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Learning Objectives
• With the aid of simple diagrams, show how different band energy ranges in solids can produce conductors, insulators, and semiconductors.
• Describe the nature and behavior of a simple PN junction.
The band theory of solids provides a clear set of criteria for distinguishing between conductors (metals), insulators and semiconductors. As we have seen, a conductor must posses an upper range of allowed levels that are only partially filled with valence electrons. These levels can be within a single band, or they can be the combination of two overlapping bands. A band structure of this type is known as a conduction band.
Band arrangements in conductors. Metallic conduction requires the presence of empty levels into which electrons can move as they acquire momentum. This can be achieved when a band is only partially occupied or overlaps an empty band (right), or when the gap between a filled band and an upper empty one is sufficiently small (left) to allow ordinary thermal energy to supply the promotion energy.
Insulators and semiconductors
An insulator is characterized by a large band gap between the highest filled band and an even higher empty band. The band gap is sufficiently great to prevent any significant population of the upper band by thermal excitation of electrons from the lower one. The presence of a very intense electric field may be able to supply the required energy, in which case the insulator undergoes dielectric breakdown. Most molecular crystals are insulators, as are covalent crystals such as diamond.
If the band gap is sufficiently small to allow electrons in the filled band below it to jump into the upper empty band by thermal excitation, the solid is known as a semiconductor. In contrast to metals, whose electrical conductivity decreases with temperature (the more intense lattice vibrations interfere with the transfer of momentum by the electron fluid), the conductivity of semiconductors increases with temperature. In many cases the excitation energy can be provided by absorption of light, so most semiconductors are also photoconductors. Examples of semiconducting elements are Se, Te, Bi, Ge, Si, and graphite.
impurity band in semiconductors
The presence of an impurity in a semiconductor can introduce a new band into the system. If this new band is situated within the forbidden region, it creates a new and smaller band gap that will increase the conductivity. The huge semiconductor industry is based on the ability to tailor the band gap to fit the desired application by introducing an appropriate impurity atom (dopant) into the semiconductor lattice. The dopant elements are normally atoms whose valance shells contain one electron more or less than the atoms of the host crystal.
Semiconductor materials have traditionally been totally inorganic, composed mostly of the lighter P-block elements. More recently, organic semiconductors have become an important field of study and development.
Thermal properties of Semiconductors
At absolute zero, all of the charge carriers reside in lower of the bands below the small band gap in a semiconductor (that is, in the valence band of the illustration on the left above, or in the impurity band of the one on the right.) At higher temperatures, thermal excitation of the electrons allows an increasing fraction jump across this band gap and populate either the empty impurity band or the conduction band as shown at the right. The effect is the same in either case; the semiconductor becomes more conductive as the temperature is raised. Note that this is just the opposite to the way temperature affects the conductivity of metals.
N- and P-type materials
For example, a phosphorus atom introduced as an impurity into a silicon lattice possesses one more valence electron than Si. This electron is delocalized within the impurity band and serves as the charge carrier in what is known as an N-type semiconductor. In a semiconductor of the P-type, the dopant might be arsenic, which has only three valence electrons. This creates what amounts to an electron deficiency or hole in the electron fabric of the crystal, although the solid remains electrically neutral overall. As this vacancy is filled by the electrons from silicon atoms the vacancy hops to another location, so the charge carrier is effectively a positively charged hole, hence the P-type designation.
Substitution of just one dopant atom into 107 atoms of Si can increase the conductivity by a factor of 100,000.
The PN junction
When P- and N-type materials are brought into contact, creating a PN junction. Holes in the P material and electrons in the N material drift toward and neutralize each other, creating a depletion region that is devoid of charge carriers. But the destruction of these carriers leaves immobile positive ions in the N material and negative ions in the P material, giving rise to an interfacial potential difference ("space charge") as depicted here.
As this charge builds up, it acts to resist the further diffusion of electrons and holes, leaving a carrier-free depletion region, which acts as a barrier at the junction interface.
9.12: The Shared-Electron Covalent Bond
More than one non-equivalent structure
It sometimes happens that the octet rule can be satisfied by arranging the electrons in different ways. For example, there are three different ways of writing valid electron dot structures for the thiocyanate ion SCN. Some of these structures are more realistic than others; to decide among them, you need to know something about the concepts of formal charge and electronegativity. These topics are discussed in the lesson that follows this one, where examples of working out such structures are given. | textbooks/chem/General_Chemistry/Chem1_(Lower)/09%3A_Chemical_Bonding_and_Molecular_Structure/9.11%3A_Bonding_in_Semiconductors.txt |
Acids and bases touch upon virtually all areas of chemistry, biochemistry, and physiology. This set of lessons will get you started by presenting the underlying concepts in a systematic way. Aside from the section on pH which presumes an elementary knowledge of logarithms. The subject of acid-base equilibrium calculations is not covered in this lesson.
