "unmixed." The dye goes from being fully dispersed in solution to returning back to a concentrated drop separate from the rest of the solution. The mixing of the dye is spontaneous, we expect it to happen, so we know that the video must be in reverse, showing a process we intuitively know is unnatural.
These two examples show processes we know from basic world experience go in one direction, and do not go in the opposite (reverse) direction. There must be an underlying reason for one direction being spontaneous while the other is not. Understanding this reason will allow one to determine which direction a process will go when it is less obvious than the explosion in the first example. Following the chemist's view of the world, we can understand what is occurring macroscopically by considering what is occurring with molecules and atoms at the microscopic level | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.01%3A_Prelude_to_Spontaneity.txt |
On a microscopic level we can easily explain why some processes occur of their own accord while others do not. A spontaneous process corresponds to rearrangement of atoms and molecules from a less-probable situation to a more-probable one. A nonspontaneous process, by contrast, corresponds to movement from a probable situation to an improbable one.
An example of what probability has to do with spontaneity is provided by expansion of a gas into a vacuum. Let us calculate the probability that the process of gas expansion from flask A into a connected flask B will reverse itself, that is, the probability that the gas molecules will all collect again in flask A. If we choose a particular molecule and label it number 1, we find that it is sometimes in flask A and sometimes in flask B. Since the molecule’s motion is random and the two flasks contain the same volume, the molecule should spend half its time in each container. The probability of finding molecule 1 in container A is therefore 1/2.
Next let us consider the probability that two molecules, labeled 1 and 2, are both in flask A. Figure $1$ shows the four possible ways these two molecules can be arranged in the two flasks. All four are equally likely, but only one has both molecules in flask A. Thus there is one chance in four that molecules 1 and 2 are both in flask A. This probability of 1/4 equals 1/2 × 1/2; i.e., it is the product of the probability that molecule 1 was in flask A times the probability that molecule 2 was in flask A. By a similar argument we can show that the probability that three given molecules are all in flask A is 1/2 × 1/2 × 1/2 = (1/2)3=1/8, and, in general, that the probability of all N gas molecules being in flask A at once is (1/2)N.
If we had 1 mol gas in the flasks, there would be 6.022 × 1023 molecules. The probability p that all of them would be in flask A at the same time would be
$p=\left( \frac{\text{1}}{\text{2}} \right)^{\text{6}\text{.022 }\times \text{ 10}^{\text{23}}}=\frac{\text{1}}{\text{2}^{\text{6}\text{.022 }\times \text{ 10}^{\text{23}}}}=\frac{\text{1}}{\text{10}^{\text{1}\text{.813 }\times \text{ 10}^{\text{23}}}} \nonumber$ This unimaginably small number could be written as 0.000 000 . . . , where there would be 1.813 × 1023 zeros and then a 1. It would take over a thousand million million years to write that many zeros! Because there are so many molecules in a mole of gas (or any other macroscopic quantity), the probability that the spontaneous expansion will reverse itself is inconceivably small. The reversal is so improbable as to be impossible in any practical sense.
Similar remarks apply to the probability of reversing other spontaneous processes. In Figure $2$a, some of the atoms in a bar of metal at uniform temperature will be vibrating more than others, but the unusually energetic atoms will be distributed fairly evenly throughout the bar.
We will not suddenly find all the energetic metal atoms on the left end of the bar and all the weakly vibrating ones on the right, as in Figure $2$b. The possibility exists that a freak series of collisions between vibrating atoms might produce a high concentration of energetic atoms on the left, but such an occurrence is inconceivably improbable.
When a falling book hits the floor, its kinetic energy is converted to heat energy. The floor warms up slightly, and the molecules there start vibrating a little more energetically. For such a process to reverse itself spontaneously, all the floor molecules under the book would suddenly have to become more energetic and vibrate in unison in a vertical direction, flinging the book into the air. As in the previous two examples, this would require a freak series of molecular collisions which is so improbable as never to occur in the entire lifetime of the universe.
These principles also apply to the processes considered in the section on spontaneous processes. In the NI3 reaction, all of the particles distributed in the explosion would all have to return back to the ring stand and reassemble themselves. Even leaving the chemical reaction unconsidered for the moment, this return of all the particles is highly improbable.
A similar argument can be put forward for the un-mixing of the dye. The probability that all of the dye molecules will return back to a single uniform drop separate from the water at one moment is another situation so improbable that it will never happen.
Simple cases we have described show how spontaneous and nonspontaneous processes can be considered from a microscopic and statistical viewpoint. In any real sample of matter there are a great many molecules jostling each other about, exchanging energy, and sometimes exchanging atoms. This constant jostle is like shuffling a gigantic deck of cards. Because the numbers involved are so large, the laws of probability are inexorable. Some probabilities are large enough to be virtual certainties, while others are small enough to be unthinkable. Invariably the reversal of a spontaneous process turns out to involve movement from an almost certain situation to one which is unimaginably improbable. Conversely, a spontaneous process occurs when a sample of matter finds itself momentarily in a highly improbable situation. As fast as possible, it will adjust on the molecular level until maximum probability is attained.
16.04: Rates of Spontaneous Processes
The phrase as fast as possible points up a major difficulty in dealing with spontaneous processes. Some of them occur quite rapidly, but others are so slow as to be imperceptible. A rapid spontaneous process occurs when 2 mol H2O is mixed with 2 mol “heavy water,” D2O, made from the isotope deuterium, ${}_{\text{1}}^{\text{2}}\text{H}$, or D. The two species start to transfer protons and deuterons (D+ ions) as soon as they are stirred together, and we rapidly obtain a mixture consisting of 2 mol H—O—D and 1 mol each of H—O—H and D—O—D. Assuming that deuterium atoms behave the same chemically as ordinary hydrogen atoms, this is what the laws of probability would predict. There are four equally likely possibilities for a randomly selected water molecule:
Two of the four possibilities have the molecular formula HDO, and so the probability of finding an HDO molecule in our mixture is 1/2. Half the molecules (2 mol) will be HDO. Similarly 1/4 of the 4 mol water will be H2O and 1/4 will be D2O.
The shift from the improbable situation of 2 mol H2O + 2 mol D2O to the more probable 2 mol HDO + 1 mol H2O + 1 mol D2O occurs rapidly because of the ease with which protons and deuterons can transfer from one water molecule to another. When such a shuffling process is slow, however, the situation is quite different. For example, we would expect that mixing 2 mol H2 with 2 mol D2 would produce 2 mol HD and a mole each of H2 and D2. At room temperature, though, nothing happens, even over a period of days, because there is no easy way for H or D atoms to swap partners. Reshuffling requires breaking an H—H or a D—D bond, and this takes some 400 kJ mol–1. The molecules are stuck in a situation of low probability because there is no pathway by which they can attain higher probability. If such a pathway is provided, by raising the temperature or adding a catalyst, the molecules start exchanging H and D and move toward the most probable situation.
The moral of this story is that saying a reaction is spontaneous is not the same as saying it will occur if the reactants are mixed. Rather, it means the reaction can occur but may be so slow that nothing seems to happen. In the case of a slow spontaneous reaction it is worthwhile to look for a catalyst, but if we know the reaction is nonspontaneous, there is no point in even mixing the reactants, let alone searching for a catalyst. A nonspontaneous reaction cannot occur of itself without outside intervention. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.03%3A_Atoms_Molecules_and_Probability.txt |
The section on atoms, molecules and probability has shown that if we want to predict whether a chemical change is spontaneous or not, we must find some general way of determining whether the final state is more probable than the initial. This can be done using a number W, called the thermodynamic probability. W is defined as the number of alternative microscopic arrangements which correspond to the same macroscopic state. The significance of this definition becomes more apparent once we have considered a few examples.
Figure $1$ a illustrates a crystal consisting of only eight atoms at the absolute zero of temperature. Suppose that the temperature is raised slightly by supplying just enough energy to set one of the atoms in the crystal vibrating. There are eight possible ways of doing this, since we could supply the energy to any one of the eight atoms. All eight possibilities are shown in Figure $1$ b.
Since all eight possibilities correspond to the crystal having the same temperature, we say that W = 8 for the crystal at this temperature. Also, we must realize that the crystal will not stay perpetually in any of these eight arrangements. Energy will constantly be transferred from one atom to the other, so that all the eight arrangements are equally probable.
Let us now supply a second quantity of energy exactly equal to the first, so that there is just enough to start two molecules vibrating. There are 36 different ways in which this energy can be assigned to the eight atoms (Figure $1$ c). We say that W = 36 for the crystal at this second temperature. Because energy continually exchanges from one atom to another, there is an equal probability of finding the crystal in any of the 36 possible arrangements.
A third example of W is our eight-atom crystal at the absolute zero of temperature. Since there is no energy to be exchanged from atom to atom, only one arrangement is possible, and W = 1. This is true not only for this hypothetical crystal, but also presumably for a real crystal containing a large number of atoms, perfectly arranged, at absolute zero.
The thermodynamic probability W enables us to decide how much more probable certain situations are than others. Consider the flow of heat from crystal A to crystal B, as shown in Figure $2$. We shall assume that each crystal contains 100 atoms. Initially crystal B is at absolute zero. Crystal A is at a higher temperature and contains 64 units of energy-enough to set 64 of the atoms vibrating. If the two crystals are brought together, the molecules of A lose energy while those of B gain energy until the 64 units of energy are evenly distributed between both crystals.
In the initial state the 64 units of energy are distributed among 100 atoms. Calculations show that there are 1.0 × 1044 alternative ways of making this distribution. Thus W1, initial thermodynamic probability, is 1.0× 1044. The 100 atoms of crystal A continually exchange energy among themselves and transfer from one of these 1.0 × 1044 arrangements to another in rapid succession. At any instant there is an equal probability of finding the crystal in any of the 1.0 × 1044 arrangements.
When the two crystals are brought into contact, the energy can distribute itself over twice as many atoms. The number of possible arrangements rises enormously, and W2, the thermodynamic probability for this new situation, is 3.6 × 1060. In the constant reshuffle of energy among the 200 atoms, each of these 3.6 × 1060 arrangements will occur with equal probability. However, only 1.0 × 1044 of them correspond to all the energy being in crystal A. Therefore the probability of the heat flow reversing itself and all the energy returning to crystal A is
$\frac{W_{\text{1}}}{W_{\text{2}}}=\frac{\text{1}\text{.0 }\times 10^{\text{44}}}{\text{3}\text{.6 }\times \text{ 10}^{\text{60}}}=\text{2}\text{.8 }\times \text{ 10}^{-\text{17}} \nonumber$ In other words the ratio of W1 to W2 gives us the relative probability of finding the system in its initial rather than its final state.
This example shows how we can use W as a general criterion for deciding whether a reaction is spontaneous or not. Movement from a less probable to a more probable molecular situation corresponds to movement from a state in which W is smaller to a state where W is larger. In other words W increases for a spontaneous change. If we can find some way of calculating or measuring the initial and final values of W, the problem of deciding in advance whether a reaction will be spontaneous or not is solved. If W2 is greater than W1, then the reaction will occur of its own accord. Although there is nothing wrong in principle with this approach to spontaneous processes, in practice it turns out to be very cumbersome. For real samples of matter (as opposed to 200 atoms in the example of Figure 2) the values of W are on the order of 101024—so large that they are difficult to manipulate. The logarithm of W, however, is only on the order of 1024, since log 10x = x. This is more manageable, and chemists and physicists use a quantity called the entropy which is proportional to the logarithm of W.
This way of handling the extremely large thermodynamic probabilities encountered in real systems was first suggested in 1877 by the Austrian physicist Ludwig Boltzmann (1844 to 1906). The equation
$S=k \text{ ln }W \label{2}$ is now engraved on Boltzmann’s tomb. The proportionality constant k is called, appropriately enough, the Boltzmann constant. It corresponds to the gas constant R divided by the Avogadro constant NA: $k=\frac{R}{N_{\text{A}}} \label{3}$ and we can regard it as the gas constant per molecule rather than per mole. In SI units, the Boltzmann constant k has the value 1.3805 × 10–23 J K–1. The symbol ln in Eq. $\ref{2}$ indicates a natural logarithm,i.e., a logarithm taken to the base e. Since base 10 logarithms and base e logarithms are related by the formula
$\text{ln } x = 2.303 \text{ log } x \nonumber$
it is easy to convert from one to the other. Equation $\ref2$, expressed in base 10 logarithms, thus becomes $S=2.303k \text{ log }W \nonumber$
Example $1$: Entropy
The thermodynamic probability W for 1 mol propane gas at 500 K and 101.3 kPa has the value 101025. Calculate the entropy of the gas under these conditions.
Solution Since
$W = 10 ^ {10^{25}}$
$\text{log } W = 10^{25}$
Thus $S = 2.303k \text{ log } W = 1.3805 \times 10^{-23} \text {J K}^{-1} \times 2.303 \times 10^{25} = 318 \text{J K}^{-1}$
Note: The quantity 318 J K–1 is obviously much easier to handle than 101025.
Note also that the dimensions of entropy are energy/temperature.
One of the properties of logarithms is that if we increase a number, we also increase the value of its logarithm. It follows therefore that if the thermodynamic probability W of a system increases, its entropy S must increase too. Further, since W always increases in a spontaneous change, it follows that S must also increase in such a change.
The statement that the entropy increases when a spontaneous change occurs is called the second law of thermodynamics. (The first law is the law of conservation of energy.) The second law, as it is usually called, is one of the most fundamental and most widely used of scientific laws. In this book we shall only be able to explore some of its chemical implications, but it is of importance also in the fields of physics, engineering, astronomy, and biology. Almost all environmental problems involve the second law. Whenever pollution increases, for instance, we can be sure that the entropy is increasing along with it.
The second law is often stated in terms of an entropy difference ΔS. If the entropy increases from an initial value of S1 to a final value of S2 as the result of a spontaneous change, then
$\Delta S = S_{2} - S_{1} \label{4}$
Since S2 is larger than S1, we can write $\Delta S >0 \label{5}$ Equation $\ref{5}$ tells us that for any spontaneous process, ΔS is greater than zero. As an example of this relationship and of the possibility of calculating an entropy change, let us find ΔS for the case of 1 mol of gas expanding into a vacuum. We have already argued for this process that the final state is 101.813 × 1023 times more probable than the initial state. This can only be because there are 101.813 × 1023 times more ways of achieving the final state than the initial state. In other words, taking logs, we have $\text{log } \frac{W_{\text{2}}}{W_{\text{1}}} = 1.813 \times 10^{23} \nonumber$ Thus
\begin{align} \Delta S=S_{2}-S_{1} & =2.303\times k\times \text{ log }W_{2}-2.303\times k\times \text{ log }W_{1} \ & = 2.303 \times k \times \text{ log } \frac{W_{\text{2}}}{W_{\text{1}}} \ & = 2.303 \times 1.3805 \times 10^{-23} \text{ J K}^{-1} \times 1.813 \times 10^{23} \end{align} \nonumber
$S = 5.76 \text{J K}^{-1} \nonumber$ As entropy changes go, this increase in entropy is quite small. Nevertheless, it corresponds to a gargantuan change in probabilities. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.05%3A_Thermodynamic_Probability_W_and_Entropy.txt |
You can experience directly the mass, volume, or temperature of a substance, but you cannot experience its entropy. Consequently you may have the feeling that entropy is somehow less real than other properties of matter. We hope to show in this section that it is quite easy to predict whether the entropy under one set of circumstances will be larger than under another set of circumstances, and also to explain why. With a little practice in making such predictions in simple cases you will acquire an intuitive feel for entropy and it will lose its air of mystery.
The entropy of a substance depends on two things: first, the state of a substance—its temperature, pressure, and amount; and second, how the substance is structured at the molecular level. We will discuss how state properties affect entropy first.
Temperature As we saw in the last section, there should be only one way of arranging the energy in a perfect crystal at 0 K. If W = 1, then S = k ln W = 0; so that the entropy should be zero at the absolute zero of temperature. This rule, known as the third law of thermodynamics, is obeyed by all solids unless some randomness of arrangement is accidentally “frozen” into the crystal. As energy is fed into the crystal with increasing temperature, we find that an increasing number of alternative ways of dividing the energy between the atoms become possible. W increases, and so does S. Without exception the entropy of any pure substance always increases with temperature.
Volume and Pressure We argued earlier that when a gas doubles its volume, the number of ways in which the gas molecules can distribute themselves in space is enormously increased and the entropy increases by 5.76 J K–1. More generally the entropy of a gas always increases with increasing volume and decreases with increasing pressure. In the case of solids and liquids the volume changes very little with the pressure and so the entropy also changes very little. Amount of Substance One of the main reasons why the entropy is such a convenient quantity to use is that its magnitude is proportional to the amount of substance. Thus the entropy of 2 mol of a given substance is twice as large as the entropy of 1 mol. Properties which behave in this way are said to be extensive properties. The mass, the volume, and the enthalpy are also extensive properties, but the temperature, pressure, and thermodynamic probability are not.
16.07: Standard Molar Entropies
Because the entropy of a substance depends on the amount of substance, the pressure, and the temperature, it is convenient to describe the entropy of a substance in terms of Sm°, its standard molar entropy, i.e., as the entropy of 1 mol of substance at the standard pressure of 1 atm (101.3 kPa) and given temperature. Values of the standard molar entropies of various substances at 298 K (25°C) are given in the table. A table like this can be used in much the same way as a table of standard enthalpies of formation in order to find the entropy change ΔSm° for a reaction occurring at standard pressure and at 298 K.
Table \(1\) The Standard Molar Entropies of Selected Substances at 298.15 K (25°C)
Compound Smo /J K-1mol-1 Compound Smo /J K-1mol-1
Solids Diatomic Gases
C (diamond) 2.377
C (graphite) 5.74 H2 130.7
Si 18.8 D2 145.0
Ge 31.1 HCl 186.9
Sn (gray) 44.1 HBr 198.7
Pb 64.8 HI 206.6
Li 29.1 N2 191.6
Na 51.2 O2 205.1
K 64.2 F2 202.8
Rb 69.5 Cl2 223.1
Cs 85.2 Br2 245.5
NaF 51.5 I2 260.7
MgO 26.9 CO 197.7
AlN 20.2 Triatomic Gases
NaCl 72.1 H2O 188.8
KCl 82.6 NO2 240.1
Mg 32.7 H2S 205.8
Ag 42.6 CO2 213.7
I2 116.1 SO2 248.2
MgH2 31.1 N2O 219.9
AgN3 99.2 O3 238.9
Liquids Polyatomic Gases( > 3)
Hg 76.0 CH4 186.3
Br2 152.2 C2H6 229.6
H2O 69.9 C3H8 269.9
H2O2 109.6 C4H10 310.2
CH3OH 126.8 C5H12 348.9
C2H5OH 160.7 C2H4 219.6
C6H6 172.8 N2O4 304.3
BCl3 206.3 B2H6 232.0
Monatomic Gases BF3 254.0
He 126.0 NH3 192.5
Ne 146.2
Ar 154.8
Kr 164.0
Xe 169.6 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.06%3A_Getting_Acquainted_with_Entropy.txt |
Since they all refer to the same temperature and pressure, and also to 1 mol of substance, any differences in the entropy values listed in the table of standard entropies must be due to differences in the molecular structure of the various substances listed. There are two aspects of the molecular structure of a substance which affect the value of its entropy: (1) The degree to which the movement of the atoms and molecules in the structure is restricted—the less restricted this movement, the greater the entropy. (2) The mass of the atoms and molecules which are moving—the greater the mass, the larger the entropy. We will consider each of these factors in turn.
Table $1$ Entropy Values
Compound Smo /J K-1mol-1 Compound Smo /J K-1mol-1
Solids Diatomic Gases
C (diamond) 2.377
C (graphite) 5.74 H2 130.7
Si 18.8 D2 145.0
Ge 31.1 HCl 186.9
Sn (gray) 44.1 HBr 198.7
Pb 64.8 HI 206.6
Li 29.1 N2 191.6
Na 51.2 O2 205.1
K 64.2 F2 202.8
Rb 69.5 Cl2 223.1
Cs 85.2 Br2 245.5
NaF 51.5 I2 260.7
MgO 26.9 CO 197.7
AlN 20.2 Triatomic Gases
NaCl 72.1 H2O 188.8
KCl 82.6 NO2 240.1
Mg 32.7 H2S 205.8
Ag 42.6 CO2 213.7
I2 116.1 SO2 248.2
MgH2 31.1 N2O 219.9
AgN3 99.2 O3 238.9
Liquids Polyatomic Gases( > 3)
Hg 76.0 CH4 186.3
Br2 152.2 C2H6 229.6
H2O 69.9 C3H8 269.9
H2O2 109.6 C4H10 310.2
CH3OH 126.8 C5H12 348.9
C2H5OH 160.7 C2H4 219.6
C6H6 172.8 N2O4 304.3
BCl3 206.3 B2H6 232.0
Monatomic Gases BF3 254.0
He 126.0 NH3 192.5
Ne 146.2
Ar 154.8
Kr 164.0
Xe 169.6
Restrictions on the Movement of Atoms and Molecules We have already encountered two examples of how the removal of a restriction on the motion of molecules allows an increase in the number of alternative ways in which the molecules can arrange themselves in space. The first of these is the case of the ring and chain forms of 1,4-butanediol discussed in the section on the molecular view of equilibrium, and displayed in figures 2 and 3 in that section. When the two ends of this molecule hydrogen bond to form a ring, the result is a fairly rigid structure. However, if the restricting influence of the hydrogen bond is removed, the more flexible chain form of the molecule is capable of many more alternative configurations. As a result, we find that W for a mole of molecules in the chain form is very much larger than W for a mole of molecules in the ring form. The chain form turns out to have a molar entropy which is 41 J K–1 mol–1 higher than that for the ring form.
A second example of the effect of relaxing a restriction on molecular motion is the expansion of a gas into a vacuum. When the molecules have more freedom of movement, many more alternative arrangements are possible, and hence W is larger. As we have already calculated, doubling the volume of a gas increases its molar entropy by 5.76 J K–1 mol–1
Mass of the Atoms and Molecules We showed that when quantum mechanics is applied, the energies of a moving particle are confined to certain specific values and not those in between. The allowed energies, or energy levels, for a particle in a box were given by equation 4 in the section on wave mechanics: $E_{k}=n^{\text{2}}\frac{h^{\text{2}}}{\text{8}md^{\text{2}}} \nonumber$ From this equation we can see that the larger the mass m of a particle, the smaller the value of its energy Ek for a given value of the quantum number n. Furthermore, the larger the value of m, the smaller the difference between Ek2 (for n = 2) and Ek1 (for n = 1). No matter what kind of energy is involved—rotational, vibrational, or translational—this is always found to be true. The greater the mass of a particle, the closer together its energy levels.