• 10.1: Introduction to Acids and Bases
The concepts of an acid, a base, and a salt are very old ones that have undergone several major refinements as chemical science has evolved. Our treatment of the subject at this stage will be mainly conceptual and qualitative, emphasizing the definitions and fundamental ideas associated with acids and bases. We will not cover calculations involving acid-base equilibria in these lessons.
• 10.2: Aqueous Solutions- pH and Titrations
As you will see in the lesson that follows this one, water plays an essential role in acid-base chemistry as we ordinarily know it. To even those who know very little about chemistry, the term pH is recognized as a measure of "acidity", so the major portion of this unit is devoted to the definition of pH and of the pH scale. But since these topics are intimately dependant on the properties of water and its ability do dissociate into hydrogen and hydroxyl ions.
• 10.3: Acid-base reactions à la Brønsted
In this lesson we develop this concept and illustrate its applications to "strong" and "weak" acids and bases, emphasizing the common theme that acid-base chemistry is always a competition between two bases for the proton. In the final section, we show how the concept of "proton energy" can help us understand and predict the direction and extent of common types of acid-base reactions without the need for calculations.
• 10.4: Acid-Base Reactions
Will this acid react with that base? And if so, to what extent? These questions can be answered quantitatively by carrying out the detailed equilibrium calculations you will learn about in another lesson. However, modern acid-base chemistry offers a few simple principles that can enable you to make a qualitative decision at a glance. More importantly, the ideas which we develop in this section are guaranteed to give you a far better conceptual understanding of proton-based acid-base reactions.
• 10.5: Lewis Acids and Bases
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry.
• 10.6: Types of Acids and Bases
You will already have noticed that not every compound that contains hydrogen atoms is acidic; .e.g, ammonia gives an alkaline aqueous solution. Similarly, some compounds containing the group -OH are basic, but others are acidic. An important part of understanding chemistry is being able to recognize what substances will exhibit acidic and basic properties in aqueous solution. Fortunately, most of the common acids and bases fall into a small number of fairly well-defined groups.
• 10.7: Acid-Base Gallery
Acids and bases are of interest not only to the chemically inclined; they play a major role in our modern industrial society — so anyone who participates in it, or who is interested in its history and development, needs to know something about them. Five of the major acids and bases fall into the "Top 20" industrial chemicals manufactured in the world.
10: Fundamentals of Acids and Bases
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above.
• Suggest simple tests you could carry out to determine if an unknown substance is an acid or a base.
• State the chemical definitions of an acid and a base in terms of their behavior in water.
• Write the formula of the salt formed when a given acid and base are combined.
The concepts of an acid, a base, and a salt are very old ones that have undergone several major refinements as chemical science has evolved. Our treatment of the subject at this stage will be mainly conceptual and qualitative, emphasizing the definitions and fundamental ideas associated with acids and bases. We will not cover calculations involving acid-base equilibria in these lessons.
1 Acids
The term acid was first used in the seventeenth century; it comes from the Latin root ac-, meaning “sharp”, as in acetum, vinegar. Some early writers suggested that acidic molecules might have sharp corners or spine-like projections that irritate the tongue or skin. Acids have long been recognized as a distinctive class of compounds whose aqueous solutions exhibit the following properties:
• A characteristic sour taste (think of lemon juice!);
• ability to change the color of litmus from blue to red;
• react with certain metals to produce gaseous H2;
• react with bases to form a salt and water.
Note: Litmas
Litmus is a natural dye found in certain lichens. The name is of Scandinavian origin, e.g. lit (color) + mosi (moss) in Icelandic. "Litmus test" has acquired a meaning that transcends both Chemistry and science to denote any kind of test giving a yes/no answer.
How oxygen got mis-named
The first chemistry-based definition of an acid turned out to be wrong: in 1787, Antoine Lavoisier, as part of his masterful classification of substances, identified the known acids as a separate group of the “complex substances” (compounds). Their special nature, he postulated, derived from the presence of some common element that embodies the “acidity” principle, which he named oxygen, derived from the Greek for “acid former”.
Note
Lavoisier had recently assigned this name to the new gaseous element that Joseph Priestly had discovered a few years earlier as the essential substance that supports combustion. Many combustion products (oxides) do give acidic solutions, and oxygen is in fact present in most acids, so Lavoisier’s mistake is understandable. In 1811 Humphrey Davy showed that muriatic (hydrochloric) acid (which Lavoisier had regarded as an element) does not contain oxygen, but this merely convinced some that chlorine was not an element but an oxygen-containing compound. Although a dozen oxygen-free acids had been discovered by 1830, it was not until about 1840 that the hydrogen theory of acids became generally accepted. By this time, the misnomer oxygen was too well established a name to be changed. The root oxy comes from the Greek word οξνς, which means "sour".
2 Acids and the hydrogen ion
The key to understanding acids (as well as bases and salts) had to await Michael Faraday's mid-nineteenth century discovery that solutions of salts (known as electrolytes) conduct electricity. This implies the existence of charged particles that can migrate under the influence of an electric field. Faraday named these particles ions (“wanderers”). Later studies on electrolytic solutions suggested that the properties we associate with acids are due to the presence of an excess of hydrogen ions in the solution. By 1890 the Swedish chemist Svante Arrhenius (1859-1927) was able to formulate the first useful theory of acids:
Arrhenius Definition
"an acidic substance is one whose molecular unit contains at least one hydrogen atom that can dissociate, or ionize, when dissolved in water, producing a hydrated hydrogen ion and an anion."
hydrochloric acid HCl → H+(aq) + Cl(aq)
sulfuric acid H2SO4→ H+(aq) + HSO4(aq)
hydrogen sulfite ion HSO3(aq) → H+(aq) + SO32–(aq)
acetic acid H3CCOOH → H+(aq) + H3CCOO(aq)
Strictly speaking, an “Arrhenius acid” must contain hydrogen. However, there are substances that do not themselves contain hydrogen, but still yield hydrogen ions when dissolved in water; the hydrogen ions come from the water itself, by reaction with the substance.
Definition of an acid
An acid is a substance that yields an excess of hydrogen ions when dissolved in water.
There are three important points to understand about hydrogen in acids:
• Although all Arrhenius acids contain hydrogen, not all hydrogen atoms in a substance are capable of dissociating; thus the –CH3hydrogens of acetic acid are “non-acidic”. An important part of knowing chemistry is being able to predict which hydrogen atoms in a substance will be able to dissociate into hydrogen ions.
• Those hydrogens that do dissociate can do so to different degrees. The strong acids such as HCl and HNO3 are effectively 100% dissociated in solution. Most organic acids, such as acetic acid, are weak; only a small fraction of the acid is dissociated in most solutions. HF and HCN are examples of weak inorganic acids.
• Acids that possess more than one dissociable hydrogen atom are known as polyprotic acids; H2SO4 and H3PO4 are well-known examples. Intermediate forms such as HPO42–, being capable of both accepting and losing protons, are called ampholytes.
H2SO4
sulfuric acid
HSO4 hydrogen sulfate ("bisulfate") ion SO42–
sulfate ion
H2S
hydrosulfuric acid
HS
hydrosulfide ion
S2–
sulfide ion
H3PO4
phosphoric acid
H2PO4
dihydrogen phosphate ion
HPO42–
hydrogen phosphate ion
PO43–
phosphate ion
HOOC-COOH
oxalic acid
HOOC-COO
hydrogen oxalate ion
OOC-COO
oxalate ion
You will find out in a later section of this lesson that hydrogen ions cannot exist as such in water, but don't panic! It turns out that chemists still find it convenient to pretend as if they are present, and to write reactions that include them.
3 Bases
The name base has long been associated with a class of compounds whose aqueous solutions are characterized by:
• a bitter taste;
• a “soapy” feeling when applied to the skin
• ability to restore the original blue color of litmus that has been turned red by acids
• ability to react with acids to form salts.
• react with certain metals to produce gaseous H2
Note
The word “alkali” is often applied to strong inorganic bases. It is of Arabic origin, from al-kali ("the ashes") which refers to the calcined wood ashes that were boiled with water to obtain potash which contains the strong base KOH, used in soap making. The element name potassium and its symbol K (from the Latin kalium) derive from these sources.
Just as an acid is a substance that liberates hydrogen ions into solution, a baseyields hydroxide ions when dissolved in water:
NaOH(s) → Na+(aq) + OH(aq)
Sodium hydroxide is an Arrhenius base because it contains hydroxide ions. However, other substances which do not contain hydroxide ions can nevertheless produce them by reaction with water, and are therefore also classified as bases. Two classes of such substances are the metal oxides and the hydrogen compounds of certain nonmetals:
Na2O(s) + H2O → [2 NaOH] → 2 Na+(aq) + 2 OH(aq)
NH3 + H2O → NH4+(aq) + OH(aq)
Defination of a base
A base is a substance that yields an excess of hydroxide ions when dissolved in water.