The effect of closeness of energy levels on the entropy is shown in Figure $1$. The more closely spaced the levels, the more different ways the same quantity of energy can be distributed among them. This applies in general for any number of particles and any quantity of energy. Therefore, the heavier the molecules of a substance, the larger its molar entropy. This effect is quite obvious among the noble gases—their molar entropies increase steadily with molar (and hence molecular) mass. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.08%3A_Dependence_of_S_on_Molecular_Structure.txt |
A close inspection of the entropy values in the Table of Molar Entropies reveals several trends which can be explained in terms of the factors of molecular mass and restriction of movement.
Table \(1\) Molar Entropy Values
Compound Smo /J K-1mol-1 Compound Smo /J K-1mol-1
Solids Diatomic Gases
C (diamond) 2.377
C (graphite) 5.74 H2 130.7
Si 18.8 D2 145.0
Ge 31.1 HCl 186.9
Sn (gray) 44.1 HBr 198.7
Pb 64.8 HI 206.6
Li 29.1 N2 191.6
Na 51.2 O2 205.1
K 64.2 F2 202.8
Rb 69.5 Cl2 223.1
Cs 85.2 Br2 245.5
NaF 51.5 I2 260.7
MgO 26.9 CO 197.7
AlN 20.2 Triatomic Gases
NaCl 72.1 H2O 188.8
KCl 82.6 NO2 240.1
Mg 32.7 H2S 205.8
Ag 42.6 CO2 213.7
I2 116.1 SO2 248.2
MgH2 31.1 N2O 219.9
AgN3 99.2 O3 238.9
Liquids Polyatomic Gases( > 3)
Hg 76.0 CH4 186.3
Br2 152.2 C2H6 229.6
H2O 69.9 C3H8 269.9
H2O2 109.6 C4H10 310.2
CH3OH 126.8 C5H12 348.9
C2H5OH 160.7 C2H4 219.6
C6H6 172.8 N2O4 304.3
BCl3 206.3 B2H6 232.0
Monatomic Gases BF3 254.0
He 126.0 NH3 192.5
Ne 146.2
Ar 154.8
Kr 164.0
Xe 169.6
Solids, Liquids, and Gases Perhaps the most obvious feature of the table of molecular entropies is a general increase in the molar entropy as we move from solids to liquids to gases. In a solid, the molecules are only capable of restricted vibrations around a fixed point, but when a solid melts, the molecules, though still hampered by their mutual attraction, are much freer to move around. Thus when a solid melts, the molar entropy of the substance increases. When a liquid vaporizes, the restrictions on the molecules’ ability to move around are relaxed almost completely and a further and larger increase in the entropy occurs. When 1 mol of ice melts, for example, its entropy increases by 22 J K–1, while on boiling the entropy increase is 110 J K–1.
Molecular Complexity A second clear trend in the table is the higher molar entropy of substances with more complex molecules. To some extent this is due to the mass since on the whole more complex molecules are heavier than simpler ones. However, we still find an increase of entropy with complexity when we compare molecules of very similar masses:
Substance Ar(g) F2(g) CO2(g) C3H8(g)
Sm°/J K–1 mol–1 155 202.7 213.6 269.9
Molar mass/g mol–1 40 38 44 44
Number of atoms 1 2 3 11
The more atoms there are in a molecule, the more ways the molecule can change its shape by vibrating. In consequence there are more ways in which the energy can be distributed among the molecules.
Strength of Bonding Another trend in entropy, most noticeable in the case of solids, is the decrease in the entropy as the forces between the atoms, molecules, or ions increases. A good example is the three solid compounds
Substance NaF(s) MgO(s) AlN(s)
Sm°(298 K)/J K–1 mol–1 51.5 26.8 20.2
Molar mass/g mol–1 42.0 40.3 41.0
which are isoelectronic with sodium fluoride. Since there is very little difference in the molar masses, the entropy decrease can only be attributed to the increase in the coulombic attraction between the ions as we move from the singly charged ions Na+ and F through the doubly charged ions Mg2+ and O2–, to the triply charged ions A13+ and N3–. (While it is true that there is a fair degree of covalent character to the bonding in AIN, the effect of this will be to increase the strength of the bonding.)
Example \(1\): Molar Entropy
From each of the following pairs of compounds choose the one with the higher standard molar entropy at 25°C. Give brief reasons for your choice.
a) HBr(g), HCl(g) c) ND3(g), Ne(g) e) C2H6(g), C2H4(g) b) Cs(s), Cs(l) d) KCl(s), CaS(s)
Solution
a) HBr and HCl are very similar except for their mass. HBr will have a higher entropy because of its greater mass. b) At the same temperature, the liquid form of a substance always has a higher entropy than the solid. c) ND3 (D = deuterium) and Ne have almost identical molar masses (20 g mol–1) However, since ND3 is more complex, it can vibrate and rotate while Ne cannot. ND3 will have the higher entropy. d) KCl and CaS are isoelectronic. Because both anion and cation are doubly charged in CaS, the ions are more tightly held to each other and can vibrate less readily. Thus KCl must have the higher entropy. e) On all counts C2H6 must have a higher entropy than C2H4. C2H6 is heavier and more complex than C2H4. In addition there is free rotation about the C—C bond in C2H6 but hindered rotation about the C=C bond in C2H4. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.09%3A_Some_Trends_In_Entropy_Values.txt |
Knowing what factors affect the magnitude of the entropy often enables us to predict whether the entropy of the products will be greater or less than that of the reactants in a given chemical reaction. This is particularly true for gaseous reactions. In a dissociation reaction like
$\ce{N2O4(g) → NO2 + NO2} \qquad \Delta S_{m}°(298 \text{K})=+176 \text{JK}^{-1} \text{mol}^{-1} \nonumber$
for instance, it is easy to see that ΔS should be positive. The two halves of the N2O4 molecule are forced to move around together before dissociation, but they can move around independently as NO2 molecules once dissociation has occurred. A similar argument applies to reactions like
$\ce{2O3 → 3O2} \qquad \Delta S°_{m}(298 \text{K})=+137 \text{JK}^{-1} \text{mol}^{-1} \nonumber$
In the form of O3, O atoms are constrained to move around in groups of three, but in the form of O2, only two atoms need move around together, a lesser restriction. Accordingly we expect ΔS to be positive for this reaction.
A further extension of this argument leads us to the general conclusion that in any reaction involving gases if the amount of substance in the gaseous phase increases, ΔS will be positive, while if it decreases, so will ΔS. For example, in the reaction
$\ce{2CO(g) + O2(g) → 2CO2(g)} \nonumber$
The amount of gas decreases from 3 to 2 mol (i.e., Δn = –1 mol). The entropy change should thus be negative for this reaction. From the Table of Standard Molar Entropies we can readily find that ΔSm°(298 K) has the value –173 J K–1 mol–1.
Table of Standard Molar Entropies
Compound Smo /J K-1mol-1 Compound Smo /J K-1mol-1
Solids Diatomic Gases
C (diamond) 2.377
C (graphite) 5.74 H2 130.7
Si 18.8 D2 145.0
Ge 31.1 HCl 186.9
Sn (gray) 44.1 HBr 198.7
Pb 64.8 HI 206.6
Li 29.1 N2 191.6
Na 51.2 O2 205.1
K 64.2 F2 202.8
Rb 69.5 Cl2 223.1
Cs 85.2 Br2 245.5
NaF 51.5 I2 260.7
MgO 26.9 CO 197.7
AlN 20.2 Triatomic Gases
NaCl 72.1 H2O 188.8
KCl 82.6 NO2 240.1
Mg 32.7 H2S 205.8
Ag 42.6 CO2 213.7
I2 116.1 SO2 248.2
MgH2 31.1 N2O 219.9
AgN3 99.2 O3 238.9
Liquids Polyatomic Gases( > 3)
Hg 76.0 CH4 186.3
Br2 152.2 C2H6 229.6
H2O 69.9 C3H8 269.9
H2O2 109.6 C4H10 310.2
CH3OH 126.8 C5H12 348.9
C2H5OH 160.7 C2H4 219.6
C6H6 172.8 N2O4 304.3
BCl3 206.3 B2H6 232.0
Monatomic Gases BF3 254.0
He 126.0 NH3 192.5
Ne 146.2
Ar 154.8
Kr 164.0
Xe 169.6
This table shows molar entropies for the standard conditions of 298.15 K (25°C) and 101.3 kPa. Such conditions need to be specificed, since entropy is propotional to substance amount, and dependent on temperature, pressure. Entropy is also dependent upon volume, but since the amount, n, temperature, and pressure are given, volume is implicitly defined. This table is taken from CoreChem:Standard Molar Entropies, and is also used on CoreChem:Dependence of S on Molecular Structure as well as CoreChem:Some Trends In Entropy Values.
16.11: Entropy Randomness and Disorder
A very useful, though somewhat rough, description of the entropy of a substance is as a measure of the randomness or disorder of the atoms and molecules which constitute that substance. In these terms the second law of thermodynamics is seen as a tendency for the disorder of the universe to increase. This way of looking at entropy is entirely compatible with the approach presented above. A situation which we intuitively recognize as being orderly is also one which can only be achieved in a limited number of ways. By contrast, situations which we recognize as disordered, random, or chaotic, can he achieved in a whole variety of ways. In other words, W, and hence S, is small for an ordered situation but large for a disordered situation.
There are limits to the lengths one can take this order-disorder approach to entropy, though. It does not lend itself to a quantitative treatment, and it is also difficult to explain some things like the effect of mass in these terms. There is nothing in our intuition about order, for example, which suggests that 1 mol Xe gas is more disordered than 1 mol He gas, even though its entropy is in fact larger. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.10%3A_Entropy_Changes_in_Gaseous_Reactions.txt |
Up to this point we have often quoted values of the entropy without giving any indication of how such values may be obtained. Alas, there is no convenient black box labeled entropy meter into which we can put a substance and read off its entropy value on a dial. Determining the entropy turns out to be both difficult and laborious. In the case of a simple gas, if we know enough about its molecular structure and enough quantum mechanics, we can actually calculate its entropy. For most substances, though, we are forced to derive the entropy from a series of calorimetric measurements, most of them at very low temperatures.
This method for determining the entropy centers around a very simple relationship between q, the heat energy absorbed by a body, the temperature T at which this absorption takes place, and ΔS, the resultant increase in entropy:
$\Delta S=\frac{q}{T} \label{1}$ It is possible to derive this relationship from our original definition of entropy, namely, S = k ln W, but the proof is beyond the level of this text.
It is easy to see how Eq. $\ref{1}$ can be used to measure the entropy. We start with our substance as close to the absolute zero of temperature as is technically feasible and heat it in many stages, measuring the heat absorbed at each stage, until we arrive at the desired temperature, say 298 K. The initial value of the entropy is zero, and we can calculate the entropy increase for each stage by means of Eq. $\ref{1}$and so the sum of all these increases is the entropy value for 298 K. In the case of simple gases, values of entropy measured in this way agree very well with those calculated from knowledge of molecular structure.
Equation $\ref{1}$ was discovered long before the statistical nature of entropy was realized. Scientists and engineers began to appreciate the importance of the quantity q/T very early in the nineteenth century because of its connection with the efficiency of steam engines. These arguments were developed by both Lord Kelvin in England and Rudolf Clausius (1822 to 1888) in Germany. It was Clausius who first formulated the second law in terms of the entropy S, but Clausius had only a vague idea that entropy was in any way connected with molecules or probability. The statistical nature of entropy was first suggested by Boltzmann in 1877 and then developed into an elegant system in 1902 by Josiah Willard Gibbs (1839 to 1903), one of the real giants among American scientists.
An important feature of Eq. $\ref{1}$ is the inverse relationship between the entropy increase and the temperature. A given quantity of heat energy produces a very large change of entropy when absorbed at a very low temperature but only a small change when absorbed at a high temperature.
Example $1$: Entropy
Calculate the increase in entropy which a substance undergoes when it absorbs 1 kJ of heat energy at the following temperatures: (a) 3 K; (b) 300 K; (c) 3000 K.
Solution
a) At 3 K we have $\Delta S=\frac{\text{1000 J}}{\text{3 K}}=\text{333}\text{.3 J K}^{-\text{1}} \nonumber$ b) At 300 K, similarly, $\Delta S=\frac{\text{1000 J}}{\text{300 K}}=\text{3}\text{.33 J K}^{-\text{1}} \nonumber$ c) At 3000K $\Delta S=\frac{\text{1000 J}}{\text{3000 K}}=\text{0}\text{.33 J K}^{-\text{1}} \nonumber$
An amusing analogy to this behavior can be drawn from everyday life. If a 10-year-old boy is allowed to play in his bedroom for half an hour, the increase in disorder is scarcely noticeable because the room is already disordered (i.e., at a higher “temperature”). By contrast, if the boy is let loose for half an hour in a neat and tidy living room (i.e., at a lower “temperature”), the effect is much more dramatic. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.12%3A_Measuring_the_Entropy.txt |
In order to determine whether a reaction is spontaneous or not, it is not sufficient just to determine ΔSm, the entropy difference between products and reactants. As an example, let us take the reaction
$\ce{2Mg(s) + O2(g) → 2MgO(s)} \qquad 1 \text{atm}, \quad 298 \text{K} \label{1}$
Since this reaction occurs at the standard pressure and at 298 K, we can find ΔS from the standard molar entropies:
\begin{align} \Delta S &= \Delta S_{m}^{º}(298\text{K})=2S_{m}^{º}(\text{Mg}) - S_{m}^{º}\text{O}_{2} \ &=(2 \times 26.8 - 2 \times 32.6 - 205.0) \frac { \text{J}}{\text{mol K}} \&=-216.6 \frac{\text{J}}{\text{mol K}} \end{align} \nonumber
This result would suggest that the reaction is not spontaneous, but in fact it is. Once ignited, a ribbon of magnesium metal burns freely in air to form solid magnesium oxide in the form of a white powder. The reaction is plainly spontaneous even though ΔS is negative. Why is this not a contradiction of the second law?
The answer is that we have failed to realize that the entropy change which the magnesium and oxygen atoms undergo as a result of the reaction is not the only entropy change which occurs. The oxidation of magnesium is a highly exothermic reaction, and the heat which is evolved flows into the surroundings, increasing their entropy as well. There are thus two entropy changes which we must take into account in deciding whether a reaction will be spontaneous or not: (1) the change in entropy of the system actually undergoing the chemical change, which we will indicate with the symbol ΔSsys; and (2) the change in entropy of the surroundings, ΔSsurr, which occurs as the surroundings absorb the heat energy liberated by an exothermic reaction or supply the heat energy absorbed by an endothermic reaction.
Of these two changes, the first is readily obtained from tables of entropy values. Thus, for the oxidation of magnesium, according to Eq. $\ref{1}$, ΔSsys has the value already found, namely, –216.6 J K–1 mol–1.
The second entropy change, ΔSsurr, can also be derived from tables, as we shall now show.
When a chemical reaction occurs at atmospheric pressure and its surroundings are maintained at a constant temperature T, then the surroundings will absorb a quantity of heat, qsurr equal to the heat energy given off by the reaction.
$q_{surr}=- \Delta H \label{4}$
(The negative sign before ΔH is needed because qsurr is positive if the surroundings absorb heat energy, but ΔH is negative if the system gives off heat energy for them to absorb.) If we now feed Eq. $\ref{4}$ into Eq. (1) in the section on measuring entropy, we obtain an expression for the entropy change of the surroundings in terms of ΔH:
$\Delta S_{\text{surr}}=\frac{q_{\text{surr}}}{T}=\frac{-\Delta H}{T} \label{5}$
Using this equation it is now possible to find the value of ΔSsurr from table of standard enthalpies of formation. In the case of the oxidation of magnesium, for example, we easily find that the enthalpy change for
$\ce{2Mg(s) + O2(g) -> 2MgO(s)} \qquad 1 \text{atm, 298 K } \nonumber$
is given by
$\Delta H = \Delta H_{m}^{º} (MgO)=2 \times -601.8 \frac{\text{J}}{\text{mol K}}=-1204 \frac{\text{J}}{\text{mol K}} \nonumber$
Substituting this result into Eq. $\ref{5}$, we then find
$\Delta S_{\text{surr}}=\frac{-\Delta H}{T}=\frac{\text{1204 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}}=\text{4040 J K}^{-\text{1}}\text{ mol}^{-\text{1}} \nonumber$
If a reaction is spontaneous, then it is the total entropy change ΔStot, given by the sum of ΔSsurr and ΔSsys, which must be positive in order to conform to the second law. In the oxidation of magnesium, for example, we find that the total entropy change is given by
$\Delta S_{tot} = \Delta S_{surr} + \Delta S_{sys} = (4040 - 216.6) \frac{\text{J}}{\text{ mol K}}=3823 \frac{\text{J}}{\text{mol K}} \nonumber$
This is a positive quantity because the entropy increase in the surroundings is more than enough to offset the decrease in the system itself, and the second law is satisfied. In the general case the total entropy change is given by
$\Delta S_{tot}= \Delta S_{surr} + \Delta S_{sys} = -\frac{\Delta H}{T} + \Delta S_{sys} \nonumber$
The second law requires that this sum must be positive; i.e.,
$\frac{-\Delta H}{T} + \Delta S_{sys}>0\qquad \label{11}$
This simple inequality gives us what we have been looking for: a simple criterion for determining whether a reaction is spontaneous or not. Since both ΔH and ΔS can be obtained from tables, and T is presumably known, we are now able to predict in advance whether a reaction will be uphill or downhill.
Example $1$: Spontaneous Reaction
Using the Table of Some Standard Enthalpies of Formation at 25°C and the Table of Standard Molar Entropies find ΔHm°(298 K) and ΔSm°(298 K) for the reaction
$\ce{N2 (g) + 3H2 (g) -> 2NH3 (g)} \qquad 1 \text{atm} \nonumber$
Predict whether the reaction will be spontaneous or not at a temperature of (a) 298 K, and (b) 1000 K.
Solution
We find from the Table of Some Standard Enthalpies of Formation that
\begin{align} \Delta H_{m}^{º} (298 \text{K}) & = \sum \Delta H_{f}^{º} \text{(products)} - \sum \Delta H_{f}^{º} \text{(reactants)} \ & = 2 \Delta H_{f}^{º} \text{(NH}_3 ) - \Delta H_{f}^{º} ( \text{N}_2 ) - 3 \Delta H_{f}^{º} ( \text{H}_2 ) \ & = (-2 \times 46.1 - 0.0 - 0.0 ) \text{kJ mol}^{-1} = -92.2 \text{kJ mol}^{-1} \end{align} \nonumber
and from the Table of Standard Molar Entropies:
\begin{align}\Delta S_m^{º} (298 \text{K}) & = \sum \Delta S_m^{º} \text{(products)} - \sum \Delta S_m^{º} \text{(reactants)} \ & = 2 \Delta S_m^º (\text{NH}_3 ) - \Delta S_m^º ( \text{N}_2 ) - 3 \Delta S_m^º ( \text{H}_2 ) \ & = (2 \times 192.2 – 191.5 – 3 \times 130.6) \text{ J K}^{-1} \text{ mol}^{-1} = –198.9 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber
a) At 298 K the total entropy change per mol N2 is given by
\begin{align} \Delta S_{tot} & = \frac{-\Delta H_{m}^{ o}}{T} + \Delta S_{sys}^{º} = \frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{298 K}} – 198.9 \text{ J K}^{-1} \text{ mol}^{-1} \ & = (309.4 – 198.9) \text{ J K}^{-1} \text{ mol}^{-1} = 110.5 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber
Since the total entropy change is positive, the reaction is spontaneous:
b) Since tables are available only for 298 K, we must make the approximate assumption that neither ΔH nor ΔS varies greatly with temperature. Accordingly we assume
\begin{align} & \Delta H_m^º (1000 \text{ K} ) = \Delta H_m^º (298 \text{ K} ) = -92.2 \text{kJ mol}^{-1} \ \text{and} ~~~~~ & \Delta S_m^º (1000 \text{ K} = \Delta S_m^º (298 \text{ K} ) = -198.9 \text{ J K}^{-1} \text{ mol}^{-1}\end{align} \nonumber
ΔHm°(1000 K) = ΔHm°(298 K) = –92.2 kJ mol–1
and ΔSm°(1000 K) = ΔSm°(298 K) = – 198.9 J K–1 mol–1
Thus for 1 mol N2 reacted,
\begin{align} \Delta S_{tot} & = \frac{-\Delta H_{m}^{ o}}{T} + \Delta S_{sys}^ º = \frac{\text{92}\text{.2 }\times \text{ 10}^{\text{3}}\text{ J mol}^{-\text{1}}}{\text{1000 K}} – 198.9 \text{ J K}^{-1} \text{ mol}^{-1} \ & = (92.2 – 198.9) \text{ J K}^{-1} \text{ mol}^{-1} = –106.7 \text{ J K}^{-1} \text{ mol}^{-1} \end{align} \nonumber
Since the total entropy change is negative at this high temperature, we conclude that N2 and H2 will not react to form NH3, but that rather NH3 will decompose into its elements.
Apart from enabling us to predict the direction of a chemical reaction from tables of thermodynamic data, the inequality [Eq. $\ref{11}$] shows that three factors determine whether a reaction is spontaneous or not: the enthalpy change ΔH, the entropy change ΔS, and the temperature T. Let us examine each of these to see what effect they have and why.