4 Neutralization
Acids and bases react with one another to yield two products: water, and an ionic compound known as a salt. This kind of reaction is called a neutralization reaction.
This "molecular" equation is convenient to write, but we need to re-cast it as a net ionic equation to reveal what is really going on here when the reaction takes place in water, as is almost always the case.
H+ + Cl + Na+ + OH Na+ + Cl + H2O
If we cancel out the ions that appear on both sides (and therefore don't really participate in the reaction), we are left with the net equation
H+(aq) + OH(aq) → H2O (1)
which is the fundamental process that occurs in all neutralization reactions.
Note
Confirmation that this equation describes all neutralization reactions that take place in water is provided by experiments indicating that no matter what acid and base are combined, all liberate the same amount of heat (57.7 kJ) per mole of H+neutralized.
In the case of a weak acid, or a base that is not very soluble in water, more than one step might be required. For example, a similar reaction can occur between acetic acid and calcium hydroxide to produce calcium acetate:
2 CH3COOH + Ca(OH)2 → CH3COOCa + 2 H2O
If this takes place in aqueous solution, the reaction is really between the very small quantities of H+ and OH resulting from the dissociation of the acid and the dissolution of the base, so the reaction is identical with Equation 1:
H+(aq) + OH(aq) → H2O
If, on the other hand, we add solid calcium hydroxide to pure liquid acetic acid, the net reaction would include both reactants in their "molecular" forms:
2 CH3COOH(l) + Ca(OH)2 (s) → 2 CH3COO + Ca2+ + 2 H2O
The “salt” that is produced in a neutralization reaction consists simply of the anion and cation that were already present. The salt can be recovered as a solid by evaporating the water. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.01%3A_Introduction_to_Acids_and_Bases.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Define the ion product of water, and know its room-temperature value.
• State the criteria, in terms of H+ and OH concentrations, of an acidic, alkaline, and a neutral solution.
• Given the effective hydrogen ion concentration in a solution, calculate the pH. Conversely, find the hydrogen- or hydroxide ion concentration in a solution having a given pH.
• Find the pH or pOH of a solution when either one is known.
• Describe the process of acid-base titration, and explain the significance of the equivalence point.
• Sketch a typical titration curve for a monoprotic or polyprotic acid.
As you will see in the lesson that follows this one, water plays an essential role in acid-base chemistry as we ordinarily know it. To even those who know very little about chemistry, the term pH is recognized as a measure of "acidity", so the major portion of this unit is devoted to the definition of pH and of the pH scale. But since these topics are intimately dependant on the properties of water and its ability do dissociate into hydrogen and hydroxyl ions, we begin our discussion with this topic. We end this lesson with a brief discussion of acid-base titration— probably the most frequently carried-out chemistry laboratory operation in the world.
Dissociation of water
The ability of acids to react with bases depends on the tendency of hydrogen ions to combine with hydroxide ions to form water:
H+(aq) + OH(aq) → H2O (1)
This tendency happens to be very great, so the reaction is practically complete— but not "completely" complete; a few stray H+ and OH ions will always be present. What's more, this is true even if you start with the purest water attainable. This means that in pure water, the reverse reaction, the "dissociation" of water
H2O → H+(aq) + OH(aq) (2)
will proceed to a very slight extent. Both reactions take place simultaneously, but (1) is so much faster than (2) that only a minute fraction of H2O molecules are dissociated.
Liquids that contain ions are able to conduct an electric current. Pure water is practically an insulator, but careful experiments show that even the most highly purified water exhibits a very slight conductivity that corresponds to a concentration of both the H+ ion and OH ions of almost exactly 1.00 × 10–7mol L–1 at 25°C.
All chemical reactions that take place in a single phase (such as in a solution) are theoretically "incomplete" and are said to be reversible.
Example
What fraction of water molecules in a liter of water are dissociated
Solution: 1 L of water has a mass of 1000 g. The number of moles in 1000 g of H2O is
(1000 g)/(18 g mol–1) = 55.5 mol. This corresponds to (55.5 mol)(6.02E23 mol-1) = 3.34E25 H2O molecules.
An average of 10-7 mole, or (10-7)(6.02E23) = 6.0E16 H2O molecules will be dissociated at any time. The fraction of dissociated water molecules is therefore (6.0E16)/(3.3E25) = 1.8E–9.
Thus we can say that only about two out of every billion (109) water molecules will be dissociated.
Ion Product of water
The degree of dissociation of water is so small that you might wonder why it is even mentioned here. The reason stems from an important relationship that governs the concentrations of H+ and OH ions in aqueous solutions:
[H+][OH] = 1.00 × 10–14 (3)
must know this!
in which the square brackets [ ] refer to the concentrations (in moles per litre) of the substances they enclose.
Note
The quantity 1.00 x 10–14is commonly denoted by Kw. Its value varies slightly with temperature, pressure, and the presence of other ions in the solution.
This expression is known as the ion product of water, and it applies to all aqueous solutions, not just to pure water. The consequences of this are far-reaching, because it implies that if the concentration of H+ is large, that of OH will be small, and vice versa. This means that H+ ions are present in all aqueous solutions, not just acidic ones.
This leads to the following important definitions, which you must know:
acidic solution [H+] > [OH]
alkaline ("basic") solution [H+] < [OH]
neutral solution [H+] = [OH] = 1.00×10–7 mol L–1
Take special note of the following definition:
A neutral solution is one in which the concentrations of H+ and OH ions are identical.
The values of these concentrations are constrained by Eq. 3. Thus, in a neutral solution, both the hydrogen- and hydroxide ion concentrations are 1.00 × 10–7 mol L–1:
[H+][OH] = [1.00 × 10–7][1.00 × 10–7] =1.00 × 10–14
Hydrochloric acid is a typical strong acid that is totally dissociated in solution:
HCl → H+(aq) + Cl(aq)
A 1.0M solution of HCl in water therefore does not really contain any significant concentration of HCl molecules at all; it is a solution in of H+ and Cl in which the concentrations of both ions are 1.0 mol L–1. The concentration of hydroxide ion in such a solution, according to Eq 2, is
[OH] = (Kw)/[H+] = (1.00 x 10–14) / (1 mol L–1) = 1.00 x 10–14 mol L–1.
Similarly, the concentration of hydrogen ion in a solution made by dissolving 1.0 mol of sodium hydroxide in water will be 1.00 x 10–14 mol L–1.
2 pH
When dealing with a range of values (such as the variety of hydrogen ion concentrations encountered in chemistry) that spans many powers of ten, it is convenient to represent them on a more compressed logarithmic scale. By convention, we use the pH scale to denote hydrogen ion concentrations:
pH = – log10 [H+] (4) must know this!
or conversely, [H+] = 10–pH.
This notation was devised by the Danish chemist Soren Sorenson (1868-1939) in 1909. There are several accounts of why he chose "pH"; a likely one is that the letters stand for the French term pouvoir hydrogène, meaning "power of hydrogen"— "power" in the sense of an exponent. It has since become common to represent other small quantities in "p-notation". Two that you need to know in this course are the following:
pOH = – log10 [OH]
pKw = – log Kw (= 14 when Kw = 1.00 × 10–14)
Note that pH and pOH are expressed as numbers without any units, since logarithms must be dimensionless.
Recall from Eq 3 that [H+][OH] = 1.00 × 10–14; if we write this in "p-notation" it becomes
pH + pOH = 14
(5) must know this!
In a neutral solution at 25°C, pH = pOH = 7.0. As pH increases, pOH diminishes; a higher pH corresponds to an alkaline solution, a lower pH to an acidic solution. In a solution with [H+] = 1 M , the pH would be 0; in a 0.00010 M solution of H+, it would be 4.0. Similarly, a 0.00010 M solution of NaOH would have a pOH of 4.0, and thus a pH of 10.0. It is very important that you thoroughly understand the pH scale, and be able to convert between [H+] or [OH] and pH in both directions.
Example
The pH of blood must be held very close to 7.40. Find the hydroxide ion concentration that corresponds to this pH.
Solution: The pOH will be (14.0 – 7.40) = 6.60.
[OH] = 10–pOH = 10–6.6 = 2.51 x 10–7 M
The pH scale
The range of possible pH values runs from about 0 to 14.
The word "about" in the above statement reflects the fact that at very high concentrations (10 M hydrochloric acid or sodium hydroxide, for example,) a significant fraction of the ions will be associated into neutral pairs such as H+·Cl, thus reducing the concentration of “available” ions to a smaller value which we will call the effective concentration. It is the effective concentration of H+ and OH that determines the pH and pOH. For solutions in which ion concentrations don't exceed 0.1 M, the formulas pH = –log [H+] and pOH = –log[OH] are generally reliable, but don't expect a 10.0 M solution of a strong acid to have a pH of exactly –1.00!
The table shown here will help give you a general feeling for where common substances fall on the pH scale. Notice especially that
• most foods are slightly acidic;
• the principal "bodily fluids" are slightly alkaline, as is seawater— not surprising, since early animal life began in the oceans.
• the pH of freshly-distilled water will drift downward as it takes up carbon dioxide from the air; CO2 reacts with water to produce carbonic acid, H2CO3.