The enthalpy change, ΔH As we well know, if ΔH is negative, heat will be released by the reaction and the entropy of the surroundings will be increased, while if ΔH is positive, the surroundings will decrease in entropy. At room temperature we usually find that this change in the entropy of the surroundings as measured by –ΔH/T is the major factor in determining the direction of a reaction since ΔS is almost always small by comparison. This explains why at room temperature most spontaneous reactions are exothermic. On the molecular level, as we saw in the sections on enthalpy, heat and energy, an exothermic reaction corresponds to a movement from a situation of weaker bonding to a situation of stronger bonding. The formation of more and/or stronger bonds is thus a big factor in tending to make a reaction spontaneous. The entropy change, ΔSsys If the system itself increases in entropy as a result of the reaction (i.e., if ΔSsys is positive), this will obviously contribute toward making the total entropy change positive and the reaction spontaneous. As we saw in the previous section of this chapter, reactions for which ΔS is positive correspond to the relaxation of some of the constraints on the motion of the atoms and molecules in the system. In particular, dissociation reactions and reactions in which the amount of substance in the gas phase increases correspond to reactions for which ΔS is positive. The temperature, T Because it alters the magnitude of –ΔH/T relative to ΔS, the temperature regulates the relative importance of the enthalpy change and the entropy change in determining whether a reaction will be spontaneous or not. As we lower the temperature, the effect of a reaction on the entropy of its surroundings becomes more and more pronounced because of the operation of the “boy in the living room” effect. As we approach absolute zero, the value of –ΔH/T begins to be an infinitely large positive or negative quantity, and ΔS becomes insignificant by comparison.
At a very low temperature, therefore, whether the reaction is spontaneous or not will depend on the sign of ΔH, i.e., on whether the reaction is exothermic or endothermic. By contrast, as we raise the temperature to very high values, the “boy in the bedroom” effect takes over and the reaction affects the entropy of its surroundings to an increasingly smaller extent until finally it is only the value of ΔSsys which determines the behavior of the reaction. In short, whether a reaction is spontaneous or not is controlled by the sign of ΔH at very low temperatures and by the sign of ΔS at very high temperatures.
Since ΔH can be positive or negative and so can ΔS, there are four possible combinations of these two factors, each of which exhibits a different behavior at high and low temperatures. All four cases are listed and described in the next table, and they are also illustrated by simple examples in Figure $1$. In this figure the surroundings are indicated by a shaded border around the reaction system. If the reaction is exothermic, the border changes from gray to pink, indicating that the surroundings have absorbed heat energy and thus increased in entropy. When the border changes from pink to gray, this indicates an endothermic reaction and a decrease in the entropy of the surroundings. In each case the change in entropy of the system should be obvious from an increase or decrease in the freedom of movement of the molecules and/or atoms.
Table $1$ Effect of Temperature on the Spontaneity of a Reaction.
Case Sign of ΔH Sign of ΔS Behavior
1
+
Spontaneous at all temperatures
2
Spontaneous only at low temperatures
3
+
+
Spontaneous only at high temperatures
4
+
Never spontaneous
Case l: Reaction is exothermic, and ΔSsys is positive The example illustrated in Figure $1$ I is the decomposition of ozone to oxygen.
$\ce{2O3(g)→3O2(g)}\qquad 1 \text{ atm} \nonumber$
for which ΔHm°(298 K) = –285 kJ mol–1 and ΔSm°(298 K) = +137 J K–1 mol–1. A reaction of this type is always spontaneous because the entropy of both the surroundings and the system are increased by its occurrence.
Case 2: Reaction is exothermic, and is ΔSsys negative The example illustrated in Figure $1$ (II) is the reaction of magnesium metal with hydrogen gas to form magnesium hydride:
$\ce{Mg(s) + H2(g) → MgH2(g)} \qquad 1 \text{ atm} \nonumber$
for which ΔHm°(298 K) = –76.1 kJ mol–1 and ΔSm°(298 K) = –132.1 J K–1 mol–1. Reactions of this type can be either spontaneous or nonspontaneous depending on the temperature. At low temperatures when the effect on the surroundings is most important, the exothermic nature of the reaction makes it spontaneous. At high temperatures the effect of ΔSsys predominates. Since ΔSsys is negative (free H2 molecules becoming fixed H ions), the reaction must become nonspontaneous at high temperatures. Experimentally, solid MgH2 will not form from its elements above 560 K, and any formed at a lower temperature will decompose.
Case 3: Reaction is endothermic, and ΔSsys is positive The example illustrated in Figure $2$ (III) is the vaporization of liquid bromine:
$\ce{Br2(l) → Br2(g)}\qquad 1 \text{ atm} \nonumber$
for which ΔHm°(298 K) = +31.0 kJ mol–1 and ΔSm°(298 K) = 93.1 J K–1 mol–1. This example is usually classified as a physical rather than as a chemical change, but such distinctions are not important in thermodynamics. As in the previous case this reaction can be spontaneous or nonspontaneous at different temperatures. At low temperatures, bromine will not boil because the entropy increase occurring in the bromine as it turns to vapor is not enough to offset the decrease in entropy, which the surroundings experience in supplying the heat energy which is needed for the change in state. At higher temperatures the entropy effect on the surroundings becomes less pronounced, and the positive value of ΔSsys makes the reaction spontaneous. At 101.3 kPa (1atm) pressure, bromine will not boil below 331 K (58°C), but above this temperature it will.
Case 4: Reaction is endothermic, and ΔSsys is negative The example illustrated in Figure $2$ IV is the reaction between silver and nitrogen to form silver azide, AgN3:
$\ce{2Ag(s) + 3N2(g) → 2AgN3(s)} \nonumber$
for which ΔHm°(298 K) = +620.6 kJ mol–1 and ΔSm°(298 K) = –461.5 J K–1 mol–1. Reactions of this type can never be spontaneous. If this reaction were to occur, it would reduce the entropy of both the system and the surroundings in contradiction of the second law. Since the forward reaction is nonspontaneous, we expect the reverse reaction to be spontaneous. This prediction is borne out experimentally. When silver azide is struck by a hammer, it decomposes explosively into its elements!
Example $2$: Classifying Reactions
Classify the following reactions as one of the four possible types (cases) just described. Hence suggest whether the reaction will be spontaneous at (i) a very low temperature, and (ii) at a very high temperature.
ΔHm°(298 K)/
kJ mol–1
ΔSm°(298 K)/
J K–1 mol–1
a) N2(g) + F2(g) → 2NF3(g) –249 –277.8
b) N2(g) + 3Cl2(g) → 2NCl3(g) +460 –275
c) N2F4(g) → 2NF2(g) +93.3 +198.3
d) C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O (g) –2044.7 +101.3
Solution
a) This reaction is exothermic, and ΔS is negative. It belongs to type 2 and will be spontaneous at low temperatures but nonspontaneous at high temperatures.
b) Since this reaction is endothermic and ΔS is negative, it belongs to type 4. It cannot be spontaneous at any temperature.
c) Since this reaction is endothermic but ΔS is positive, it belongs to type 3. It will be spontaneous at high temperatures and nonspontaneous at ow temperatures. (All dissociation reactions belong to this class.)
d) This reaction belongs to type 1 and is spontaneous at all temperatures.
Table of Some Standard Enthalpies of Formation at 25°C
Compound
ΔHf/kJ
mol–1
ΔHf/kcal
mol–1
Compound
ΔHf/kJ
mol–1
ΔHf/kcal
mol–1
AgCl(s) –127.068 –30.35 H2O(g) –241.818 –57.79
AgN3(s) +620.6 +148.3 H2O(l) –285.8 –68.3
Ag2O(s) –31.0 –7.41 H2O2(l) –187.78 –44.86
Al2O3(s) –1675.7 –400.40 H2S(g) –20.63 –4.93
Br2(l) 0.0 0.00 HgO(s) –90.83 –21.70
Br2(g) +30.907 +7.385 I2(s) 0.0 0.0
C(s), graphite 0.0 0.00 I2(g) +62.438 +14.92
C(s), diamond +1.895 +0.453 KCl(s) –436.747 –104.36
CH4(g) –74.81 –17.88 KBr(s) –393.798 –94.097
CO(g) –110.525 –26.41 MgO(s) –601.7 –143.77
CO2(g) –393.509 –94.05 NH3(g) –46.11 –11.02
C2H2(g) +226.73 +54.18 NO(g) +90.25 +21.57
C2H4(g) +52.26 +12.49 NO2(g) +33.18 +7.93
C2H6(g) –84.68 –20.23 N2O4(g) +9.16 +2.19
C6H6(l) +49.03 +11.72 NF3(g) –124.7 –29.80
CaO(s) –635.09 –151.75 NaBr(s) –361.062 –86.28
CaCO3(s) –1206.92 –288.39 NaCl(s) –411.153 –98.24
CuO(s) –157.3 –37.59 O3(g) +142.7 +34.11
Fe2O3(s) –824.2 –196.9 SO2(g) –296.83 –70.93
HBr(g) –36.4 –8.70 SO3(g) –395.72 –94.56
HCl(g) –92.307 –22.06 ZnO(s) –348.28 –83.22
HI(g) +26.48 +6.33
Above is a table of standard enthalpies of formation at 25°C. This is the enthalpy change which occurs when a compound is formed from its constituent elements. For instance, the enthalpy of formation value of ΔHm = –285.8 kJ mol–1 is the enthalpy change for the reaction: H2(g) + ½O2(g) → H2O(l). It is important to note that the enthalpy of formation for an element in its most stable state is 0. Using these enthalpies of formation and Hess' Law the enthalpy change of any reaction involving the compounds with a known enthalpy of formation can be determined.
This table is found on CoreChem:Standard Enthalpies of Formation | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.13%3A_Including_the_Surroundings.txt |
In the previous section, we were careful to differentiate between the entropy change occurring in the reaction system ΔSsys, on the one hand, and the entropy change occurring in the surroundings, ΔSsurr, given by –ΔH/T, on the other. By doing this we were able to get a real insight into what controls the direction of a reaction and why. In terms of calculations, though, it is a nuisance having to look up both entropy and enthalpy data in order to determine the direction of a reaction. For reasons of convenience, therefore, chemists usually combine the entropy and the enthalpy into a new function called the Gibbs free energy, or more simply the free energy, which is given the symbol G. If free-energy tables are available, they are all that is needed to predict the direction of a reaction at the temperature for which the tables apply.
In order to introduce free energy, let us start with the inequality
$-\frac{\Delta H}{T}+ {\Delta S}_{sys} > 0 \nonumber$
This inequality must be true if a reaction occurring at constant pressure in surroundings at constant temperature is to be spontaneous. It is convenient to multiply this inequality by T; it then becomes
$-\Delta H + T \Delta S > 0 \nonumber$
(From now on we will abandon the subscript "sys".) If –ΔH + T ΔS is greater than zero, it follows that multiplying it by –1 produces a quantity which is less than zero, that is,
$\Delta H - T \Delta S < 0 \label{3}$
This latest inequality can be expressed very neatly in terms of the free energy G, which is defined by the equation
$G=H-TS \nonumber$
When a chemical reaction occurs at constant temperature, the free energy will change from an initial value of G, given by $G_{1}=H_{1}=TS_{1} \nonumber$
to a final value
$G_{2}=H_{2}-TS_{2} \nonumber$
The change in free energy ΔG will thus be $\Delta G = G_{2} - G_{1} = H_{2} - H_{1} - T(S_{2}-S_{1}) \nonumber$
or
$\Delta G = \Delta H - T\Delta S \nonumber$
Feeding this result back into inequality $\ref{3}$ gives the result
$\Delta G = \Delta H - T \Delta S <0 \nonumber$
$\Delta G < 0 \label{10}$
This very important and useful result tells us that when a spontaneous chemical reaction occurs (at constant temperature and pressure), the free-energy change is negative. In other words a spontaneous change corresponds to a decrease in the free energy of the system.
If we have available the necessary free-energy data in the form of tables, it is now quite easy to determine whether a reaction is spontaneous or not. We merely calculate ΔG for the reaction using the tables. If ΔG turns out to be positive, the reaction is nonspontaneous, but if it turns out to be negative, then by virtue of Eq. $\ref{10}$ we can conclude that it is spontaneous. Data on free energy are usually presented in the form of a table of values of standard free energies of formation. The standard free energy of formation of a substance is defined as the free-energy change which results when 1 mol of substance is prepared from its elements at the standard pressure of 1 atm and a given temperature, usually 298 K. It is given the symbol ΔGf°. A table of values of ΔGf° (298 K) for a limited number of substances is given in the following table.
Table $1$: Some Standard Free Energies of Formation at 298.15 K (25°C)
Compound ΔGfo /kJ mol-1 Compound ΔGfo /kJ mol-1
AgCl(s) -109.789 H2O(g) -228.572
AgN3(s) 591.0 H2O(l) -237.129
Ag2O(s) -11.2 H2O2(l) -120.35
Al2O3(s) -1582.3 H2S(g) -33.56
Br2(l) 0.0 HgO(s) -58.539
Br2(g) 3.110 I2(s) 0.0
CaO(s) -604.03 I2(g) 19.327
CaCO3(s) -1128.79 KCl(s) -409.14
C--graphite 0.0 KBr(s) -380.66
C--diamond 2.9 MgO(s) -569.43
CH4(g) -50.72 MgH2(s) 76.1
C2H2(g) 209.2 NH3(g) -16.45
C2H4(g) 68.15 NO(g) 86.55
C2H6(g) -32.82 NO2(g) 51.31
C6H6(l) 124.5 N2O4(g) 97.89
CO(g) -137.168 NF3(g) -83.2
CO2(g) -394.359 NaCl(s) -384.138
CuO(s) -129.7 NaBr(s) -348.983
Fe2O3(s) -742.2 O3(g) 163.2
HBr(g) -53.45 SO2(g) -300.194
HCl(g) -95.299 SO3(g) -371.06
HI(g) 1.7 ZnO(s) -318.3
This table is used in exactly the same way as a table of standard enthalpies of formation. This type of table enables us to find ΔG values for any reaction occurring at 298 K and 1 atm pressure, provided only that all the substances involved in the reaction appear in the table. The two following examples illustrate such usage.
Example $1$: Spontaneous Reactions
Determine whether the following reaction is spontaneous or not:
$\ce{4NH3(g) + 5 O2(g) → 4NO(g) + 6H2O(l)}\qquad 1 \text{ atm, 298K} \nonumber$
Solution
Following exactly the same rules used for standard enthalpies of formation, we have
$\Delta G_{m}^{\circ} = \Sigma \Delta G_{f}^{\circ}\text{ (products)} - \Sigma \Delta G_{f}^{\circ} \text{ (reactants)}$
$\qquad = \ce{4 \Delta G_{f}^{\circ}(NO) + 6 \Delta G_{f}^{\circ}(H2O) - 4\Delta G_{f}^{\circ}(NH3) - 5\Delta G_{f}^{\circ}(O2)}$
Inserting values from the table of free energies of formation, we then find
$\Delta G_{m}^{\circ} = [4 \times 86.7 + 6 \times (-273.3) - 4 \times (-16.7) - 5 \times 0.0]\frac{ \text{kJ }}{\text{mol}}$
$\qquad = -1010\frac{\text{ kJ}}{\text{ mol}}$
Since $\Delta G_{m}^{\circ}$ is very negative, we conclude that this reaction is spontaneous.
The reaction of NH3 with O2 is very slow, so that when NH3 is released into the air, no noticeable reaction occurs. In the presence of a catalyst, though, NH3 burns with a yellowish flame in O2. This reaction is very important industrially, since the NO produced from it can be reacted further with O2 and H2O to form HNO3:
$\ce{2NO + \frac{3}{2}O2 + H2O→2HNO3} \nonumber$
Nitric acid, HNO3 is used mainly in the manufacture of nitrate fertilizers but also in the manufacture of explosives.
Example $2$: Spontaneous Reactions
Determine whether the following reaction is spontaneous or not:
$\ce{2NO(g) + 2CO(g) → 2CO2(g) + N2 (g)}\qquad 1 \text{ atm, 298K} \nonumber$
Solution Following previous procedure we have
$(\Delta G_{m}^{\circ} = (-2\times 394.4 + 0.0 - 2 \times 86.7 + 2 \times 137.3)\frac{\text{ kJ}}{\text{ mol}}$
$\qquad = -687.6\frac{ \text{ kJ}}{\text{ mol}}$
The reaction is thus spontaneous.
This example is an excellent illustration of how useful thermodynamics can be. Since both NO and CO are air pollutants produced by the internal-combustion engine, this reaction provides a possible way of eliminating both of them in one reaction, killing two birds with one stone. A thee-way catalytic converter is able to perform the equivalent of this reaction. The reduction step coverts NOx to O2 and N2. Then, in the oxidation step, CO and O2 are converted to CO2. If ΔGm° had turned out be +695 kJ mol–1, the reaction would be nonspontaneous and there would be no point at all in developing such a device.[1]
We quite often encounter situations in which we need to know the value of ΔGm° for a reaction at a temperature other than 298 K. Although extensive thermodynamic tables covering a large range of temperatures are available, we can also obtain approximate values for ΔG from the relationship
$\Delta G_{m}^{\circ}=\Delta H_{m}^{\circ}-T\Delta S_{m}^{\circ} \nonumber$
If we assume, as we did previously, that neither ΔHm° nor ΔSm° varies much as the temperature changes from 298 K to the temperature in question, we can then use the values of ΔHm°(298 K) obtained from the Table of Some Standard Enthalpies of Formation at 25°C and ΔSm°(298 K) obtained from the Table of Standard Molar Entropies to calculate ΔGm° for the temperature in question.
Example $3$: Spontaneous at Different Temperatures
Using the enthalpy values and the entropy values, calculate ΔHm° and ΔSm° for the reaction
$\ce{CH4(g) + H2O(g) → 3 H2(g) + CO(g)} \qquad 1 \text{ atm} \nonumber$
Calculate an approximate value for $\Delta G_{m}^{º}$ for this reaction at 600 and 1200 K and determine whether the reaction is spontaneous at either temperature.
Solution From the tables we find
$\Delta H_{m}^{\circ}(298 \text{ K})= 3\Delta H_{f}^{\circ}(\text{H}_{2})+\Delta H_{f}^{\circ}\text{(CO)} - \Delta H_{f}^{º}( \text{CH}_{4}) - \Delta H_{f}^{\circ}(\text{H}_{2}\text{O})$
$\qquad = (3 \times 0.0 - 110.6 + 74.8 + 241.8) \frac{\text{ kJ}}{\text{ mol}} = +206.1\frac{\text{kJ}}{\text{ mol}}$
and similarly
$\Delta S_{m}^{\circ}(298K) = (3 \times 130.6 + 197.6 - 187.9 - 188.7)\frac{\text{ J}}{\text{ mol K}} = +212.8\frac{\text{ J}}{\text{ mol K}}$
At 600 K we estimate
$\Delta G_{m}^{\circ}=\Delta H^{\circ}(298 \text{ K}) - T\Delta S^{\circ}(298 \text{ K})$
$\qquad =206.1\frac{ \text{ kJ}}{\text{ mol}} - 600 \times 212.8\frac{\text{ J}}{\text{ mol}}$
$\qquad = (206.1 - 127.7)\frac{\text{kJ}}{\text{ mol}} = +78.4\frac{\text{ kJ}}{\text{ mol}}$
Since $\Delta G$ is positive, the reaction is not spontaneous at this temperature
At 1200 K by contrast
$\Delta G_{m}^{\circ}=206.1\frac{\text{ kJ}}{\text{ mol}} - 1200 \times 212.8\frac{\text{ J}}{\text{ mol}}$
$\qquad = (206.1 - 255.4)\frac{\text{ kJ}}{\text{ mol}} = -49.3\frac{\text{ kJ}}{\text{ mol}}$
At this higher temperature, therefore, the reaction is spontaneous.
From more extensive tables we find that accurate values of the free-energy change are $\Delta G_{m}^{\circ}(600 \text{ K}) = +72.6\frac{\text{ kJ}}{\text{ mol}}$ and $\Delta G_{m}^{\circ}(1200 \text{ K})=-77.7\frac{\text{ kJ} }{\text{ mol}}$. Our approximate value at 1200 K is thus about 50 percent in error. Nevertheless it predicts the right sign for $\Delta G$, a result which is adequate for most purposes.
1. Baird, C., Cann, M. Environmental Chemistry. 3rd edition. 2005. W. H. Freeman and Company. 83-85.
16.15: Maximum Useful Work
The Gibbs free energy has another very useful property. When a spontaneous chemical reaction occurs, the decrease in free energy,–ΔG corresponds to the maximum possible quantity of useful work, wmax, which can be obtained. Symbolically,
$-\Delta G = w_{max} \nonumber$
For a reaction which is not spontaneous ΔG is positive and wmax is negative. This means that work must be done on the system (through some outside intervention) to force the nonspontaneous reaction to occur. The minimum work that must be done is given by ΔG. As an example of the utility of this interpretation of ΔG, consider the recovery of Al from Al2O3 ore:
$\ce{Al2O3→2Al + \frac{3}{2}O2}\qquad \Delta G^{\circ}_{m}(298 \text{ K}) = 1576.4 \frac{\text{ kJ}}{\text{ mol}} \nonumber$
The positive ΔG tells us that at least 1576.4 kJ of work must be done on 1 mol Al2O3 to effect this change. In a modern aluminum manufacturing plant this work is supplied electrically, and the electricity is often provided by burning coal. Assuming coal to be mainly carbon, we can write
$\ce{C(s) + O2(g)→CO2(g)}\qquad \Delta G^{\circ}_{m}(298 \text{ K}) = -394.4\frac{\text{ kJ}}{\text{ mol}} \nonumber$
Thus 1 mol C can do almost exactly one-quarter the work required to decompose 1 mol Al2O3 and we must burn at least 4 mol C to process each 1 mol Al2O3 ore. (In practice the aluminum smelting process is only 17 percent efficient, so it is necessary to burn nearly 6 times the theoretical 4 mol C.)
In the context we have just described, free energy is energy that is available to do useful work, not energy that we can get for nothing. When a spontaneous process occurs and there is a free energy decrease, it is the availability of useful energy which decreases. According to the first law of thermodynamics, energy cannot be consumed in any process, but according to the second law, free (or available) energy is always consumed in a spontaneous process.
When we talk about consuming energy resources by burning fossil fuels, it is the availability of energy that is used up. The energy originally stored in a fuel is converted to heat energy and dispersed to the surroundings. Once this has happened its usefulness is lost. There is no way of abstracting this energy from the surroundings and using it to lift a weight or do other useful work, because that would correspond to the reversal of a spontaneous process. The second law thus adds a very important qualification to the first law. While the first law tells us that we cannot destroy energy, the second law tells as that we cannot recycle it either. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.14%3A_The_Free_Energy.txt |
In discussing entropy and spontaneity, we have tended to treat chemical reactions as though they had no other alternatives than either going to completion or not occurring at all. It is only when both reactants and products are pure solids or pure liquids, however, that we find this all-or-nothing-at-all type of behavior. Reactions which involve gases or solutions are governed by an equilibrium constant, and as we saw in the sections on chemical equilibrium, this means that there is always some reactant and some product in the equilibrium mixture. Consequently such reactions must always occur to some extent, no matter how minute, and can never go quite to completion.