• the pH of water that occurs in nature varies over a wide range. Groundwaters often pick up additional CO2 respired by organisms in the soil, but can also become alkaline if they are in contact with carbonate-containing sediments. "Acid" rain is by definition more acidic than pure water in equilibrium with atmospheric CO2, owing mainly to sulfuric and nitric acids that originate from fossil-fuel emissions of nitrogen oxides and SO2.
pH indicators
The colors of many dye-like compounds depend on the pH, and can serve as useful indicators to determine whether the pH of a solution is above or below a certain value.
Natural indicator dyes
The best known of these is of course litmus, which has served as a means of distinguishing between acidic and alkaline substances since the early 18th century.
Many flower pigments are also dependent on the pH. You may have noticed that the flowers of some hydrangea shrub species are blue when grown in acidic soils, and white or pink in alkaline soils.
Red cabbage is a popular make-it-yourself indicator.
Universal indicators
Most indicator dyes show only one color change, and thus are only able to determine whether the pH of a solution is greater or less than the value that is characteristic of a particular indicator. By combining a variety of dyes whose color changes occur at different pHs, a "universal" indicator can be made. Commercially-prepared pH test papers of this kind are available for both wide and narrow pH ranges.
Titration
Since acids and bases readily react with each other, it is experimentally quite easy to find the amount of acid in a solution by determining how many moles of base are required to neutralize it. This operation is called titration, and you should already be familiar with it from your work in the Laboratory.
We can titrate an acid with a base, or a base with an acid. The substance whose concentration we are determining (the analyte) is the substance being titrated; the substance we are adding in measured amounts is the titrant. The idea is to add titrant until the titrant has reacted with all of the analyte; at this point, the number of moles of titrant added tells us the concentration of base (or acid) in the solution being titrated.
Example \(1\):
36.00 ml of a solution of HCl was titrated with 0.44 M KOH. The volume of KOH solution required to neutralize the acid solution was 27.00 ml. What was the concentration of the HCl?
Solution: The number of moles of titrant added was
(.027 L)(.44 mol L–1) = .0119 mol. Because one mole of KOH reacts with one mole of HCl, this is also the number of moles of HCl; its concentration is therefore
(.0119 mol) ÷ (.036 L) = 0.33 M .
Titration curves
The course of a titration can be followed by plotting the pH of the solution as a function of the quantity of titrant added. The figure shows two such curves, one for a strong acid (HCl) and the other for a weak acid, acetic acid, denoted by HAc. Looking first at the HCl curve, notice how the pH changes very slightly until the acid is almost neutralized. At that point, which corresponds to the vertical part of the plot, just one additional drop of NaOH solution will cause the pH to jump to a very high value— almost as high as that of the pure NaOH solution.
Compare the curve for HCl with that of HAc. For a weak acid, the pH jump near the neutralization point is less steep. Notice also that the pH of the solution at the neutralization point is greater than 7. These two characteristics of the titration curve for a weak acid are very important for you to know.
If the acid or base is polyprotic, there will be a jump in pH for each proton that is titrated. In the example shown here, a solution of carbonic acid H2CO3 is titrated with sodium hydroxide. The first equivalence point (at which the H2CO3 has been converted entirely into bicarbonate ion HCO3) occurs at pH 8.3. The solution is now identical to one prepared by dissolving an identical amount of sodium bicarbonate in water.
Addition of another mole equivalent of hydroxide ion converts the bicarbonate into carbonate ion and is complete at pH 10.3; an identical solution could be prepared by dissolving the appropriate amount of sodium carbonate in water.
Finding the equivalence point: indicators
When enough base has been added to react completely with the hydrogens of a monoprotic acid, the equivalence point has been reached. If a strong acid and strong base are titrated, the pH of the solution will be 7.0 at the equivalence point. However, if the acid is a weak one, the pH will be greater than 7; the “neutralized” solution will not be “neutral” in terms of pH. For a polyprotic acid, there will be an equivalence point for each titratable hydrogen in the acid. These typically occur at pH values that are 4-5 units apart, but they are occasionally closer, in which case they may not be readily apparent in the titration curve.
The key to a successful titration is knowing when the equivalence point has been reached. The easiest way of finding the equivalence point is to use an indicator dye; this is a substance whose color is sensitive to the pH. One such indicator that is commonly encountered in the laboratory is phenolphthalein; it is colorless in acidic solution, but turns intensely red when the solution becomes alkaline. If an acid is to be titrated, you add a few drops of phenolphthalein to the solution before beginning the titration. As the titrant is added, a local red color appears, but quickly dissipates as the solution is shaken or stirred. Gradually, as the equivalence point is approached, the color dissipates more slowly; the trick is to stop the addition of base after a single drop results in a permanently pink solution.