Figure $1$ illustrates how the free energy G varies as the reaction proceeds in the two cases. If only pure solids and liquids are involved, then a plot of G against the extent of the reaction is a straight line, as shown in part a of the figure for the reaction
$\ce{Hg(l) + HgBr2(s)→Hg2Br2(s)}\qquad 1 \text{ atm, 298K} \nonumber$
In such a case, if ΔGm°, the free-energy difference between reactants and products, is negative, then the reaction will attain the lowest value of G possible by going to completion.
When gases and solutions are involved in the reaction, a plot of G against the extent of the reaction is no longer a straight line but exhibits a “sag,“ as shown in Figure $1$ b for the reaction
$\ce{2NO2(g)→N2O4(g)}\qquad 1 \text{ atm}\ \label{2}$
at two different temperatures. In such a case, even though ΔGm° is negative, the reaction will not go to completion but will end up at the lowest point of the curve. If the value of ΔGm° is quite small, below about 10 or 20 kJ mol–1, this results in an eventual equilibrium mixture containing an appreciable proportion of both reactants and products. This is the case for Eq. $\ref{2}$ at 298 K, when the reaction attains only 81 percent completion at equilibrium. A more usual situation is that shown for this same reaction at 200 K. Because the free-energy difference is larger at this temperature, the “sag” in the curve is much smaller, and as a result the minimum value of G lies extremely close to 100 percent completion (actually 99.9 percent). As a rough rule of thumb, therefore, we can say that if ΔGm° is negative and numerically greater than 20 kJ mol–1, the reaction will go virtually to completion; if ΔGm° is positive and larger than 20 kJ mol–1, the reaction will scarcely occur at all. Between the limits ΔGm° ± 20 kJ mol–1, we can expect measurable quantities of both reactant and product to be present at equilibrium. The above discussion suggests that there must be some relationship between the free-energy change ΔGm° and the equilibrium constant. The derivation of this relationship is too complex to produce here, so that only the result will be given. For gases the relationship has the form
${\Delta G_{m}^{ o}}={-RT}{\text{ ln } K^{o}} \label{3}$
or, if base 10 logarithms are used,
${\Delta G_{m}^{ o}}={-2.303 RT}{\text{ ln } K^{o}} \label{4}$
Where K° (K standard) is called the standard equilibrium constant. K° is closely related to Kp and differs from it only by being a dimensionless number. Recall from the section on the equilibrium constant in terms of pressure that Kp is defined for the general chemical reaction
$a \text{A} + b \text{B} + \cdots \rightleftharpoons c \text{C} + d \text{D} + \cdots \nonumber$
by the equation
$K_{p}=\dfrac{p_{\text{C}}^{c}\text{ }\times \text{ }p_{\text{D}}^{d}\text{ }\times \text{ }\cdots }{p_{\text{A}}^{a}\text{ }\times \text{ }p_{\text{B}}^{b}\text{ }\times \text{ }\cdots } \nonumber$
etc., are partial pressures. The definition of K° is identical except that it involves pressure ratios (i.e., pure numbers) rather than pressures:
$K\text{ }\!\!{}^\circ\!\!\text{ }=\dfrac{\left( \dfrac{p_{\text{C}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{c}\text{ }\times \text{ }\left( \dfrac{p_{\text{D}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{d}\text{ }\times \text{ }\cdots }{\left( \dfrac{p_{\text{A}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{a}\text{ }\times \text{ }\left( \dfrac{p_{\text{B}}}{p\text{ }\!\!{}^\circ\!\!\text{ }} \right)^{b}\text{ }\times \text{ }\cdots } \nonumber$
Where p° is a standard pressure—almost always 1 atm (101.325 kPa). Thus if the partial pressures pa, pb, etc., are expressed in atmospheres, K° involves the same number as Kp.
Example $1$: ΔGm°
The equilibrium constant Kp for the reaction
$\ce{2NO2 <=> N2O4} \nonumber$
is 0.0694 kPa–1 at 298 K. Use this value to find ΔGm°(298 K) for the reaction.
Solution
We must first express Kp in atmospheres:
$K_{p}=\text{0}\text{.0694 kPa}^{-\text{1}}=\dfrac{\text{0}\text{.0694}}{\text{kPa}}\text{ }\times \text{ }\dfrac{\text{101}\text{.3 kPa}}{\text{1 atm}}=\text{7}\text{.03 atm}^{-\text{1}} \nonumber$
Thus
$K^\circ = 7.03 \qquad \text{ a pure number}$
From Eq. $\ref{4}$ we now have
\begin{align} \Delta G_m^{\circ} & = –2.303RT \text{ log } K^{\circ} \ &= – 8.3143 \text{ J K}^{-1} \text{ mol}^{-1} \times 298 \text{ K} \times 2.303 \times \text{ log } 7.03 \ & = –4833 \text{ J mol}^{-1} = –4.833 \text{ kJ mol}^{-1} \end{align}
Note: You may also use ln x as well as log x as long as you use Eq. $\ref{3}$ directly:
$\Delta G_m^{\circ} = –RT \text{ ln } K^{\circ} = – RT \times \text{ ln } 7.03 = – RT \times 1.950 = –4.833 \nonumber$
In the section on the molecular view of equilibrium, we argued that the value of an equilibrium constant is the product of two factors:
${K}=\text{energy factor}\times\text{probability factor} \label{14}$
The energy factor takes account of the fact that a higher-energy species is less likely to occur in an equilibrium mixture than a lower-energy species, especially at low temperatures. The probability factor reflects the fact that if there are a larger number of ways in which a molecule can arrange itself in space, a molecule is more likely to occur in that state than in one for which a smaller number of spatial arrangements is possible.
We are now in a position to make quantitative the qualitative argument presented. Combining
${\Delta G_{m}^{o}}={\Delta H_{m}^{ o}}-{T\Delta S_{m}^{ o}} \nonumber$
with
${\Delta G_{m}^{o}}=\text{RT}{\text{ ln }{K^o}} \nonumber$
we find
${-}\text{RT}{\text{ ln }{K^o}}={\Delta H_{m}^{ o}}-{T\Delta S_{m}^{ o}} \nonumber$
giving us
${ \text{ ln }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{RT} + \dfrac{\Delta S_{m}^{ o}}{R} \label{18}$
or, in base 10 logarithms
${ \text{ log }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} + \dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R} \label{19}$
If we now take the logarithm of each side of Eq. $\ref{14}$, we have
$\text{ log }{K}=\text{ log }(\text{energy factor})+{\text{ log }(\text{probability factor})}\, \nonumber$
This equation has the same form as Eq. $\ref{19}$, and if we confine ourselves to the standard equilibrium constant K°, we can say that
$\text{ log } (\text{energy factor})=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} \label{21}$
and
$\text{ log } (\text{probability factor})=\dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R} \label{22}$
Although we did not formally derive Eq. $\ref{3}$, on which these results are based, we can examine the last two equalities [Eqs. $\ref{21}$ and $\ref{22}$] to see that they agree with our qualitative expectations. If ΔHm° is negative, we know that the products of a reaction have a lower enthalpy (and hence lower energy) than the reactants. Thus we would predict that products would he favored by the energy factor, and this is what Eq. $\ref{21}$ says—the more negative ΔHm°, the more positive the logarithm of the energy factor and the larger the standard equilibrium constant. Also, since T appears in the denominator of the right-hand side of Eq. $\ref{21}$, the smaller the value of T, the larger (and hence more important) the energy factor for a given value of ΔHm°. We have also seen in the section on thermodynamic probability and entropy that an increase in entropy of a system corresponds to an increase in thermodynamic probability. This is precisely what Eq. $\ref{22}$ says. The larger the value of ΔSm°, the larger the probability factor and hence the larger the standard equilibrium constant. Thus our qualitative description of the two factors affecting the equilibrium constant has been refined to the point where macroscopic quantities, ΔHm° and ΔSm°, can be related to what is happening on the microscopic level.
In addition to the feature we have just mentioned, Eq. $\ref{19}$ is useful in another way. If we can measure or estimate ΔHm° and ΔSm° at temperatures other than the usual 298.15 K, we can obtain K° and calculate the extent of reaction as shown in the next example.
Example $2$: Concentration
Find the concentration of NO in equilibrium with air at 1000 K and 1 atm pressure.
$\ce{N2 (g) + O2 (g) -> 2NO (g) } \Delta H_m^{\circ} = 181 \text{kJ mol}^{-1} \nonumber$
$\Delta S_m^{\circ} = 35.6 \text{J K}^{-1} \text{ mol}^{-1} \nonumber$
Solution
We first find K° :
${\text{ log }{K^o}}=\dfrac{-\Delta H_{m}^{ o}}{\text{2}\text{.303}RT} + \dfrac{\Delta S_{m}^{ o}}{\text{2}\text{.303}R}= \text{–9.45 + 1.34}= \text{–8.11} \nonumber$
Thus
$K^{\circ} = 10^{-8.11} = 7.71 \times 10^{-9} \nonumber$
However, since Δn = 0, $K^{\circ} = K_p = 7.71 \times 10^{-9}$
Assuming the air to be 80% N2 and 20% O2, we can write
$p_{\text{N2}} = 0.8 \text{ atm} ~~~~~\text{and}~~~~~ p_{\text{O2}} = 0.2 \text{ atm} \nonumber$
Thus
${K}_{p}=\dfrac{p_{\text{NO}_{\text{2}}}^{\text{2}}}{\text{(}p_{\text{N}_{\text{2}}}\text{)(}p_{\text{O}_{\text{2}}}\text{)}} = \dfrac{p_{\text{NO}_{\text{2}}}^{\text{2}}}{\text{0}\text{.8 atm }\times \text{ 0}\text{.2 atm}} = {\text{7.71}}\times{\text{10}}^{-\text{9}} \nonumber$
Giving
$p_{\text{NO}_{\text{2}}}^{\text{2}} = {\text{7.71}}\times{\text{10}}^{-\text{9}}\times {\text{0.16 atm}}^{\text{2}}={\text{1.23}}\times{\text{10}}^{-\text{9}}{\text{atm}}^{\text{2}}$
${p}_{\text{NO}_2}=\sqrt{\text{1}\text{.23 }\times \text{ 10}^{-9}\text{ atm}^{\text{2}}}= {\text{3.5}}\times{\text{10}}^{- \text{5}}{\text{atm}} ={\text{3.5}}\times{\text{10}}^{- \text{3}}{\text{kPa}}$
But
${c}_{\text{NO}_2}=\dfrac{c_{\text{NO}_{\text{2}}}}{V} = \dfrac{p_{\text{NO}_{\text{2}}}}{RT}$
Thus ${c}_{\text{NO}_2}=\dfrac{\text{3}\text{.5 }\times \text{ 10}^{-\text{3}}\text{ kPa}}{\text{8}\text{.3143 J K}^{-\text{1}}\text{ mol}^{-\text{1}}}\text{ }\times \text{ }\dfrac{\text{1}}{\text{1000 K}} = {\text{4.3}}\times{\text{10}}^{- \text{7}}{\text{dm}}^{- \text{3}} \nonumber$
Almost all combustion processes heat N2 and O2 to high temperatures, producing small concentrations of NO. Under most circumstances this is not of great consequence, but in the presence of sunlight and partially burned gasoline, NO2 can initiate a form of air pollution called photochemical smog. The presence of minute concentrations of NO in the upper atmosphere from high-flying supersonic jet airplanes can also deplete the ozone layer there. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/16%3A_Entropy_and_Spontaneous_Reactions/16.16%3A_Equilibrium_Constants_Revisited.txt |
When an electrical current flows through matter, permanent chemical changes often occur. In some cases electrical energy supplied from an outside source can cause a chemical reaction to take place. Such a process is called electrolysis, and the system to which electricity is supplied is called an electrolytic cell. It is also possible to produce a flow of electricity as a result of a spontaneous chemical reaction. A chemical system which can cause a current to flow in this way is called a galvanic cell or a voltaic cell. Since an electrical current is a flow of electrons or other charged particles, it should come as no surprise that both electrolytic and galvanic cells involve redox reactions.
17: Electrochemical Cells
When an electrical current flows through matter, permanent chemical changes often occur. In some cases electrical energy supplied from an outside source can cause a chemical reaction to take place. Such a process is called electrolysis, and the system to which electricity is supplied is called an electrolytic cell. A typical example of electrolysis is the laboratory preparation of H2(g) and O2(g) by passing electrical current through-water. Electrolysis is quite important in chemical industry, being involved in manufacture of aluminum, chlorine, copper, and numerous other substances.
It is also possible to produce a flow of electricity as a result of a spontaneous chemical reaction. A chemical system which can cause a current to flow in this way is called a galvanic cell or a voltaic cell. An example of a galvanic cell with which you are almost certainly familiar is a flashlight battery. Since an electrical current is a flow of electrons or other charged particles, it should come as no surprise that both electrolytic and galvanic cells involve redox reactions.
In an electrolytic cell electric energy supplied from an outside source causes a nonspontaneous reaction to occur. A galvanic (or voltaic) cell, on the other hand. harnesses a spontaneous reaction to produce electric current. In either kind of cell the electrode at which oxidation occurs is called the anode and the electrode at which reduction occurs is the cathode.
Electrolytic cells have numerous commercial applications. Chlorine, sodium hydroxide, hydrogen, aluminum, magnesium, sodium, calcium, and high-purity copper are some of the more important chemicals produced by electrolysis. Electroplating of metals such as chromium, silver, nickel, zinc, and tin is also quite important. In any electrolysis reaction the amount of substance consumed or produced can be related to the electric charge which passes through the cell by means of the Faraday constant F, which equals 9.649 × 104 C mol–1
A galvanic cell may be represented by an abbreviated notation such as
$\text{Zn}│\text{Zn}^{2+} (1\; M)║ \text{Ag}^{2+}(1\; M)│\text{Ag} \label{1}$
When a cell is written this way, $\dfrac{1}{2}$it is always assumed that the left-hand electrode is an anode and an oxidation half-equation occurs there. The right-hand electrode must then be taken as the cathode and a reduction half-equation is assumed to occur there. The cell reaction is the sum of these two half-equations. If it is spontaneous, our assumptions about anode on the left and cathode on the right were correct. Electrons will be forced into an external circuit on the left and the cell emf is taken to be positive. If the cell reaction written according to the above convention turns out to be nonspontaneous, then its reverse will be spontaneous. Our assumptions about which electrode is the anode and which the cathode must also be reversed, and the cell emf is given a negative sign.
Equation $\ref{1}$ is lots of fun. Because cell emf values indicate whether a process is spontaneous, they are quite useful. They are additive and are conventionally reported as standard electrode potentials. These refer to the emf of a cell with a hydrogen-gas electrode on the left and the electrode whose potential is reported on the right. The standard electrode potential is directly related to the standard free energy change for a reaction, thus allowing direct determination of ΔG°.
Many galvanic cells are of commercial importance. These include dry cells, mercury cells, rechargeable Ni-Cd batteries, and lead storage cells. Fuel cells, in which a continuous supply of both oxidizing and reducing agent is supplied, may eventually become important because of their high efficiencies. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.01%3A_Prelude_to_Electrochemistry.txt |
A typical electrolytic cell can be made as shown in Figure $1$. Two electrical conductors (electrodes) are immersed in the liquid to be electrolyzed. These electrodes are often made of an inert material such as stainless steel, platinum, or graphite. The liquid to be electrolyzed must be able to conduct electricity, and so it is usually an aqueous solution of an electrolyte or a molten ionic compound. The electrodes are connected by wires to a battery or other source of direct current. This current source may be thought of as an “electron pump” which takes in electrons from one electrode and forces them out into the other electrode. The electrode from which electrons are removed becomes positively charged, while the electrode to which they are supplied has an excess of electrons and a negative charge.
The negatively charged electrode will attract positive ions (cations) toward it from the solution. It can donate some of its excess electrons to such cations or to other species in the liquid being electrolyzed. Hence this electrode is in effect a reducing agent. In any electrochemical cell (electrolytic or galvanic) the electrode at which reduction occurs is called the cathode.
The positive electrode, on the other hand, will attract negative ions (anions) toward itself. This electrode can accept electrons from those negative ions or other species in the solution and hence behaves as an oxidizing agent. In any electrochemical cell the anode is the electrode at which oxidation occurs. An easy way to remember which electrode is which is that anode and oxidation begin with vowels while cathode and reduction begin with consonants.
The following video shows this process taking place in a neutral solution of water with some electrolytes present.
As an example of how electrolysis can cause a chemical reaction to occur, suppose we pass a direct electrical current through 1 M HCl. The H3O+ ions in this solution will be attracted to the cathode, and the $\ce{Cl^{–}}$ ions will migrate toward the anode. At the cathode, H3O+ will be reduced to H2 gas according to the half-equation
$\text{2H}^{+} + \text{2e}^{-} \rightarrow \text{H}_2\label{1}$
(As seen in other sections, we shall write H+ instead of H3O+ in half-equations to save time.) At the anode, electrons will be accepted from $\ce{Cl^{–}}$ ions, oxidizing them to Cl2:
$\text{2Cl}^{-} \rightarrow \text{Cl}_2 + \text{2e}^{-} \label{2}$
During electrolysis $\ce{H2(g)}$ and $\ce{Cl2(g)}$ bubble from the cathode and anode, respectively. The overall equation for the electrolysis is the sum of Eqsuations \ref{1} and $\ref{2}$ :
$\text{2H}^{+}(aq) + \text{2Cl}^{-}(aq) \rightarrow \text{H}_2(g) + \text{Cl}_2(g)\label{3}$
or
$\text{2H}_3\text{O}^{+}(aq) + \text{2Cl}^{-}(aq) \rightarrow \text{H}_2(g) + \text{Cl}_2(g) + \text{2H}_2\text{O}(l) \nonumber$
The net reaction in Equation \ref{3} is the reverse of the spontaneous combination of $\ce{H2(g)}$ with C$\ce{Cl2(g)}$ to form $\ce{HCl(aq)}$. Such a result is true of electrolysis in general: electrical current supplied from outside the system causes a non-spontaneous chemical reaction to occur.
Although electrolysis always reverses a spontaneous redox reaction, the result of a given electrolysis may not always be the reaction we want. In an aqueous solution, for example, there are always a great many water molecules in the vicinity of both the anode and cathode. These water molecules can donate electrons to the anode or accept electrons from the cathode just as anions or cations can. Consequently the electrolysis may oxidize and/or reduce water instead of causing the dissolved electrolyte to react. An example of this problem is electrolysis of lithium fluoride, $\ce{LiF}$. We might expect reduction of $\ce{Li^{+}}$ at the cathode and oxidation of $\ce{F^{–}}$ at the anode, according to the half-equations
$\text{Li}^{+}(aq) + \text{e}^{-} \rightarrow \text{Li}(s)\label{5}$
$\text{2F}^{-}(aq) \rightarrow \text{F}_2(g) + \text{2e}^{-} \nonumber$
However, $\ce{Li^{+}}$ is a very poor electron acceptor, and so it is very difficult to force Equation \ref{5} to occur. Consequently, excess electrons from the cathode are accepted by water molecules instead:
$\text{2H}_2\text{O}(l) + \text{2e}^{-} \rightarrow \text{2OH}^{-}(aq) + \text{H}_2(g)\label{7}$
A similar situation arises at the anode. F ions are extremely weak reducing agents—much weaker than H2O molecules—so the half-equation is
$\text{2H}_2\text{O}(l) \rightarrow \text{O}_2(g) + \text{4H}^{+}(aq) + \text{4e}^{-}\label{8}$
The overall equation can be obtained by multiplying Equation $\ref{7}$ by 2, adding it to Equation $\ref{8}$ and combining H+ with OH to form H2O:
$\text{2H}_2\text{O}(l) \rightarrow \text{2H}_2(g) + \text{O}_2(g) \nonumber$
The following video shows the electrolysis of water taking place, using sulfuric acid as a bridge to allow for the transfer of charge. After the electrolysis is complete, the identities of the gases formed are verified using burning splint tests.
Thus this electrolysis reverses the spontaneous combination of H2 and O2 to form H2O. In discussing redox reactions we mention several oxidizing agents, such as which are strong enough to oxidize H2O. At the same time we describe reducing agents which are strong enough to reduce H2O such as the alkali metals and the heavier alkaline earths. As a general rule such substances cannot be produced by electrolysis of aqueous solutions because H2O is oxidized or reduced instead. Substances which undergo spontaneous redox reaction with H2O are usually produced by electrolysis of molten salts or in some other solvent. There are some exceptions to this rule, however, because some electrode reactions are slower than others. Using Table 11.5, for example, we would predict that H2O is a better reducing agent than $\ce{Cl^{–}}$.
Hence we would expect O2, not Cl2, to be produced by electrolysis of 1 M HCl, in contradiction of Equation $\ref{1}$. It turns out that O2 is produced more slowly than Cl2, and the latter bubbles out of solution before the H2O can be oxidized. For this reason Table 1 found in the Redox Couples section cannot always be used to predict what will happen in an electrolysis. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.02%3A_Electrolysis.txt |
Three important chemicals, NaOH, Cl2, H2, can be obtained by electrolyzing an aqueous NaCl solution (brine). This forms the basis of the chlor-alkali industry. The diaphragm cell (also called a Hooker cell) in which the electrolysis is carried out is shown schematically in Figure $1$. At the cathode, water is reduced:
$\ce{2H2O + 2} e^{-} \ce{ -> H2 + 2OH^-} \nonumber$
Chlorine is produced at the anode:
$\text{2Cl}^{-} \rightarrow \text{Cl}_2 + 2e^{-} \nonumber$
Thus the overall reaction is
$\text{2H}_2\text{O}(l) + \text{2Cl}^{-}(aq) \rightarrow \text{H}_2(g) + \text{Cl}_2(g) + \text{2OH}^{-}(aq)\label{3}$
Since the H2(g) and Cl2(g) might recombine explosively should they come in contact, the cathode must be entirely surrounded by a porous diaphragm of asbestos. Hence the name of this type of cell.
Both the H2(g) and Cl2(g) produced in Eq. $\ref{3}$ are dried, purified, and compressed into cylinders. Fresh brine is continually pumped into the cell, and the solution which is forced out contains about 10% NaOH together with a good deal of NaCl. [Remember that the spectator ions, Na+(aq), are not included in a net ionic equation such as Eq. $\ref{3}$. H2O is allowed to evaporate from this solution until the concentration of the solution reaches 50% NaOH, by which time most of the NaCl has crystallized out and can be recycled to the electrolysis. The NaOH is sold as a 50% solution or further dried to give crystals whose approximate formula is NaOH•H2O.
The considerable effort required to concentrate the NaOH solution obtained from diaphragm cells can be avoided by using mercury cells. The cathode in such a cell is mercury, and the cathode reaction is
$\text{Na}^{+}(aq) + e^{-} + \text{xHg}(l) \rightarrow \text{NaHg}_x(l) \nonumber$
The sodium metal produced in this reaction dissolves in the liquid mercury, producing an amalgam. The liquid amalgam is then transferred to an- other part of the cell and reacted with water:
$\text{NaHg}_x(l) + \text{2H}_2\text{O}(l) \rightarrow \text{2Na}^{+}(aq) + \text{2OH}^{-}(aq) + \text{H}_2(g) + \text{xHg}(l) \nonumber$
The 50% sodium hydroxide solution produced by this reaction contains no sodium chloride and can be sold directly, without being concentrated further. Up until 1970, however, chlor-alkali plants using mercury cells did not have adequate controls to prevent losses of mercury to the environment. About 100 to 200 g mercury was lost for each 1000 kg chlorine produced-apparently a small quantity until one realizes that 2 500 000 kg chlorine was produced by mercury cells every day during 1960 in the United States. Thus every 2 to 4 days 1000 kg mercury entered the environment, and by 1970 sizable quantities were being found in fish. Since 1970 adequate controls have been installed on mercury cells and most new alkali plants use diaphragm cells, but the very large quantities of mercury introduced into rivers and lakes prior to 1970 are expected to remain for a century or more.
17.04: Aluminum Production
Aluminum is easily oxidized, and so its ore, Al2O3, is difficult to reduce. In fact water is reduced rather than Al3+(aq), and so electrolysis must be carried out in a molten salt. Even this is difficult because the melting point of Al2O3 is above 2000°C—a temperature which is very difficult to maintain.
The first successful method for reducing Al2O3 is the one still used today. It was developed in the United States in 1886 by Charles Hall (1863 to 1914), who was then 23 years old and fresh out of Oberlin College. Hall realized that if Al2O3 were dissolved in another molten salt, the melting point of the mixture would be lower than for either pure substance. The substance Hall used was cryolite, Na3AlF6, in which the Al2O3 can be dissolved at just over 1000°C.
The electrolytic cell used for the Hall process. (Figure $1$) consists of a steel box lined with graphite. This contains the molten Na3AlF6 and Al2O3 and also serves as the cathode. The anode is a large cylinder of carbon. Passage of electrical current maintains the high temperature of the cell and causes the following half-equations to occur:
$\text{Al}^{3+} + \text{3}e^{-} \rightarrow \text{Al}(l) \label{1}$
$\text{2O}^{2-} + \text{C}(s) \rightarrow \text{CO}_2(g) + \text{4}e^{-}\label{2}$
Since the carbon anode is consumed by the oxidation half-equation, it must be replaced periodically.
Aluminum production requires vast quantities of electrical energy, both to maintain the high temperature and to cause half-equations $\ref{1}$ and $\ref{2}$ to occur. Currently about 5 percent of the total electrical energy produced in the United States goes into the Hall process. Much of this energy comes from combustion of fossil fuels and hence consumes a valuable, nonrenewable resource. Since Al is protected from oxidation back to Al2O3 by a surface coating of oxide, it is a prime candidate for recycling, as well as for applications such as house siding, where it is expected to remain for a long time. Throwing away aluminum beverage cans, on the other hand, is a tremendous waste of energy.
Several other easily oxidized metals are currently produced by electrolysis, but not in such large quantities as Al. Mg is obtained by electrolyzing molten MgCl2 which is derived from seawater, and Na and Ca are produced together from a molten mixture of NaCl and CaCl2 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.03%3A_Electrolysis_of_Brine.txt |
Unrefined or “blister” copper is about 99 percent pure when obtained from the ore, but it is desirable to increase this to 99.95 percent if the copper is to be used in electrical wiring. Even small concentrations of impurities noticeably lower copper’s electrical conductivity. Such a high degree of purity can be obtained by electrolytic refining in a cell similar to that shown in Figure $1$.
In such a cell a thin sheet of high-purity Cu serves as the cathode, and the anode is the impure Cu which is to be refined. The electrolyte is a solution of copper(II) sulfate. Some of the impurities are metals such as Fe and Zn which are more easily oxidized than Cu. When current passes through the cell, these impurities go into solution from the anode, along with Cu:
$\text{Cu}(s) \rightarrow \text{Cu}^{2+}(aq) + \text{2}e^{-} \nonumber$
$\text{Fe}(s) \rightarrow \text{Fe}^{2+}(aq) + \text{2}e^{-} \nonumber$
$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{2}e^{-} \nonumber$
These ions all migrate toward the cathode, but Cu2+(aq) is more readily reduced than Fe2+(aq) or Zn2+(aq) and so it is the only one that plates out. The impurity ions remain in solution. Other impurities, such as Ag, Au, and Pt, are less easily oxidized than Cu. These remain in metallic form and fall to the bottom of the cell, forming “anode sludge” from which they can later be recovered. The great value of Ag, Au, and Pt helps to offset the cost of refining.
17.06: Electroplating
An important industrial application of electrolysis is the plating of one metal on top of another. A typical example is the bumper of a car. This is made from steel and then plated with a thin layer of chromium to make it resistant to rusting and scratching. Many other metal objects, such as pins, screws, watchbands, and doorknobs, are made of one metal with another plated on the surface.
An electroplating cell works in much the same way as the cell used to purify copper. The object to be plated is used as the cathode, and the electrolyte contains some ionic compound of the metal to be plated. As current flows, this compound is reduced to the metal and deposits on the surface of the cathode. In chromium plating, for instance, the electrolyte is usually a solution of potassium dichromate, K2Cr2O7, in fairly concentrated sulfuric acid. In this very acidic solution CrO72 ions are completely protonated, and so the reduction half-equation is
$\text{H}_2\text{Cr}_2\text{O}_7(aq) + \text{12H}^{+}(aq) + \text{12}e^{-} \rightarrow \text{2Cr}(s) + \text{7 H}_2\text{O}(l) \nonumber$
Other metals which are often electroplated are silver, nickel, tin, and zinc.
In the case of silver the electrolyte must contain the polyatomic ion Ag(CN)2 rather than Ag+. Otherwise the solid silver will be deposited as jagged crystals instead of a shiny uniform layer.
17.07: Quantitative Aspects of Electrolysis
Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electrical charge Q which passes through the cell. For example, the half-equation
$\text{Ag}^{+} + e^{-} \rightarrow \text{Ag} \nonumber$
tells us that when 1 mol Ag+ is plated out as 1 mol Ag, 1 mol e must be supplied from the cathode. Since the negative charge on a single electron is known to be 1.6022 × 10–19 C, we can multiply by the Avogadro constant to obtain the charge per mole of electrons. This quantity is called the Faraday constant, symbol F:
F = 1.6022 × 10–19 C × 6.0221 × 1023 mol–1 = 9.649 × 104 C mol–1
Thus in the case of Eq. (1), 96 490 C would have to pass through the cathode in order to deposit 1 mol Ag. For any electrolysis the electrical charge Q passing through an electrode is related to the amount of electrons ne– by
$\text{F}=\frac{Q}{n_{e^{-}}} \nonumber$
Thus F serves as a conversion factor between ne– and Q.
Example $1$: Electrical Charge
Calculate the quantity of electrical charge needed to plate 1.386mol Cr from an acidic solution of K2Cr2O7 according to half-equation $\ce{H2Cr2O7}(aq) + \text{12H}^{+}(aq) + \text{12}e^{-} \rightarrow \ce{2Cr}(s) + \ce{7H2O}(l)\label{3}$.
Solution
According to Eq. $\ref{3}$, 12 mol e is required to plate 2 mol Cr, giving us a stoichiometric ratio S(e/Cr). Then the Faraday constant can be used to find the quantity of charge. In road-map form
nCr $\xrightarrow{S\text{(}e^{-}\text{/Cr)}}$ ne– $\xrightarrow{F}$ Q
Q = 1.386 mol Cr × $\frac{\text{12 mol }e^{-}}{\text{2 mol Cr}}$ × $\frac{\text{9}\text{.649 }\times \text{ 10}^{\text{4}}\text{ C}}{\text{ 1 mol }e^{-}}$ = 8.024 × 105 C
Often the electrical current rather than the quantity of electrical charge is measured in an electrolysis experiment. Since a coulomb is defined as the quantity of charge which passes a fixed point in an electrical circuit when a current of one ampere flows for one second, the charge in coulombs can be calculated by multiplying the measured current (in amperes) by the time (in seconds) during which it flows:
$\text{Q} = \text{It} \nonumber$
In this equation I represents current and t represents time. If you remember that
coulomb = 1 ampere × 1 second 1 C = 1 A s
you can adjust the time units to obtain the correct result.
Example $2$ : Mass of Hydrogen Peroxide
Hydrogen peroxide, H2O2, can be manufactured by electrolysis of cold concentrated sulfuric acid. The reaction at the anode is
$\ce{2H2SO4 -> H2S2O8 + 2H}^{+} + \text{2}e^{-}\label{5}$
When the resultant peroxydisulfuric acid, H2S2O8, is boiled at reduced pressure, it decomposes:
$\ce{2H2O + H2S2O8 -> 2H2SO4 + H2O2}\label{6}$
Calculate the mass of hydrogen peroxide produced if a current of 0.893 flows for 1 h.
Solution
he product of current and time gives us the quantity of electricity, Q. Knowing this we easily calculate the amount of electrons, ne–. From half-equation $\ref{5}$ we can then find the amount of peroxydisulfuric acid. Equation $\ref{6}$ then leads to nH2O2 and finally to mH2O2. The road map to describe this logic is as follows:
$I\xrightarrow{t}Q\xrightarrow{F}n_{e^{-}}\xrightarrow{S_{e}}n_{\text{H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\xrightarrow{S}n_{\text{H}_{\text{2}}\text{O}_{\text{2}}}\xrightarrow{M}m_{\text{H}_{\text{2}}\text{O}_{\text{2}}} \nonumber$
so that
$m_{\text{H}_{\text{2}}\text{O}_{\text{2}}}=\text{0}\text{.893 A }\times \text{ 3600 s }\times \text{ }\frac{\text{1 mol }e^{-}}{\text{96 490 C}}\text{ }\times \text{ }\frac{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}{\text{2 mol }e^{-}} \nonumber$
$=\frac{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{S}_{\text{2}}\text{O}_{\text{8}}}\text{ }\times \text{ }\frac{\text{34}\text{.01 g H}_{\text{2}}\text{O}_{\text{2}}}{\text{1 mol H}_{\text{2}}\text{O}_{\text{2}}} \nonumber$
= 05666 $\frac{\text{A s}}{\text{C}}$ × g H2O2 = 0.5666 g H2O2 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.05%3A_Refining_of_Copper.txt |
In an electrolytic cell electrical energy is consumed and an otherwise spontaneous redox reaction is reversed. A galvanic cell, on the other hand, produces electrical energy as a result of a spontaneous redox process. The electron transfer characteristic of such a process is made to occur in two separate half-cells. Electrons released during an oxidation half-equation must flow through a wire or other external circuit before they can be accepted in a reduction half-equation. Consequently an electrical current is made to flow.
A typical galvanic cell, the Daniell cell, was used to power telegraphs 100 years ago. This cell is based on the spontaneous redox reaction
$\text{Zn}(s) + \text{Cu}^{2+}(aq) \rightarrow \text{Zn}^{2+}(aq) + \text{Cu}(s)\label{1}$
(You can verify that this reaction is spontaneous by dipping a piece of zinc metal in a copper sulfate solution. In a short time the surface of the zinc will become plated with red-brown copper metal.) The half-equations
$\text{Zn}(s) \rightarrow \text{Zn}^{2+}(aq) + \text{2}e^{-}\label{2}$
$\text{Cu}^{2+}(aq) + \text{2}e^{-} \rightarrow \text{Cu}(s)\label{3}$
indicate that for each mole of zinc which is oxidized and goes into solution as zinc ions, 2 mol electrons are transferred to copper ions, converting them to copper atoms.
To produce electrical current we must prevent the Zn(s) from contacting the Cu2+(aq) ions and transferring the electrons directly. This is done in the Daniell cell by pouring a concentrated copper sulfate solution into the bottom of a glass jar and then carefully pouring a layer of less concentrated zinc sulfate solution above it. Because it contains less solute per unit volume, the zinc sulfate solution is less dense. It floats on the copper sulfate and does not mix with it. Therefore a copper electrode placed in the bottom of the jar contacts only Cu2+(aq) ions, and a zinc electrode suspended in the zinc sulfate solution contacts only Zn2+(aq) ions.
In the laboratory it is more convenient to set up a cell based on the Zn + Cu2+ reaction, as shown in Figure $1$. Two electrodes or half-cells are separated by a salt bridge. This contains an electrolyte, KCl, so that current can flow from one half-cell to the other, but the contents of the two half-cells cannot mix. The left-hand electrode in Figure $1$ is a Zn rod dipping in a solution of ZnSO4. Thus both components of the Zn2+/Zn redox couple are present, and the metal electrode can conduct electrons produced by Eq. $\ref{2}$ to the wire in the external circuit. Since oxidation of Zn to Zn2+ occurs at the left-hand electrode, this electrode is the anode.
The right-hand electrode is a strip of Cu dipping in a solution of CuSO4. Here both components of the Cu2+/Cu redox couple are present, and Eq. $\ref3$ can occur. Electrons supplied by the external circuit are conducted through Cu to the electrode surface, where they combine with Cu2+ ions to produce more Cu. Since reduction occurs at this right-hand electrode, this electrode is the cathode. The net effect of the two half-cells is that electrons are forced into the external circuit at the anode and withdrawn from it at the cathode. This will cause current to flow, or, if current is prevented from flowing by a device such as the voltmeter in Figure $1$, it will cause an electrical potential difference (voltage) to build up.
The components of the redox couples at the electrodes in a galvanic cell need not always be a solid and a species in solution. This is evident from Figure $2$. In this case the spontaneous redox reaction
$\text{2Fe}^{2+}(aq) + \text{Cl}_2(g) \rightarrow \text{2Fe}^{3+}(aq) + \text{2Cl}^{-}(aq)\label{4}$
is involved. The oxidation half-equation at the anode is
$\text{Fe}^{3+}(aq) \rightarrow \text{Fe}^{2+}(aq) + e^{-}\label{5}$
Thus at the right-hand electrode in Figure $2$ both components of the redox couple are in aqueous solution. Reaction $\ref{5}$ occurs at the surface of the platinum wire, which conducts the released electrons to the external circuit.
The left-hand electrode in Figure $2$ is a gas electrode. It consists of a platinum strip dipping in a solution which contains chloride ions. The electrode is surrounded by a glass tube through which chlorine gas can be pumped. At this electrode the reaction is a reduction:
$\text{Cl}_2(g) + \text{2}e^{-} \rightarrow \text{2Cl}^{-}(aq)\label{6}$
Therefore the left-hand electrode is the cathode. Since electrons are forced into the external circuit at the anode and withdrawn at the cathode, electrons flow from right to left in this cell. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.08%3A_Galvanic_Cells.txt |
Rather than drawing a complete diagram like the figures in the Galvanic Cells section, it is convenient to specify a galvanic cell in shorthand form. The two cells we have just described would be written as
$\ce{Zn} | \ce{Zn^{2+}(1 \,M)}\, || \, \ce{Cu^{2+}(1\,M)} | \ce{Cu}\label{1}$
and
$\ce{Pt(s)} | \ce{Cl_{2}(g)} | \ce{Cl^{–}(1\, M)}\, ||\, \ce{Fe^{2+}(1 \,M), Fe^{3+}(1\, M)} | \ce{Pt(s)} \label{2}$
The components of the cell are written in order, starting with the left-hand and moving across the salt bridge to the right. A single vertical line indicates a phase boundary, such as that between the solid $\ce{Zn}$ electrode and $\ce{Zn^{2+}(aq)}$, or between $\ce{Cl2(g)}$ and $\ce{Cl^{–}(aq)}$. The double vertical line represents a salt bridge. Spectator ions, like $\ce{SO4^{2–}(aq)}$ in the $\ce{Zn–Cu}$ cell, are usually omitted.
By convention, the electrode written to the left of the salt bridge in this cell notation is always taken to be the anode, and the associated half-equation is always written as an oxidation. The right-hand electrode is therefore always the cathode, and the half-equation is always written as a reduction. This is easy to remember, because reading from left to right gives anode and cathode in alphabetical order. The cell reaction corresponding to a given shorthand description is obtained by summing the half-equations after multiplying by any factors necessary to equalize the number of electrons lost at the anode with the number gained at the cathode. Unless otherwise stated, standard conditions of 1 M are usually implied. However, non-standard cells can be created.
Example $1$: Half-Reactions
Write the half-equations and cell reactions for each of the following cells:
1. $\ce{Ag} \mid \ce{Ag^{+}} \parallel \ce{H^{+}} \mid \ce{H_{2} \mid Pt(s)}$
2. $\ce{Pt(s)} \mid \ce{Cr_{2}O_{7}^{2–}, Cr^{3+}, H^{+}} \parallel \ce{Br^{–}} \mid \ce{Br_{2}(l) \mid Pt(s)}$
Solution
a) Since the half-equation at the left electrode is taken by convention to be oxidation, we have
$\ce{Ag} \rightarrow \ce{Ag^{+} + }e^– \nonumber$
At the right-hand electrode, then, we must have a reduction:
$\ce{2H^{+} + 2}e^{–} \ce{\rightarrow H_{2}} \nonumber$
Multiplying the first half-equation by 2 and summing gives the cell reaction
$\ce{2Ag + 2H^{+} \rightarrow 2Ag^{+} + H_{2} (g)} \nonumber$
b) Following the same procedure as in part a, we obtain
$\ce{2Cr^{3+} + 7H_{2}O \rightarrow Cr_{2}O_{7}^{2–} + 14H^{+} + 6}e^{-} \nonumber$ Left electrode
$\ce{(Br_{2}(l) + 2}e^– \ce{\rightarrow 2Br^{–}) \times 3} \nonumber$ Right electrode
$\ce{2Cr^{3+} + 7H_{2}O + 3Br_{2}(l) \rightarrow Cr_{2}O_{7}^{2–} + 14H^{+} + 6Br^{-}} \nonumber$ Cell reaction
Note: The procedures described in other sections for balancing redox equations were used in arriving at the above results.
Example $2$: Galvanic Cell
Describe in shorthand notation a galvanic cell for which the cell reaction is
$\ce{Cu(s) + 2Fe^{3+}(aq) \rightarrow Cu^{2+}(aq) + 2Fe^{2+}(aq)} \nonumber$
Solution
First divide the cell reaction into half-equations:
• Oxidation: $\ce{Cu(s) \rightarrow Cu^{2+}(aq) + 2}e^{–}$
• Reduction: $\ce{Fe^{3+} + }e^– \ce{\rightarrow Fe^{2+}}$
Then write the oxidation as the left-hand electrode and the reduction on the right:
$\ce{Cu} \mid \ce{Cu^{2+}} \parallel \ce{Fe^{2+}, Fe^{3+}} \mid \ce{Pt} \nonumber$
(Since both Fe2+ and Fe3+ are in solution, a Pt electrode is used.)
The conventions we have developed can be used to decide whether the cell reaction is actually spontaneous. If it is, an oxidation will release electrons to the external circuit at the left-hand electrode. If a voltmeter is placed in the circuit, these electrons will make its left-hand terminal negative. Since the right-hand electrode corresponds to reduction, electrons will be withdrawn from the right-hand terminal of the voltmeter. This is shown for the Zn–Cu cell in Figure 1 from the Galvanic Cells section. You can readily confirm that the spontaneous cell reaction (Eq. (1) from Galvanic Cells) corresponds to the shorthand cell notation of Equation $\ref{1}$.
For the cell shown in Figure 1 in Galvanic Cells, the shorthand notation is
$\ce{Pt(s)} | \ce{Cl_2(g)} \mid \ce{Cl^{–}(1 M)} \parallel \ce{Fe^{2+}(1 M), Fe^{3+}(1 M)} \mid \ce{Pt(s)} \nonumber$
According to the conventions we have just developed, this corresponds to the cell reaction
$\ce{2Cl^{–}(aq) + 2Fe^{3+}(aq) \rightarrow Cl_2(g) + 2Fe^{2+}(aq)} \nonumber$
But this is the of the spontaneous cell reaction we described before (Eq. (2) from Galvanic Cells). Since the cell reaction is nonspontaneous, electrons will not be forced into the external circuit at the left-hand electrode, and they will not be withdrawn at the right. In fact the reverse will actually occur. Thus if a voltmeter is connected to this cell, its right-hand terminal will become more negative and its left-hand terminal will become more positive. This is shown in Figure 2 from Galvanic Cells.
In general, if a galvanic cell is connected to a voltmeter, the electrode connected to the negative terminal of the meter must be the anode. If our shorthand cell notation shows that electrode on the left, then the corresponding cell reaction must be spontaneous. Electrons will be released by the oxidation half-equation on the left and accepted by the reduction on the right. If, on the other hand, the voltmeter shows that the right-hand electrode is releasing electrons, then we must have written our shorthand notation backwards. This means that the reverse of the cell reaction obtained by our rules must actually be occurring, and it is that reverse reaction which is spontaneous. Thus by simply observing which electrode in the cell releases electrons and which accepts them, that is, by finding which electrode is negative and which positive, we can determine whether the cell reaction is spontaneous. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.09%3A_Cell_Notation_and_Conventions.txt |
Using a voltmeter to measure the electrical potential difference (commonly called voltage) between two electrodes provides a quantitative indication of just how spontaneous a redox reaction is. The potential difference is measured in volts (V), an SI unit which corresponds to one joule per ampere-second (1V = 1 J A–1 s–1). The voltage indicates the tendency for current to flow in the external circuit, that is, it shows how strongly the anode reaction can push electrons into the circuit and how strongly the cathode reaction can pull them out. The potential difference is greatest when a large electrical resistance in the external circuit prevents any current from flowing. The maximum potential difference which can be measured for a given cell is called the electromotive force, abbreviated emf and represented by the symbol E
By convention, when a cell is written in shorthand notation, its emf is given a positive value if the cell reaction is spontaneous. That is, if the electrode on the left forces electrons into the external circuit and the electrode on the right withdraws them, then the dial on the voltmeter gives the cell emf. On the other hand, if the half-cell on the right side of the shorthand cell notation is releasing electrons, making the right-hand terminal of the voltmeter negative, the cell emf is minus the reading of the meter. This corresponds to a nonspontaneous cell reaction, written in the conventional way.