Different indicators change color at different pH values. Since the pH of the equivalence point varies with the strength of the acid being titrated, one tries to fit the indicator to the particular acid. One can titrate polyprotic acids by using a suitable combination of several indicators, as is illustrated above for carbonic acid. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.02%3A_Aqueous_Solutions-_pH_and_Titrations.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Explain the difference between the Arrhenius and Bronsted-Lowry concepts of acids and bases, and give examples of an acid and a base in which the Arrhenius concept is inapplicable.
• Explain why a hydrogen ion cannot exist in water.
• Given the formula of an acid or base, write the formula of its conjugate.
• State the fundamental difference between a strong acid and a weak acid.
• Describe the leveling effect, and explain its origin.
• State the factors that determine whether a solution of a salt will be acidic or alkaline.
• Write the equation for the autoprotolysis of a given ampholyte.
In this lesson we develop this concept and illustrate its applications to "strong" and "weak" acids and bases, emphasizing the common theme that acid-base chemistry is always a competition between two bases for the proton. In the final section, we show how the concept of "proton energy" can help us understand and predict the direction and extent of common types of acid-base reactions without the need for calculations.
Proton donors and acceptors
The older Arrhenius theory of acids and bases viewed them as substances which produce hydrogen ions or hydroxide ions on dissociation. As useful a concept as this has been, it was unable to explain why NH3, which contains no OH ions, is a base and not an acid, why a solution of FeCl3 is acidic, or why a solution of Na2S is alkaline. A more general theory of acids and bases was developed by Franklin in 1905, who suggested that the solvent plays a central role. According to this view, an acid is a solute that gives rise to a cation (positive ion) characteristic of the solvent, and a base is a solute that yields a anion (negative ion) which is also characteristic of the solvent. The most important of these solvents is of course H2O, but Franklin's insight extended the realm of acid-base chemistry into non-aqueous systems as we shall see in a later lesson.
Brønsted acids and bases
In 1923, the Danish chemist J.N. Brønsted, building on Franklin's theory, proposed that an acid is a proton donor; a base is a proton acceptor. In the same year the English chemist T.M. Lowry published a paper setting forth some similar ideas without producing a definition; in a later paper Lowry himself points out that Brønsted deserves the major credit, but the concept is still widely known as the Brønsted-Lowry theory.
Brønsted-Lowry Acids and Bases
An acid is a proton donor and a base is a proton acceptor.
These definitions carry a very important implication: a substance cannot act as an acid without the presence of a base to accept the proton, and vice versa. As a very simple example, consider the equation that Arrhenius wrote to describe the behavior of hydrochloric acid:
$HCl \rightarrow H^+ + A^–$
This is fine as far as it goes, and chemists still write such an equation as a shortcut. But in order to represent this more realistically as a proton donor-acceptor reaction, we now depict the behavior of HCl in water by
in which the acid HCl donates its proton to the acceptor (base) H2O.
"Nothing new here", you might say, noting that we are simply replacing a shorter equation by a longer one. But consider how we might explain the alkaline solution that is created when ammonia gas NH3 dissolves in water. An alkaline solution contains an excess of hydroxide ions, so ammonia is clearly a base, but because there are no OH ions in NH3, it is clearly not an Arrhenius base. It is, however, a Brønsted base:
In this case, the water molecule acts as the acid, donating a proton to the base NH3 to create the ammonium ion NH4+.
The foregoing examples illustrate several important aspects of the Brønsted-Lowry concept of acids and bases:
• A substance cannot act as an acid unless a proton acceptor (base) is present to receive the proton;
• A substance cannot act as a base unless a proton donor (acid) is present to supply the proton;
• Water plays a dual role in many acid-base reactions; H2O can act as a proton acceptor (base) for an acid, or it can serve as a proton donor (acid) for a base (as we saw for ammonia.
• The hydronium ion H3O+ plays a central role in the acid-base chemistry of aqueous solutions.
Brønsted
Brønsted (1879-1947) was a Danish physical chemist. Although he is now known mainly for his proton donor-acceptor theory of acids and bases (see his original article), he published numerous earlier papers on chemical affinity, and later on the catalytic effects of acids and bases on chemical reactions. In World War II he opposed the Nazis, and this led to his election to the Danish parliament in 1947, but he was unable to take his seat because of illness and he died later in that year.
Lowry
Lowry (1874-1936) was the first holder of the chair in physical chemistry at Cambridge University. His extensive studies of the effects of acids and bases on the optical behavior of camphor derivatives (specifically, how they rotate the plane of polarized light) led him to formulate a theory of acids and bases similar to and simultaneously with that of Brønsted.