Example $1$ : Galvanic Cell EMF
When the galvanic cell shown in Figure 2 from Galvanic Cells is connected to a voltmeter, the reading is 0.59 V. The shorthand notation for this cell is
$\ce{Pt , Cl_2(g)} \mid \ce{Cl^{–}(1 M)} \parallel \ce{Fe^{2+}(1 M), Fe^{3+}(1 M)} \mid \ce{Pt} \nonumber$
What is the value of the cell emf?
Solution:
We have already seen that this cell as written corresponds to a nonspontaneous reaction. Therefore the emf must be negative and E = – 0.59 V.
Example $2$ : Voltmeter Readings
If the voltmeter in Figure 17.5 reads 1.10 V, what is the emf for the cell
$\ce{Cu} \mid \ce{Cu^{2+}(1 M)} \parallel \ce{Zn^{2+}(1 M)} \mid \ce{Zn} \nonumber$
Solution
In this case the shorthand notation corresponds to the reverse of Eq. (1) in Galvanic Cells; that is, it refers to the nonspontaneous cell reaction
$\ce{Cu + Zn^{2+} \rightarrow Cu^{2+} + Zn} \nonumber$
Consequently the emf for this cell must be negative and E = –1.10 V.
Example $2$ shows that if the cell notation is written in reverse, the cell emf changes sign, since for the spontaneous reaction shown in Eq. (2) from Galvanic Cells the emf would have been +1.10 V.
Experimentally measured cell emf's are found to depend on the concentrations of species in solution and on the pressures of gases involved in the cell reaction. Consequently it is necessary to specify concentrations and pressures when reporting an emf, and we shall only consider cells in which all concentrations are 1mol dm–3 and all pressures are 1 atm (101.3 kPa).
The emf of such a cell is said to be its standard electromotive force and is given the symbol E°.
The electromotive forces of galvanic cells are found to be additive. That is, if we measure the emf’s of the two cells
$\ce{Zn} \mid \ce{Z^{2+}(1 M)} \parallel \ce{H^{+}(1 M)} \mid \ce{H_{2}(1 atm), Pt}\label{4}$ E° = 0.76 V
$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Cu^{2+}(1 M)} \mid \ce{Cu}\label{5}$ E° = 0.34 V
the sum of the E° values corresponds to the measured emf for a third cell with which we discuss in the section on Cell Notation and Conventions:
$\ce{Zn \mid Zn^{2+} (1 M)} \parallel \ce{Cu^{2+}(1 M)} \mid \ce{Cu E^{o} = 1.10}\label{6}$
Whenever the right-hand electrode of one cell is identical to the left-hand electrode of another, we can add the emf's in this way, canceling the electrode which appears twice. This additivity makes it possible to store a large amount of emf data in a small table. By convention such data are tabulated as standard reduction potentials. These refer to the emf of a cell whose left-hand electrode is the hydrogen-gas electrode and whose right-hand electrode is the electrode whose emf is being sought. Table $1$ contains a number of useful standard reduction potentials.
As an example of the use of the table, the entry corresponding to the electrode $Cu^{2+}(1 M)\mid Cu is + 0.34 V$. Thus when this electrode is written
TABLE $1$: Standard Reduction Potentials at 298.15 K.
Acidic Solution Standard Reduction Potential, E° (volts)
$\ce{F_{2}(g) + 2}e^– \ce{/rightarrow 2F^{–} (aq)}$ $2.87$
$\ce{Co^{3+}(aq) + }e^– \ce{/rightarrow Co^{2+}(aq)}$ $1.92$
$\ce{Au^{+}(aq) + }e^– \ce{\rightarrow Au(s)}$ $1.83$
$\ce{H_{2}O_{2}(aq) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow 4H_{2}O(l)}$ $1.763$
$\ce{Ce^{4+}(aq) + }e^– \ce{\rightarrow Ce^{3+}(aq)}$ $1.72$
$\ce{Pb^{4+}(aq) + 2}e^– \ce{\rightarrow Pb^{2+}(aq)}$ $1.69$
$\ce{PbO_{2}(s) + SO_{4}^{2−}(aq) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow PbSO_{4}(s) + 6H_{2}O(l)}$ $1.690$
$\ce{NiO_{2}(s) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow Ni^{2+}(aq) + 6H_{2}O(l)}$ $1.68$
$\ce{2HClO(aq) + 2H_\ce{3}O^\ce{+}(aq) + 2}e^– \ce{\rightarrow Cl_{2}(g) + 4H_{2}O(l)}$ $1.63$
$\ce{Au^{3+}(aq) + 3}e^− \ce{\rightarrow Au(s)}$ $1.52$
$\ce{MnO_4^{−}(aq) + 8H_{3}O^{+}(aq) + 5}e^– \ce{\rightarrow Mn^{2+}(aq) + 12H_{2}O(l)}$ $1.51$
$\ce{BrO_{3}^{−}(aq) + 6H_{3}O^{+}(aq) + 5}e^− \ce{\rightarrow \frac{1}{2} Br_{2}(aq) + 9H_{2}O(l)}$ $1.478$
$\ce{2ClO_{3}^{−}(aq) + 12H_{3}O^{+}(aq) + 10}e^– \ce{\rightarrow Cl_{2}(g) + 18H_{2}O(l)}$ $1.47$
$\ce{Cr_{2}O_{7}^{2−}(aq) + 14H_{3}O^{+}(aq) + 6}e^– \ce{\rightarrow 2Cr^{3+}(aq) + 21H_{2}O(l)}$ $1.36$
$\ce{Cl_{2} (g) + 2}e^− \ce{\rightarrow 2Cl^{−}(aq)}$ $1.358$
$\ce{N_{2}H_{5}^+(aq) + 3H_{3}O^{+}(aq) + 2}e^- \ce{\rightarrow 2NH_{4}^{+}(aq) + 3H_{2}O(l)}$ $1.275$
$\ce{MnO_{2}(s) + 4H_{3}O^+(aq) + 2}e^– \ce{\rightarrow Mn^{2+}(aq) + 6H_{2}O(l)}$ $1.23$
$\ce{O_{2}(g) + 4H_{3}O^{+}(aq) + 4}e^– \ce{\rightarrow 6H_{2}O(l)}$ $1.229$
$\ce{ClO_{4}^{−}(aq) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow ClO_{3}^{−}(aq) + 3H_{2}O(l)}$ $1.201$
$\ce{IO_{3}^{−}(aq) + 6H_{3}O^+(aq) + 5}e^– \ce{\rightarrow \frac{1}{2} I_{2}(aq) + 9H_{2}O(l)}$ $1.195$
$\ce{Pt^{2+}(aq) + 2}e^− \ce{\rightarrow Pt(s)}$ $1.188$
$\ce{Br_{2}(l) + 2}e^− \ce{\rightarrow 2Br^{-}(aq)}$ $1.066$
$\ce{AuCl_{4}^{−}(aq) + 3}e^- \ce{\rightarrow Au(s) + 4Cl^{−}(aq)}$ $1.00$
$\ce{NO_{3}^{−}(aq) + 4H_{3}O^{+}(aq) + 3}e^{–} \ce{\rightarrow NO(g) + 6H_{2}O(l)}$ $0.96$
$\ce{NO_{3}^{−}(aq) + 3H_{3}O^+(aq) + 2}e^– \ce{\rightarrow HNO_{2}(aq) + 4H_{2}O(l)}$ $0.94$
$\ce{Pd^{2+}(aq) + 2}e^− \ce{\rightarrow Pd(s)}$ $0.915$
$\ce{2Hg^{2+}(aq) +2}e^− \ce{\rightarrow Hg_{2}^{2+}(aq)}$ $0.9110$
$\ce{Hg^{2+}(aq) +2}e^− \ce{\rightarrow Hg(l)}$ $0.8535$
$\ce{SbCl_{6}^{−}(aq) + 2}e^− \ce{\rightarrow SbCl_{4}^{−}(aq) + 2Cl^{−}(aq)}$ $0.84$
$\ce{Ag^{+}(aq) + }e^− \ce{\rightarrow Ag(s)}$ $0.7991$
$\ce{Hg_{2}^{2+}(aq) + 2}e^- \ce{\rightarrow 2Hg(l)}$ $0.7960$
$\ce{Fe^{3+}(aq) + }e^− \ce{\rightarrow Fe^{2+}(aq)}$ $0.771$
$\ce{[PtCl_{4}]^{2−}(aq) + 2}e^- \ce{\rightarrow Pt(s) + 4Cl^{–}(aq)}$ $0.758$
$\ce{[PtCl_{6}]^{2−}(aq) + 2}e^− \ce{\rightarrow [PtCl_4]^{2−}(aq) + 2Cl^{–}(aq)}$ $0.726$
$\ce{O_{2}(g) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}O_{2}(aq) + 2H_{2}O(l)}$ $0.695$
$\ce{TeO_{2}(s) + 4H_{3}O^+(aq) + 4}e^– \ce{\rightarrow Te(s) + 6H_{2}O(l)}$ $0.604$
$\ce{H_{3}AsO_{4}(aq) + 2H_{3}O^{+}(aq) + 2}e^− \ce{\rightarrow HAsO_{2}(aq) + 4H_{2}O(l)}$ $0.560$
$\ce{I_{2}(s) + 2}e^− \ce{\rightarrow 2I^{−}(aq)}$ $0.535$
$\ce{Cu^{+}(aq) + }e^− \ce{\rightarrow Cu(s)}$ $0.521$
$\ce{[RhCl_{6}]^{3−}(aq) + 3}e^− \ce{\rightarrow Rh(s) + 6Cl^{–}(aq)}$ $0.5$
$\ce{Cu^{2+}(aq) + 2}e^− \ce{\rightarrow Cu(s)}$ $0.340$
$\ce{Hg_{2}Cl_{2}(s) + 2}e^− \ce{\rightarrow 2Hg(l) + 2Cl^{−}(aq)}$ $0.27$
$\ce{AgCl(s) + }e^− \ce{\rightarrow Ag(s) + Cl^{−}(aq)}$ $0.222$
$\ce{Cu^{2+}(aq) + }e^− \ce{\rightarrow Cu^{+}(aq)}$ $0.159$
$\ce{SO_{4}^{2−}(aq) + 4H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}SO_{3}(aq) + 5H_{2}O(l)}$ $0.158$
$\ce{Sn^{4+}(aq) + 2}e^− \ce{\rightarrow Sn^{2+}(aq)}$ $0.15$
$\ce{S(s) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}S(aq) + 2H_{2}O(l)}$ $0.144$
$\ce{AgBr(s) + }e^− \ce{\rightarrow Ag(s) + Br^{−}(aq)}$ $0.0713$
$\ce{2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow 2H_{2}(g) + 2H_{2}O(l )}$ (reference electrode) $0.0000$
$\ce{N_{2}O(g) + 6H_{3}O^{+}(aq) + 4}e^– \ce{\rightarrow 2NH_{3}OH^{+}(aq) + 5H_{2}O(l)}$ $– 0.05$
$\ce{HgS(s, black) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow Hg(l) + H_{2}S(g) + 2H_{2}O(l)}$ $– 0.085$
$\ce{Se(s) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow H_{2}Se(aq) + 2H_{2}O(l)}$ $– 0.115$
$\ce{Pb^{2+}(aq) + 2}e^− \ce{\rightarrow Pb(s)}$ $– 0.125$
$\ce{Sn^{2+}(aq) + 2}e^− \ce{\rightarrow Sn(s)}$ $– 0.1375$
$\ce{AgI(s) + }e^− \ce{\rightarrow Ag(s) + I^{−}(aq)}$ $– 0.1522$
$\ce{[SnF_6]^{2–}(aq) + 4}e^− \ce{\rightarrow Sn(s) + 6F^{−}(aq)}$ $– 0.200$
$\ce{Ni^{2+}(aq) + 2}e^− \ce{\rightarrow Ni(s)}$ $– 0.25$
$\ce{Co^{2+}(aq) + 2}e^− \ce{\rightarrow Co(s)}$ $-0.277$
$\ce{Tl^{+}(aq) + }e^− \ce{\rightarrow Tl(s)}$ $– 0.3363$
$\ce{PbSO_{4}(s) + 2}e^− \ce{\rightarrow Pb(s) + SO_{4}^{2−}(aq)}$ $– 0.3505$
$\ce{Cd^{2+}(aq) + 2}e^− \ce{\rightarrow Cd(s)}$ $– 0.403$
$\ce{Cr^{3+}(aq) + }e^− \ce{\rightarrow Cr^{2+}(aq)}$ $– 0.424$
$\ce{Fe^{2+}(aq) + 2}e^− \ce{\rightarrow Fe(s)}$ $– 0.44$
$\ce{2CO_{2}(g) + 2H_{3}O^{+}(aq) + 2}e^– \ce{\rightarrow (COOH)_{2}(aq) + 2H_{2}O(l)}$ $– 0.481$
$\ce{Ga^{3+}(aq) + 3}e^− \ce{\rightarrow Ga(s)}$ $– 0.53$
$\ce{Cr^{3+}(aq) + 3}e^− \ce{\rightarrow Cr(s)}$ $– 0.74$
$\ce{Zn^{2+}(aq) + 2}e^− \ce{\rightarrow Zn(s)}$ $– 0.763$
$\ce{Cr^{2+}(aq) + 2}e^− \ce{\rightarrow Cr(s)}$ $– 0.90$
$\ce{V^{2+}(aq) + 2}e^− \ce{\rightarrow V(s)}$ $– 1.13$
$\ce{Mn^{2+}(aq) + 2}e^− \ce{\rightarrow Mn(s)}$ $– 1.18$
$\ce{Zr^{4+}(aq) + 4}e^− \ce{\rightarrow Zr(s)}$ $– 1.55$
$\ce{Al^{3+}(aq) + 3}e^− \ce{\rightarrow Al(s)}$ $– 1.676$
$\ce{H_{2}(g) + 2}e^− \ce{\rightarrow 2H^{−}(aq)}$ $– 2.25$
$\ce{Mg^{2+}(aq) + 2}e^− \ce{\rightarrow Mg(s)}$ $– 2.356$
$\ce{Na^{+}(aq) + }e^− \ce{\rightarrow Na(s)}$ $- 2.714$
$\ce{Ca^{2+}(aq) + 2}e^− \ce{\rightarrow Ca(s)}$ $– 2.84$
$\ce{Sr^{2+}(aq) + 2}e^− \ce{\rightarrow Sr(s)}$ $– 2.89$
$\ce{Ba^{2+}(aq) + 2}e^- \ce{\rightarrow Ba(s)}$ $– 2.92$
$\ce{Rb^{+}(aq) + }e^− \ce{\rightarrow Rb(s)}$ $– 2.925$
$\ce{K^{+}(aq) + }e^− \ce{\rightarrow K(s)}$ $– 2.925$
$\ce{Li^{+}(aq) + }e^− \ce{\rightarrow Li(s)}$ $– 3.045$
Basic Solution Standard Reduction Potential, E° (volts)
$\ce{ClO^{–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow Cl^{–}(aq) + 2OH^{–}(aq)}$ $0.89$
$\ce{OOH^{-}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow 3OH^{–}(aq)}$ $0.867$
$\ce{2NH_{2}OH(aq) + 2}e^– \ce{\rightarrow N_{2}H_{4}(aq) + 2OH^{-}(aq)}$ $0.73$
$\ce{ClO_{3}^{–}(aq) + 3H_{2}O(l) + 6}e^– \ce{\rightarrow Cl^{–}(aq) + 6OH^{–}(aq)}$ $0.622$
$\ce{ClO_{3}^{–}(aq) + 3H_{2}O(l) + 6}e^– \ce{\rightarrow Cl^{–}(aq) + 6OH^{–}(aq)}$ $0.622$
$\ce{MnO_{4}^{–}(aq) + 2H_{2}O(l) + 3}e^– \ce{\rightarrow MnO_{2}(s) + 4OH^{–}(aq)}$ $0.60$
$\ce{MnO_{4}^{–}(aq) + }e^– \ce{\rightarrow MnO_{4}^{2–}(aq)}$ $0.56$
$\ce{NiO_{2}(s) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow Ni(OH)_{2}(s) + 2OH^{–}(aq)}$ $0.49$
$\ce{Ag_{2}CrO_{4}^{–}(s) + 2}e^– \ce{\rightarrow 2Ag(s) + CrO_{4}^{2–}(aq)}$ $0.4491$
$\ce{O_{2}(g) + 2H_{2}O(l) + 4}e^– \ce{\rightarrow 4OH^{–}(aq)}$ $0.401$
$\ce{ClO_{4}^{–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow ClO_{3}^{–}(aq) + 2OH^{–}(aq)}$ $0.374$
$\ce{Ag_{2}O(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow 2Ag(s) + 2OH^{–}(aq)}$ $0.342$
$\ce{2NO_{2}^{–}(aq) + 3H_{2}O(l) + 4}e^– \ce{\rightarrow N_{2}O(g) + 6OH^{–}(aq)}$ $0.15$
$\ce{[Co(NH_{3})_{6}]^{3+}(aq) + }e^- \ce{\rightarrow [Co(NH_{3})_{6}]^{3+}(aq)}$ $0.058$
$\ce{HgO(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow Hg(l) + 2OH^{–}(aq)}$ $0.0977$
$\ce{O_{2}(g) + H_{2}O(l) + 2}e^– \ce{\rightarrow OOH^{–}(aq) + OH^{–}(aq)}$ $0.0649$
$\ce{NO_{3}^{-}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow NO_{2}^{–}(aq) + 2OH^{–}(aq)}$ $0.01$
$\ce{MnO_{2}(s) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow Mn(OH)_{2}(s) + 2OH^{–}(aq)}$ $- 0.05$
$\ce{CrO_{4}^{2–}(aq) + 4H_{2}O(l) + 3}e^– \ce{\rightarrow Cr(OH)_{3}(s) + 5OH^{–}(aq)}$ $-0.11$
$\ce{Cu_{2}O(s) + H_{2}O(l) + 2}e^– \ce{\rightarrow 2Cu(s) + 2OH^{–}(aq)}$ $-0.365$
$\ce{FeO_{2}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow HFeO_{2}^{–}(aq) + OH^{–}(aq)}$ $-0.69$
$\ce{2H_{2}O(l) + 2}e^– \ce{\rightarrow H_{2}(g) + 2OH^{–}(aq)}$ $-0.8277$
$\ce{2NO_{3}^{–}(aq) + 2H_{2}O(l) + 2}e^– \ce{\rightarrow N_{2}O_{4}(g) + 4OH^{–}(aq)}$ $-0.86$
$\ce{HFeO_{2}^{-}(aq) + 2}e^– \ce{\rightarrow Fe(s) + 3OH^{–}(aq)}$ $-0.8$
$\ce{SO_{4}^{2–}(aq) + H_{2}O(l) + 2}e^– \ce{\rightarrow SO_{3}^{2–}(aq) + 2OH^{–}(aq)}$ $-0.936$
$\ce{N_{2}(g) + 4H_{2}O(l) + 4}e^– \ce{\rightarrow N_{2}H_{4}(aq) + 4OH^{–}(aq)}$ $-1.16$
$\ce{[Zn(OH)_{4}]^{2–}(aq) + 2}e^– \ce{\rightarrow Zn(s) + 4OH^{–}(aq)}$ $-1.285$
$\ce{Zn(OH)_{2}(s) + 2}e^{–} \ce{\rightarrow Zn(s) + 2OH^{–}(aq)}$ $-1.246$
$\ce{[Zn(CN)_{4}]^{2–}(aq) + 2}e^– \ce{\rightarrow Zn(s) + 4CN^{–}(aq)}$ $-1.34$
$\ce{Cr(OH)_{3}(s) + 3}e^– \ce{\rightarrow Cr(s) + 3OH^{–}(aq)}$ $-1.33$
$\ce{SiO_{3}^{2–}(aq) + 3H_{2}O(l) + 4}e^– \ce{\rightarrow Si(s) + 6OH^{–}(aq)}$ $-1.69$
contacts here
to the right of Pt, H2(1 atm)│H+(1 M), as in Eq. $\ref{5}$ above, the E° is + 0.34 V. For the Zn2+│Zn redox couple, we find E° = – 0.76 V in Table 17.1. This means that for the cell
$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Zn^{2+} (1 M)} \mid \ce{Zn E^{o} = – 0.76 V} \nonumber$
Since Eq. $\ref{4}$ shows this cell in reverse, we change the sign of E°, obtaining + 0.76 V. Thus we can combine standard reduction potentials from Table 1 to obtain emf's for cells like Eq. $\ref{6}$ so long as both electrodes are given in the table.
Example $3$ : Find the standard emf for the cell
$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{Br^{–}} \mid \ce{Br_{2}(l), Pt} \nonumber$
Solution
From Table $1$ we have;
$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Hg^{2+} (1 M)} \mid \ce{Hg(l)E^{o} = + 0.85 V} \nonumber$
Since we want to be the left-hand electrode, this must be reversed and the sign of E° must be changed:
$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{H^{+}(1 M)} \mid {H_{2}(1 atm), Pt E^{o} = – 0.85 V}\label{10}$
For the other electrode Table $1$ gives
$\ce{Pt, H_{2}(1 atm)} \mid \ce{H^{+}(1 M)} \parallel \ce{Br^{–}(1 M)} \mid \ce{Br_{2}(l), Pt E^{o} = +1.07 V}\label{11}$
Adding the cells of Eqs. $\ref{10}$ and $\ref{11}$, we obtain
$\ce{Hg(l)} \mid \ce{Hg^{2+} (1 M)} \parallel \ce{Br^{–}(1 M)} \mid \ce{Br_{2}(l), Pt E^{o} = (1.07 – 0.85) V =+ 0.22 V} \nonumber$
The positive value of the standard emf obtained in Example $3$ indicates that the corresponding cell reaction is spontaneous:
$\ce{Hg(l) + Br_{2}(l) \rightarrow Hg^{2+} (1 M) + 2Br^{–}(1 M)} \nonumber$
In other words, bromine is a strong enough oxidizing agent to convert mercury metal to mercury(II) ions in aqueous solution, assuming the concentrations of mercury(II) and bromide ions to be 1 mol dm–3. This corresponds to the observations made where liquid mercury combined with liquid bromine to form mercury(II) bromide. Thus the standard reduction potentials in Table $1$ can be used to predict whether a particular reaction will take place, just as Table 1 in Redox Couples was used in our earlier discussion of redox reactions. The advantage of Table $1$ is that it gives quantitative as well as qualitative information. It not only tells us that Br2(l) is a stronger oxidizing agent than Hg2+ (1 M) [because Br2(l) is above Hg2+ (1 M), but it also tells us how much stronger, in terms of the cell emf of + 0.22 V. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.10%3A_Electromotive_Force_of_Galvanic_Cells.txt |
A great disadvantage of the two cells we describe in Commercial Galvanic Cells is that they must be discarded once the cell reaction has gone to completion. Since they contain valuable metals like mercury and manganese whose supplies will soon become more limited, we can expect that batteries like this will become progressively more expensive and the impetus will grow to recycle the materials they contain, perhaps through a system of returnable deposits. A much more satisfactory solution to this problem, though, is the increased use of batteries which are rechargeable. Such batteries are called storage batteries, and they have the property that once the cell reaction has gone to completion, it can easily be reversed by electrolysis.
$\text{Pb}, \text{PbSO}_4(s)\mid \text{H}_2\text{SO}_4(5 M)\mid \text{PbSO}_4(s), \text{PbO}_2(s), \text{Pb} \nonumber$ and the electrode reactions are and $\ce{Pb(s) + SO4^{2–}(aq) -> PbSO4(s) + 2}e^– \nonumber$
and
$\ce{2}e^– \ce{+ 4H^{+} + SO_{4}^{2–} + PbO_{2} \rightarrow PbSO_{4}(s) + 2H_{2}O} \nonumber$
giving an overall cell reaction $\ce{Pb(s) + PbO_{2}(s) + 4H^{+} + 2SO_{4}^{2–} \rightarrow 2PbSO_{4} + 2H_{2}O} \nonumber$ Both electrodes are similarly constructed of a flat grid of lead. For the electrode, the interstices are filled with finely divided spongy Pb of high surface area, while for the right-hand electrode chocolate-brown lead dioxide powder, PbO2, is used.
As the cell discharges, both electrodes become plated with white PbSO4. Some storage batteries are made with transparent polystyrene cases. When this is the case, it is very apparent when the battery has been discharged since both electrodes begin to acquire a similar white appearance. The discharge of the battery also results in the consumption of H2SO4 [see Eq. (1)]. As the H2SO4 becomes more dilute, its density decreases, and so by measuring the density of the acid, we can decide whether the battery needs recharging or not.
Once a lead storage battery has been discharged, it can be recharged by electrolyzing it with a source of direct current. This results in the cell reaction reversing itself. The PbSO4 disappears from both electrodes, and the concentration of H2SO4 increases. Even when the cell becomes fully charged, little damage is done by continued electrolysis. H2 gas is evolved at one electrode and O2 at the other. Lead storage cells do not last indefinitely, however. Some of the PbSO4 formed at the electrodes becomes dislodged and falls to the bottom of the container where it is no longer available to take part in the recharging electrolysis. If lead storage batteries are allowed to discharge completely, this loss of PbSO4 particularly liable to occur. Batteries which are not mistreated in this way inevitably last longer. In a car battery three or six lead cells are connected in series. Since each produces 2.0 V when fully charged, the resultant potential difference is 6 or 12 V.
A second everyday example of a storage battery is the nickel-cadmium battery now commonly used in electronic calculators. These cells have the following construction:
$\ce{Cd(s), CdO(s)} \mid \ce{KOH(1 M)} \mid \ce{Ni(OH)_{3}(s), Ni(OH)_{2}(s), Ni} \nonumber$ with the following electrode half-equations: $\ce{Cd(s) + 2OH^{–} \rightarrow CdO + H_{2}O + 2}e^– \nonumber$
$\ce{2}e^– \ce{+ 2Ni(OH)_{3} \rightarrow 2Ni(OH)_{2} + 2OH^{–}} \nonumber$
This cell produces a potential difference of 1.25 V and can be used in place of the dry cell for many purposes. A simple contribution anyone can make toward conserving the world’s supply of zinc is to invest in a simple battery charger and use nickel-cadmium storage batteries to replace dry cells.
17.12: Fuel Cells
A type of galvanic cell which promises to become increasingly important in the future is the fuel cell. By contrast to a conventional cell, where only limited quantities of oxidizing agent and reducing agent are available, a continuous supply of both is provided to a fuel cell, and the reaction product is continually removed. A somewhat oversimplified diagram of a fuel cell in which the cell reaction is the production of water from hydrogen and oxygen is shown in Figure $1$.
Hydrogen enters the cell through a porous carbon electrode which also contains a platinum catalyst. Oxygen is supplied to a similar electrode except that the catalyst is silver. The electrolyte is usually a warm solution of potassium hydroxide, and the two electrode reactions can be written as
$\text{H}_2(g) + 2 \text{OH}^-(aq) \rightarrow 2 \text{H}_2\text{O}(l) + 2e^– \nonumber$
and
$\frac{1}{2} \text{O}_2(g) + \text{H}_2\text{O} + 2e^– \rightarrow 2\text{OH}^–(aq) \nonumber$
giving the overall result
$\text{H}_2(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{H}_2\text{O} \nonumber$
Unless it is removed, water produced by the reaction will gradually dilute the potassium hydroxide, rendering the cell inoperative. Hence the electrolyte is kept warm enough that water evaporates just as fast as it is produced by the cell reaction. A fuel cell like this will continue to operate and produce electrical energy as long as a supply of hydrogen and oxygen are available.
Fuel cells have an important advantage over all other devices which burn fuel to obtain useful energy: their efficiency. While an internal-combustion engine is only about 25% efficient and a steam engine about 35% efficient, the H2–O2 cell just described can already operate at an efficiency of 45%. The theoretically highest possible efficiency of such a cell, set by the second law of thermodynamics, is 83%. Because of this high efficiency many possible uses and developments for fuel cells have been proposed. One of these scenarios for the future envisions large nuclear power plants floating on the sea producing hydrogen gas by the decomposition of water rather than producing electrical power. This hydrogen gas could then be piped to individual homes where it could either be burned for heat or converted to electricity with the aid of a fuel cell. A second scenario involves automobiles powered by cells fueled by conventional gasoline or perhaps hydrogen. These automobiles would run virtually noiselessly without any pollution problems and deliver twice as many kilometers per liter of fuel as a conventional vehicle. Alas for such scenarios, many technological problems still intervene, but further development of fuel cells is certainly one approach to our current energy problems that should be thoroughly investigated. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.11%3A_Storage_Batteries.txt |
We see in the section on the Electromotive Force of Galvanic Cells that the emf of a galvanic cell can tell us whether the cell reaction is spontaneous. In other sections we show that the free-energy change ΔG of a chemical process also indicates whether that process is spontaneous. It is quite reasonable, then, to expect some relationship between ΔG and E, and indeed one exists.
In the section on Free Energy we stated that the free-energy change corresponds to the maximum quantity of useful work which can be obtained when a chemical reaction occurs. In other words,
$\triangle G = –w_{max} \nonumber$ where the minus sign is necessary because the free energy decreases as the chemical system does useful work on its surroundings. If we are referring to a redox reaction, that work can be obtained in electrical form by means of an appropriate galvanic cell. It can be measured readily, because when a quantity of charge Q moves through a potential difference ΔV, the work done is given by $w = Q \triangle V \nonumber$ Thus if one coulomb passes through a potential difference of one volt, the work done is $w = 1 \text{ C} \times 1 \text{ V} = 1 \cancel{\text{As}} × 1 \text{ J} \cancel{\text{A}^{-1}\text{ s}^{-1}} = 1 \text{ J} \nonumber$ Now suppose we construct a Zn-Cu cell of the type described earlier: $\ce{Zn} \mid \ce{Zn^2+ (1 M)} \parallel \ce{Cu^2+ (1 M)} \mid \ce{Cu} \nonumber$ and suppose we make the cell large enough that the concentrations of Cu2+ and Zn2+ will not change significantly even though 1 mol Zn is oxidized to 1 mol Zn2+ according to the cell reaction
$\ce{Zn (s) + Cu^2+ (aq) -> Zn^2+ (aq) + Cu(s)} \label{5}$
If this cell is discharged through a large enough resistance, the potential difference will have its maximum value, namely, the cell emf, E°; so if we know how much charge is transferred, we can calculate the electrical work done. For the oxidation of 1 mol Zn [that is, for the occurrence of 1 mol of reaction $\ref{5}$], there must be 2 mol e transferred according to the half-equation $\ce{Zn (s) -> Zn^2+ (aq) + 2}e^– \nonumber$ Therefore the quantity of electrical charge transferred per mole of reaction is $Q_m = \ce{2 * F = 2 * 9.649 × 10^4 C mol^{-1} = 1.930 * 10^5 C mol^{-1}} \nonumber$ (The Faraday constant, F, is the quantity of charge per mole of electrons. It has the value 96,485 C/mol.) The maximum useful work per mole of reaction which the cell can perform while discharging is thus $w_{max} = Q_{m}E^{o} = \ce{2FE^{o} = 2 * 9.649 * 10^{4} C mol^{-1} * 1.10 V = 212} \frac{\text{ kJ}}{\text{ mol}} \nonumber$ The standard molar free energy change for the cell reaction is thus
$\Delta G_{m}^{o} = – w_{max} = – 2FE^{o} = – 212 \frac{\text{ kJ}}{\text{ mol}} \nonumber$
A similar argument can be applied to any cell in which the reactants and products are all at their standard concentrations or pressures. If the standard emf of such a cell is E°, while ΔGm° is the standard molar free energy change for the cell reaction,these two quantities are related by the equation
$\Delta G_{m}^{o} = – zFE^{\circ} \nonumber$ where z (a dimensionless number) corresponds to the number of moles of electrons transferred per mole of cell reaction. A similar relationship holds even when reactants and products are not at standard concentrations and pressure:
$\Delta G_{m} = – zFE \nonumber$
This connection between cell emf and free-energy change provides a means of measuring ΔGm, directly, rather than by determining ΔHm, and ΔSm, and then combining them.
Example $1$ : The emf of a Cell
$\ce{Pt, H_{2} (1 atm)} \mid \ce{H^{+} (1 M)} \mid \ce{O_{2} (1 atm), Pt} \nonumber$ is 1.229 V at 298.15 K. Calculate ΔGf° for liquid water at this temperature.
Solution: The half-equations for the cell are
$\ce{2H2 (g) -> 4H+ (aq) + 4}e^{-}$
$\ce{4}e^{-} \ce{+ 4H+ (aq) + O2 (g) -> 2H2O (l)}$
so that the cell reaction is $\ce{2 H2 (g) + O_{2} (g) -> 2 H_{2}O (l)} \quad \text{1 atm, 298.15 K} \nonumber$ Since there are 4 mol e transferred per mol cell reaction, z = 4 and $\Delta G_{m} = - zFE = - 4 \times \frac{9.649 * 10^{4} \text{C}}{ \text{1 mol}} \times 1.229 \text{V} = - 474.3 \frac{ \text{ kJ}}{ \text{ mol}} \nonumber$ The reaction produces liquid water at standard pressure and the desired temperature, but 1 mol reaction produces 1 mol 2H2O(l), that is, 2 mol H2O. Therefore $\triangle G_{f}^{o}[\text{H}_2 \text{O} (l) \text{, 298 K}] = \frac{1}{2} \triangle G_{m} = – 237.2 \frac{\text{ kJ}}{ \text{ mol}} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.13%3A_Galvanic_Cells_and_Free_Energy.txt |
As shown in the section on cell notation, a galvanic cell in shorthand form can be written:
$\text{Zn}│\text{Zn}^{2+} ~(1 M)~║~ \text{Cu}^{2+} ~~(1 M)~|~ \text{Cu} \nonumber$
The parenthetical notes of (1 M) are frequently omitted because 1 M is at the standard state. However, cells can be created which use concentrations other than 1 M. In such a case, one must always indicate the concentrations as shown above. In fact, as the reaction for the cell written above takes place (that is, as the cell provides electric current), Cu2+ will be used up while Zn2+ will be generated. The reactant concentations will decrease and the product concentrations will increase until the solution has reached a state of equilibrium. These equilibrium concentrations are not likely to be at the standard 1 M solutions indicated above.
The voltage of a cell at non-standard state is modified by the relative concentrations of the reactants and products. That is, the cell emf depends on the reaction quotient, $Q$. The equation for the reaction in the cell is
$\text{Zn} + \text{Cu}^{2+} \rightleftharpoons \text{Cu}+\text{Zn}^{2+} \nonumber$
We can determine the reaction quotient, Q, as follows. (Note that when denoting actual concentrations, which may or may not be equilibrium concentrations, curly brackets are used.)
$Q=\dfrac{\left\{\text{Zn}^{2+}\right\}}{\left\{\text{Cu}^{2+}\right\}} \nonumber$
Notice that only the aqueous states are included in the reaction quotient. Using an equation from Galvanic Cells and Free Energy, we can see that the electromotive force is related to ΔG° (at standard states). (Recall that z is the number of electrons transferred and F is the Faraday constant.) This holds true even when the cell is not at standard conditions.
$\Delta G^{\circ} = – zFE^{\circ} ~~~ \text{and} ~~~ \Delta G = – zFE \nonumber$
We will not derive it here, but there is a relationship for ΔG for non-standard conditions:
$\Delta G = \Delta G^{\circ} + RT \text{ ln } Q \nonumber$
Combining these equations gives a single equation for the Electromotive force of a non-standard galvanic cell: The Nernst Equation.
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{RT}{zF}\ln Q \nonumber$
For ease of calculation, we can change the base of the logarithm to base 10 (although most standard calculators have a natural logarithm button), the simple formula below is another commonly used form of the Nernst Equation:
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{2.303 RT}{zF}\log Q \nonumber$
As most of the Galvanic cells in use will be at room temperature (T = 25 °C), we can substitute all of the constant values (R = 8.314 J mol–1 K–1 and F = 96,485 J V–1 mol–1), yielding an even simpler formula:
$E_{\text{cell}} = E_{\text{cell}}^{\circ} - \dfrac{0.0592 \text{V}}{z}\log Q \qquad (T = 298 \text{ K}) \nonumber$
If you calculate Q for the standard-state concentration of 1 M (or pressure of 1 bar), note how the second term on the right-hand-side disappears and you recover $E_{\text{cell}} = E_{\text{cell}}^{\circ}$.
Although the Nernst equation is useful to predict the actual voltage of a cell under non-standard conditions, it is frequently more useful to use the measured voltage to detect the concentration of one of the species. For instance, if we use a standard H2/Pt half-cell, the detected voltage of that half reaction coupled with an unknown concentration of Fe2+ can be used to determine the concentration of Fe2+.
Example $1$ : Voltage of a Galvanic Cell
Determine the voltage measured for this galvanic cell. You may need to use the Table of Standard Reduction Potentials.
$\text{Sn}^{2+} (aq) (1 M│\text{Sn}║\text{Ag}│\text{Ag}^{+} (0.8 M) \nonumber$
Solution:
The strategy for solving these problems is to first find the standard electromotive force, as shown in another section, and then write the balanced chemical equation, from which you can derive Q and z.
$E_{\text{cell}}^{\circ} = 0.7991\text{ V} - (- 0.1375)\text{ V} = 0.9366\text{ V} \nonumber$
$2\text{Ag}^{+}(aq)+\text{Sn(s)} \rightarrow 2\text{Ag}(\text{s}) + \text{Sn}^{2+}(aq) \nonumber$
The expression for Q can be derived only from the balanced chemical equation. Thus, using the Nernst Equation, we have,
\begin{align*}E_{\text{cell}} &= E_{\text{cell}}^{\circ} - \dfrac{0.0592 \text{ V}}{z}\log Q\&= 0.9366\text{ V} - \dfrac{0.0592\text{ V}}{2}\log \dfrac{\left\{\text{Sn}^{2+}(aq)\right\}}{\left\{\text{Ag}^{+}(aq)\right\}^2}\&= 0.9366\text{ V} - \dfrac{0.0592\text{ V}}{2}\log \dfrac{\left(1\text{ }M\right)}{\left(0.8\text{ }M\right)^2}\E_{\text{cell}}&=0.9308\text{ V}\end{align*} \nonumber | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/17%3A_Electrochemical_Cells/17.14%3A_Cells_at_Non-Standard_Conditions.txt |
Chemical kinetics is concerned with the rates of chemical reactions, that is, whether reactions proceed quickly or slowly. As we have already mentioned, some spontaneous reactions are extremely slow. Chemical kinetics is concerned with the rates of chemical reactions, the dependence of those rates on temperature, concentration, and catalysts, and the microscopic mechanisms by which reactions occur.
18: Chemical Kinetics
Chemical kinetics is concerned with the rates of chemical reactions, that is, whether reactions proceed quickly or slowly. As we have already mentioned, some spontaneous reactions are extremely slow. An example is the Haber-process synthesis of ammonia:
$\text{ N}_{\text{2}}+ \text{ 3H}_{\text{2}}\rightarrow\text{ 2NH}_{\text{3}} \nonumber$
with ΔGm°(298 K) = – 33.27 kJ mol–1
Even though a negative ΔGm° predicts that this reaction can occur at room temperature, it is of little value unless chemists can find some way to speed it up. On the other hand we often want to slow down undesirable reactions, such as spoilage of food or decomposition of wood. Hence it is quite useful to know how factors such as temperature, concentrations of reactants and products, and catalysts will affect the rates of reactions. Moreover, studying these factors gives valuable information about the sequence of microscopic events by which a reaction occurs. Knowledge of when and where bonds are formed and broken as well as how molecular structures change during a reaction can be very useful in helping us to devise ways to speed up or slow down that reaction.
Chemical kinetics is concerned with the rates of chemical reactions, the dependence of those rates on temperature, concentration, and catalysts, and the microscopic mechanisms by which reactions occur. The rate of a reaction is defined in terms of the change in concentration of a reactant or product per unit time, and it usually decreases as the reaction progresses. The reaction rate ordinarily is proportional to the concentrations of reactants and/or products, each raised to a power called the order with respect to that reactant or product. When the concentration of a species which is not a reactant or product in the overall reaction affects the rate, that species is called a catalyst. An equation expressing the dependence of reaction rate on concentrations is called a rate equation or rate law.
On a microscopic level, a reaction usually involves unimolecular processes, in which a single molecule changes structure, or bimolecular processes, in which two molecules collide. One determinant of rate in this situation is frequency of collisions as seen in the animation on this page. Collision of three or more molecules simultaneously is much less probable. In both uni- and bimolecular elementary processes an activation energy barrier must be surmounted before product molecules can be produced. The species at the top of a graph of energy versus reaction coordinate is called the activated complex or transition state. Because of this energy barrier only a small fraction of the molecules are energetic enough to reach the transition state, but that fraction increases rapidly as temperature increases, and so reaction rates are strongly dependent on temperature. The temperature dependence of the rate constant can be used to obtain the activation energy by means of an Arrhenius plot.
This animation displays the number of collisions that occur between the dark blue particle and a set of lighter blue particles. When hit, a lighter blue particle changes to black. A second hit changes a black particle to a red particle, so that the number of collisions can be easily counted. The frequency at which the dark blue particles hits light blue particles is an important determinant of how quickly a reaction will go.
Most reactions occur in two or more steps. Such a sequence of elementary processes is called a reaction mechanism, and the overall rate is determined by the slowest, or rate-limiting, step. The experimental rate law tells us the composition of the activated complex for the rate-limiting step, but often several mechanisms are possible which agree with the rate law. Other evidence must then be used to decide among these mechanisms.
A catalyst speeds up a reaction by changing the mechanism so that the activation energy is lowered. Many heterogeneous catalysts are of great industrial importance, but the most efficient catalysts known are the enzymes in living organisms. An enzyme operates by adsorbing a substrate molecule at an active site whose structure is exactly right to stretch bonds which are to be broken or to hold atoms in position while new bonds form. This almost ideal structure of the active site makes enzymes highly specific and extremely efficient catalysts. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.01%3A_Prelude_to_Kinetics.txt |
A chemist who investigates the rate of a reaction often wants to see how the rate differs at two different temperatures. In the video below, this situation is explicitly investigated with the glucose oxidase enzyme reaction of glucose to gluconolactone, which is visualized by a dye which turns to a blue-green color as the reaction proceeds. One reaction is run at 22°C and the other at 4°C. The reaction clearly proceeds more quickly at 22°C, which gains a darker blue-green color more quickly than the reaction run at 4°C.
Maybe instead of being run at two different temperatures, a reaction has an added catalyst, which is a compound which does not change in the reaction, but causes the reaction to go through a different mechanism. The following video shows two cases of the reaction:
$\text{ 2MnO}_{\text{4}}^{-}(aq) + \text{ 5H}_{\text{2}}\text{C}_2\text{O}_{\text{4}}(aq) + \text{ 6H}_{\text{3}}\text{O}^{+}(aq)\rightarrow \text{2Mn}^{2+}(aq) + \text{ 10CO}_{\text{2}}(aq) + \text{ 14H}_{\text{2}}\text{O} \nonumber$
Manganese(II) sulfate is added as a catalyst to the solution on the right, which increases the rate of reaction. This solution changes from a dark pink solution to a clear solution far more quickly than the solution on the left.
In order think meaningfully about these different rates of reaction, the rate needs to be quantified. This can be done for reactions in solution by looking at the dependence of concentration of a reactant or product when considering time. Figure 1 displays a decomposition reaction of dye into colorless chemicals at two different temperatures, giving dye concentration at different times.
As the dye decomposes, its concentration decreases and the color of the solution becomes fainter. This happens in a shorter time at 80°C than at 50°C, indicating that the dye decomposes faster at the higher temperature. By convention, then, the rate of the reaction is described in terms of the change in concentration per unit time for a reactant or product. The faster that concentration changes, the faster the reaction is going.
As an example of the use of this definition of reaction rate, consider the first 10 s during decomposition of the dye at 50°C. According to Figure 1 the concentration of dye drops to 0.70 mol dm–3 from an initial value of 1.0 mol dm–3. If we represent the change in the concentration of the reactant dye by ΔcR, then
ΔcR = c1c2 = (0.70 – 1.00) mol dm–3 = – 0.30 mol dm–3
In other words the concentration of dye has decreased by 0.30 mol dm–3. We can calculate an average rate of reaction by dividing this concentration decrease by the time interval (Δt = t2t1 = 10 s – 0 s = 10 s) during which it occurred:
Average rate = 0.30 mol dm–3 10–1 s–1 = 0.03 mol dm–3 s–1
Clearly, no matter what reactant we observe, its concentration will decrease with time and ΔcR will be negative. On the other hand, the concentration of a reaction product will always increase, and so ΔcP will always be positive. Since we want the average reaction rate always to be positive, we define it as
$\text{Rate = }\dfrac{-\Delta c_{R}}{\Delta t} = \dfrac{\Delta c_{P}}{\Delta t} \nonumber$
Example $1$ : Rate of Decomposition
Calculate the average rate of decomposition of the dye in Figure 1 at 80°C during the time interval(a) 0 to 10 s; (b) 10 to 20 s; (c) 20 to 30 s; (d) 0 to 30 s.
Solution
a) In the interval 0 to 10 s,
ΔcR = 0.49 mol dm–3 – 1.00 mol dm–3 = – 0.51 mol dm–3
Δt = 10 s – 0 s = 10 s
$\text{Rate = }\dfrac{-\Delta c_{R}}{\Delta t}$ = – $\dfrac{-\text{(}-\text{0}\text{.51 mol dm}^{-\text{3}}\text{)}}{\text{10 s}}$ = 0.051 mol dm–3 s–1
b) $\text{Rate = }\dfrac{-\Delta c_{R}}{\Delta t}$ = $\dfrac{-\text{(0}\text{.24 }-\text{0}\text{.49) mol dm}^{-\text{3}}}{\text{(20}-\text{10) s}}$ = 0.025 mol dm–3 s–1
c) $\text{Rate = }\dfrac{-\Delta c_{R}}{\Delta t}$ = $\dfrac{-\text{(0}\text{.11 }-\text{0}\text{.24) mol dm}^{-\text{3}}}{\text{(30}-2\text{0) s}}$ = 0.013 mol dm–3 s–1
d) $\text{Rate = }\dfrac{-\Delta c_{R}}{\Delta t}$ = $\dfrac{-\text{(0}\text{.11 }-1.\text{00) mol dm}^{-\text{3}}}{\text{(30}-\text{0) s}}$ = 0.030 mol dm–3 s–1
This example illustrates two important points about reaction rates. The first is that the rate of a reaction usually decreases with time, reaching a value of zero when the reaction is complete. This is usually because the rate depends in some way on the concentrations of one or more reactants, and as those concentrations decrease, the rate also decreases. The second point is a corollary to the first. Because the reaction rate changes with time, the rate we measure depends on the time interval used. For example, the average rate over the period to 30 s was 0.030 mol dm–3 s–1, but the average rate over the middle 10 s of that period was 0.025 mol dm–3 s–1. The different average for the 30-s interval reflects the fact that the rate dropped from 0.051 to 0.013 mol dm–3 s–1 during that period.
To measure a reaction rate as accurately as possible at a particular time, we need to make the time interval Δt as small as we can so that there is the least possible change in rate over the interval. If you are familiar with the branch of mathematics known as the differential calculus, you will recognize that as Δt becomes smaller, Δct approaches a limit known as the derivative. Thus the rate of reaction can be defined exactly as
$\text{Rate = }\dfrac{-dc_{R}}{dt} = \dfrac{dc_{P}}{dt} \nonumber$
The derivative also gives the slope of the tangent to a graph of versus t. Such a graph is shown in Figure $1$, with a tangent line drawn at t = 15 s. The slope of this line is – 0.0245 dm–3 s–1, giving
$\text{Rate = }\dfrac{-dc_{R}}{dt}$ = – slope of tangent = 0.0245 mol dm–3 s–1
This exact value is quite close to the average rate of 0.025mol mol dm–3 s–1 calculated for the interval 10 to 20 s, but it is farther from the 0.030 mol dm–3 s–1 calculated over the 0 to 30-s interval. As expected, the larger the time interval, the less accurate the calculated rate.
Reaction Rate and Equation Coefficients
The stoichiometry of the reaction in which the dye decomposes has a simple equation
Dye → colorless products
in which the coefficient of the reactant dye is one. When some coefficients are larger than one, there is one additional aspect of defining the rate of reaction. Consider the reaction
$\text{ 2N}_{\text{2}}\text{O}_5(g)\rightarrow \text{ 4NO}_{\text{2}}(g) + \text{O}_{\text{2}}(g) \nonumber$
According to stoichiometry, two molecules of N2O5 must disappear for every one molecule of O2 that is formed. And for every mole of O2 formed, there must be four moles of NO2 formed. Thus NO2 is formed four times as fast as O2, and N2O5 disappears twice as fast as O2 appears. If we define the rate in terms of the change in concentration of NO2 per unit time, then the rate will be four times faster than if we define the rate in terms of the change in concentration of O2. Similarly, the rate of disappearance of N2O5 will be twice as great as the rate of appearance of O2
Having the rate of reaction differ depending on which substance we consider could cause a lot of confusion. Therefore, by convention, the rate is defined in a way that always gives the same value. This is done by multiplying the appropriate derivative by the reciprocal of the coefficient in the balanced chemical equation. For the reaction above, the rate is defined as
$Rate = -\dfrac{1}{2}\dfrac{d[N_2 O_5]}{dt}=\dfrac{1}{4}\dfrac{d[NO_2 ]}{dt}=\dfrac{1}{1}\dfrac{d[O_2]}{dt} \nonumber$
Although N2O5 is disappearing twice as fast as O2 is forming (that is, d[N2O5]/dt is twice d[O2]/dt), multiplying d[N2O5]/dt by the factor 1/2 and changing its sign insures that the rate has the same value. Similarly, even though d[NO2]/dt is four times as great as d[O2]/dt, the factor of 1/4 makes the rate the same. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.02%3A_The_Rate_of_Reaction.txt |
As was mentioned in the section on reaction rate, the rate of reaction depends upon the concentrations of the reactants. Let us now look at the iodine clock reaction as an example. Two reactions actually occur in the iodine clock:
$\text{ H}_{\text{2}}\text{O}_2 + \text{3I}^{-}\rightarrow \text{I}_{\text{3}}^{-} + \text{ 2H}_{\text{2}}\text{O} \nonumber$
and
$\text{I}_{\text{3}}^{-} + \text{ 2S}_{\text{2}}\text{O}_3^{2-}\rightarrow \text{3I}^{-} + \text{ S}_{\text{4}}\text{O}_6^{2-} \nonumber$
The solution changes color as soon as all of the thiosulfate has been consumed. In the following video, we see how the concentration of the iodide ion affects the rate of reaction.
In this video, four iodine clock reactions are run with iodide concentrations in ratios of 1 : 0.75 : 0.5 : 0.375 from left to right. All other reactants are the same concentration. The video demonstrates that higher iodide ion concentrations increase the rate of the first reaction, meaning that thiosulfate is depleted more quickly. This causes the change in color signaling the end of the reaction to occur more quickly.
The iodide ion is not the only chemical which will effect the rate of reaction though. In the next video, H2SO4 concentration is varied to demonstrate this reagent's affect on the reaction rate:
In this video, four iodine clock reaction are run in four vials, with concentration ratios of H2SO4 going as 1 : 0.83 : 0.67 : 0.51 from left to right. Again, all other concentrations are kept constant including the iodide concentration. We see that, as was shown for the iodide ion before, the concentration of H2SO4 affects the reaction rate, with higher concentrations of H2SO4 producing a quicker reaction.
We now see that the concentration of two reactants in the iodine clock can modulate the rate of the reaction. Does hydrogen peroxide, H2O2, have an affect upon the iodine clock reaction rate? The following video tests whether this is true:
As in the first two videos, four iodine clock reactions are run, with H2O2 concentrations in the ratio of 1 : 0.8 : 0.6 : 0.4 respectively. For a third time, we see that increasing the concentration of one of the reactants, in this case, H2O2 increases the rate of the first reaction, and thus cause the thiosulfate to be more quickly.
These three examples show that the rate of a reaction can depend on the concentration of one or more of the reactants. The rate of reaction for the Iodine Clock reaction is dependent upon the concentration of hydrogen peroxide, iodide ions, as well as sulfuric acid. The rate can be related mathematically to the concentrations of the reactants. Before we look at a reaction dependent upon multiple reactants, let us look at a simpler example. A good choice is the decomposition of the dye that we have already seen in Figure 1 and Figure 2 in the section on the rate of reaction. In this case it can be shown that the rate is directly proportional to the concentration of dye, ΔcD. That is,
$\text{Rate} = k_{1}c_D\label{3}$
The constant of proportionality k1 is called the rate constant. It is independent of the concentration of dye, but increases as temperature increases. This accounts for the greater rate of reaction at 80°C than at 50°C, as well as the greater rate of the iodine clock reaction at 20°C compared to 8°C.
Example $1$: Concentration
In the example in the rate of reaction section average rates were calculated for three 10-s time intervals for the decomposition of a dye at 80°C. Using the following graph from that example, obtain concentrations of the dye at various times, show that Eq. $\ref{3}$ is valid for this dye.
Solution
Our previous discussion has indicated that the average reaction rate over a small time interval is very close to the actual rate at the midpoint of that interval. Thus we obtain the following table:
Average Rate/mol dm–3 s–1 Time Interval Time from Start to Midpoint of Interval/s Concentration of Dye/mol dm–3
.051
0 – 10 s
5
0.70
0.025
10 – 20 s
15
0.34
0.013
20 – 30 s
25
0.18
If Eq. $\ref{3}$ is valid, k1 should not depend on the concentration of dye. Rearranging Eq. (1) to calculate the rate constant, we find that at 5 s,
$k_{\text{1}}=\dfrac{\text{rate}}{c_{D}}=\dfrac{\text{0}\text{.051 mol dm}^{-\text{3}}\text{ s}^{-\text{1}}}{\text{0}\text{.70 mol dm}^{-\text{3}}}=\text{0}\text{.073 s}^{-\text{1}}$
At 15 s, $k_{\text{1}}=\dfrac{\text{rate}}{c_{D}}=\dfrac{\text{0}\text{.025 mol dm}^{-\text{3}}\text{ s}^{-\text{1}}}{\text{0}\text{.34 mol dm}^{-\text{3}}}=\text{0}\text{.074 s}^{-\text{1}}$ At 25 s, $k_{\text{1}}=\dfrac{\text{rate}}{c_{D}}=\dfrac{\text{0}\text{.013 mol dm}^{-\text{3}}\text{ s}^{-\text{1}}}{\text{0}\text{.18 mol dm}^{-\text{3}}}=\text{0}\text{.072 s}^{-\text{1}}$ Thus the rate constant is constant to two significant digits. This is within the accuracy of the measurements.
The rate of a reaction may depend on the concentration of one or more products as well as reactants. In some cases it may even be influenced by the concentration of a substance which is neither a reactant nor a product in the overall stoichiometric equation for the reaction. An example of this latter situation is provided by the conversion of cis-2-butene to trans-2-butene:
(2)
If some iodine is present, this reaction goes faster, and the rate law is found to be
Rate = k2(ccis-2-butene)(cI2)1/2 (3)
Although iodine does not appear in Eq. (2), its concentration does affect the reaction rate. Consequently iodine is called a catalyst for the reaction. The section on catalysis shows how this catalytic effect of iodine actually works.
It should be clear from the examples we have just given that we cannot tell from a stoichiometric equation like Eq. (2) which reactants, products, or catalysts will affect the rate of a given reaction. Such information must be obtained from experiments in which the concentrations of various species are altered, and the effects of those alterations on the rate of reaction are served. The results of such experiments are usually summarized in a rate equation or rate law for a given reaction. Equations (1) and (3), for example, are the rate laws for decomposition of a dye and for cis-trans isomerization of 2-butene (in the presence of the catalyst, I2). In general a rate equation has the form
Rate = k(cA)a(cB)b . . . (4)
It is necessary to determine experimentally which substances (A, B, etc.) affect the rate and to what powers (a, b, etc.) their concentrations must be raised. Then the rate constant k can be calculated. The exponents a, b, etc., are usually positive integers, but they may sometimes be fractions [as in Eq. (3)] or negative numbers. We say that the order of the reaction with respect to component A is a, while the order with respect to B is b. The overall order of the reaction is a + b.
Example $2$: Order of Reactions
For each reaction and experimentally measured rate equation listed below determine the order with respect to each reactant and the overall order:
a) $\ce{14H3O^+ + 2HCrO4^- + 6I^- -> 2Cr^{3+} + 3I2 + 22H2O}$
Rate = k(CHCrO4-)(CI-)2(CH3O+)2
b) $\ce{2I^- + H2O2 + 2H3O^+ -> I2 + 4H2O}$
Rate = k(CH2O2)(CI) 3 ≤ pH ≤ 5
Solution
a) The reaction is second order in H3O+ ion, second order in I ion, and first order in HCrO4. The overall order is 2 + 2 + 1 = 5.
b) The rate law is first order in I and first order in H2O2, and so the overall order is 2. The concentration of H3O+ does not appear at all. When this happens the reaction is said to be zero order in H3O+, because (cH3O+)0 = 1, and a factor of 1 in the rate law has no effect.
How is the order of a reaction experimentally obtained? In the next example, we will return to the iodine clock and determine the order of each reactant considered at the beginning of this section. Remember that in this reaction the same quantity of thiosulfate ions was added to each reaction mixture so the time taken for the color change to occur is inversely proportional to the rate of the reaction. Therefore 1/t—the reciprocal of the time for the concentration of thiosulfate ions to go to zero—is proportional to the initial rate of the reaction. Then each concentration can be plotted against 1/t to find the relationship between rate and concentration.
Example $3$: Order of Reactants
The next table gives the time(t) and 1/t for the completion of the iodine clock reactions seen at the beginning of the section for each of the experimental conditions in those three videos. Determine the order of each reactant by plotting initial concentration versus 1/t (which is proportional to the initial rate).
Reactant Relative
Concentration
Time to
Complete Reaction (s)
1/t (s-1)
I-
1
24
0.041
0.75
33
0.031
0.5
46
0.022
0.375
59
0.017
H2SO4
1
17
0.058
0.83
20
0.050
0.67
24
0.042
0.51
28
0.035
H2O2
1
24
0.041
0.8
31
0.032
0.6
43
0.023
0.4
68
0.015
I-
1
24
0.041
Solution
irst, determine how plotting concentration versus 1/t will determine the order of the reactant in question. If we break down equation 4 into a proportionality, and think of the rate as 1/t, we can find the relationship:
$frac{1}{t}\propto {c_A}^a \nonumber$
Thus, if a = 1, the graph will be linear, if a=2, the graph will look like a quadratic function, and so on. In the graph below, the data from above is plotted.
As can bee seen, all three reactants are linear on the concentration versus 1/t plot. Thus all three reactants, I-, H2SO4, and H2O2 have an order of 1.
In the last example, it was possible to obtain the order of the three reactants, but not the full rate law. How can a rate law be obtained from experimental measurements? One way has already been illustrated in Example $1$. There we guessed a rate law and then used it to calculate the rate constant k. Since k remained the same at various points during the reaction, we concluded that the rate equation was correct. Another way of obtaining a rate equation is illustrated in the next example.
Example $4$: Order and Rate Constant
The rate of the reaction
$\ce{2NO + O2 -> 2NO2} \nonumber$
was measured at 25°C, with various initial concentrations of NO and O2. The following results were obtained for the initial rate of the reaction:
Experiment Number cNO/mol dm-3 cO2/mol dm-3 Initial rate/mol dm-3 s-1
1 0.0020 0.0010 0.028
2 0.0020 0.0020 0.057
3 0.0020 0.0040 0.114
4 0.0040 0.0020 0.227
5 0.0010 0.0020 0.014
Find the order of this reaction with respect to each reactant, the overall order, and the rate constant.
Solution
In experiments 1, 2, and 3 the concentration of NO remains constant while the concentration of O2 increases from 0.010 to 0.020 to 0.040 mol dm–3. Each doubling of cO2 also doubles the rate (from 0.028 to 0.057 to 0.114 mol dm–3 s–1). Therefore the rate is proportional to cO2.
In experiments 2, 4, and 5 the concentration of O2 remains constant. Comparing experiments 2 and 4, we find that doubling cNO increases the rate from 0.057 to 0.227 mol dm–3 s–1. This is a factor of 4, or 22. The same factor is observed comparing experiment 5 to experiment 2. Thus we conclude that the rate is proportional to the square of cNO, and the rate equation must be
Rate = k(cO2)(cNO)2
The reaction is third-order overall—first order in O2 and second order in NO.
The rate constant can be calculated by rearranging the rate equation and substituting the concentrations and rate from any one of the five experiments. From experiment 3
$k = \frac{\text{rate}}{\text{(}c_{\text{O}_{\text{2}}}\text{)(}c_{\text{NO}}\text{)}^{\text{2}}} = \frac{\text{0}\text{.114 mol dm}^{-\text{3}}\text{ s}^{-\text{1}}}{\text{(0}\text{.040 mol dm}^{-\text{3}}\text{)(0}\text{.020 mol dm}^{-\text{3}}\text{)}^{\text{2}}} = \frac{\text{0}\text{.114 mol dm}^{-\text{3}}\text{ s}^{-\text{1}}}{\text{1}\text{.6 }\times \text{ 10}^{-\text{5}}\text{ mol}^{\text{3}}\text{ dm}^{-\text{9}}} \nonumber$ = 7.1 × 103 mol–2 dm6 s–1 A better value of the rate constant could be obtained by calculating for all five experiments and averaging the results.
The reaction described in the previous example is of considerable importance in the air above cities. As we mentioned in the section on Group VA elements, automobile engines emit NO which is then oxidized to brown NO2, an important precursor of photochemical air pollution. As we have just seen, this oxidation is second order in NO, and so when heavy traffic increases the concentration of NO by a factor of 10, the rate of production of NO2 goes up by the considerably larger factor of 102, or 100. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.03%3A_18.2-The_Rate_Equation.txt |
Now that we know something about how reaction rates are defined, measured, and related to the concentrations of substances which participate in a reaction, we would like to be able to interpret these macroscopic observations in terms of some microscopic model. On the microscopic level, a chemical reaction involves transformation of reactant atoms, ions, and/or molecules into product atoms, ions, and/or molecules. This requires that some bonds be broken, other bonds be formed, and some nuclei be moved to new locations. There are a limited number of categories into which such microscopic transformations can be classified, and each of these can be related to a macroscopic rate law. Therefore studies of reaction rates provide some insight into what the atoms and molecules of reactants and products are doing as a reaction occurs.
18.05: Unimolecular Processes
A reaction is said to be unimolecular if, on the microscopic level, rearrangement of the structure of a single molecule produces the appropriate product molecules. An example of a unimolecular process is conversion of cis-2-butene to trans-2-butene (in the absence of any catalyst):
All that is required for this reaction to occur is a twist or rotation around the double bond, interchanging the methyl group with the hydrogen atom on the right-hand side. Only one cis-2-butene molecule need be involved as a reactant in this process.
Rotating part of a molecule about a double bond is not easy, however, -because it involves a distortion of the electron clouds forming the double bond. This barrier to rotation was described in orbital descriptions of multiple bonds. A considerable increase in energy is required to twist one end of cis-2-butene around the other. This is shown in Figure 1, where the energy has been plotted as a function of the angle of rotation. The maximum energy is reached when one end of the molecule has been rotated by 90° with respect to the other.
This conformation is 262 kJ mol–1 higher than the energy of the original planar molecule. From this maximum it is downhill energetically on either side; so if the molecule has twisted this far, it should keep on twisting, eventually becoming trans-2-butene when the angle of rotation reaches 180°. Trans-2-butene is slightly lower in energy than cis-2-butene, as indicated the enthalpy change of – 4 kJ mol–1 for the overall reaction.
Figure $1$ shows that the barrier to rotation around a double bond is an energy barrier. At least 262 kJ mol–1 must be supplied to transform cis-2-butene into trans-2-butene by a rotation such as we have described. The minimum quantity of energy required to surmount an energy barrier during a chemical reaction is called the activation energy, and the molecular species at the top of the barrier is called the activated complex or the transition state. Quantities associated with this activated complex are usually denoted by a double dagger (‡). For example, the activation energy is given the symbol E.
In the sample of gaseous cis-2-butene at room temperature, only a tiny fraction of molecules have enough energy to surmount the activation-energy barrier. (Recall from figure 2 in the section on molecular speed distribution in gases that only a very small fraction of all gas molecules are traveling at very high speeds and hence have large kinetic energies. The same applies to the energy a molecule has because it is vibrating or rotating.) Not only do few molecules have enough energy to overcome the activation-energy barrier, but fewer still have that energy concentrated so that it can cause the atomic movements needed for the reaction to occur. In the case of cis-2-butene for example, very few of the high-energy molecules have their energy distributed so that most of it is causing a twist around the double bond. Thus over a given period of time only a very small fraction of the cis-2-butene molecules will be converted to trans-2-butene.
Now suppose that we double the concentration of a sample of cis-2-butene. This means that there will be twice as many molecules in each cubic decimeter. At a given temperature the fraction of the molecules which can react during a given time interval will be the same, but with twice as many molecules there will be twice as many conversions to trans-2-butene. Therefore in a period of 1 s the change in the amount of substance per unit volume will be twice as great, and this means that the reaction rate is twice as great.
What we have just said applies to any unimolecular process. The reaction rate must always be directly proportional to the concentration of the reacting species. That is, for a general unimolecular process, A → products, the rate equation must be first order in A:
$Rate = kc_A \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/18%3A_Chemical_Kinetics/18.04%3A_Microscopic_View_of_Chemical_Reactions.txt |
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