The Hydronium ion
There is another serious problem with the Arrhenius view of an acid as a substance that dissociates in water to produce a hydrogen ion. The hydrogen ion is no more than a proton, a bare nucleus. Although it carries only a single unit of positive charge, this charge is concentrated into a volume of space that is only about a hundred-millionth as large as the volume occupied by the smallest atom. (Think of a pebble sitting in the middle of a sports stadium!) The resulting extraordinarily high charge density of the proton strongly attracts it to any part of a nearby atom or molecule in which there is an excess of negative charge. In the case of water, this will be the lone pair (unshared) electrons of the oxygen atom; the tiny proton will be buried within the lone pair and will form a shared-electron (coordinate) bond with it, creating a hydronium ion, H3O+. In a sense, H2O is acting as a base here, and the product H3O+ is the conjugate acid of water:
Owing to the overwhelming excess of $H_2O$ molecules in aqueous solutions, a bare hydrogen ion has no chance of surviving in water.
Although other kinds of dissolved ions have water molecules bound to them more or less tightly, the interaction between H+ and H2O is so strong that writing “H+(aq)” hardly does it justice, although it is formally correct. The formula H3O+ more adequately conveys the sense that it is both a molecule in its own right, and is also the conjugate acid of water.
The equation "HA → H+ + A" is so much easier to write that chemists still use it to represent acid-base reactions in contexts in which the proton donor-acceptor mechanism does not need to be emphasized. Thus it is permissible to talk about “hydrogen ions” and use the formula H+ in writing chemical equations as long as you remember that they are not to be taken literally in the context of aqueous solutions.
Interestingly, experiments indicate that the proton does not stick to a single H2O molecule, but changes partners many times per second. This molecular promiscuity, a consequence of the uniquely small size and mass the proton, allows it to move through the solution by rapidly hopping from one H2O molecule to the next, creating a new H3O+ ion as it goes. The overall effect is the same as if the H3O+ ion itself were moving. Similarly, a hydroxide ion, which can be considered to be a “proton hole” in the water, serves as a landing point for a proton from another H2O molecule, so that the OH ion hops about in the same way.
Because hydronium- and hydroxide ions can “move without actually moving” and thus without having to plow their way through the solution by shoving aside water molecules as do other ions, solutions which are acidic or alkaline have extraordinarily high electrical conductivities.
Acid-base reactions à la Brønsted
According to the Brønsted concept, the process that was previously written as a simple dissociation of a generic acid HA ("HA → H+ + A)" is now an acid-base reaction in its own right:
$HA + H_2O \rightarrow A^- + H_3O^+$
The idea, again, is that the proton, once it leaves the acid, must end up somewhere; it cannot simply float around as a free hydrogen ion.
Conjugate pairs
A reaction of an acid with a base is thus a proton exchange reaction; if the acid is denoted by AH and the base by B, then we can write a generalized acid-base reaction as
$AH + B \rightarrow A^- + BH^+$
Notice that the reverse of this reaction,
$BH^+ + A^- \rightarrow B + AH^+$
is also an acid-base reaction. Because all simple reactions can take place in both directions to some extent, it follows that transfer of a proton from an acid to a base must necessarily create a new pair of species that can, at least in principle, constitute an acid-base pair of their own.
In this schematic reaction, base1 is said to be conjugate to acid1, and acid2 is conjugate to base2. The term conjugate means “connected with”, the implication being that any species and its conjugate species are related by the gain or loss of one proton. The table below shows the conjugate pairs of a number of typical acid-base systems.
Some common conjugate acid-base pairs
acid
base
hydrochloric acid HCl Cl chloride ion
acetic acid CH3CH2COOH CH3CH2COO acetate ion
nitric acid HNO3 NO3 nitrate ion
dihydrogen phosphate ion H2PO4 HPO4 monohydrogen phosphate ion
hydrogen sulfate ion HSO4 SO42– sulfate ion
hydrogen carbonate ("bicarbonate") ion HCO3 CO32– carbonate ion
ammonium ion NH4+ NH3 ammonia
iron(III) ("ferric") ion Fe(H2O)63+ Fe(H2O)5OH2+
water H2O OH hydroxide ion
hydronium ion H3O+ H2O water
Strong acids and weak acids
We can look upon the generalized acid-base reaction
as a competition of two bases for a proton:
If the base H2O overwhelmingly wins this tug-of-war, then the acid HA is said to be a strong acid. This is what happens with hydrochloric acid and the other common strong "mineral acids" H2SO4, HNO3, and HClO4:
hydrochloric acid HCl + H2O → Cl + H3O+
sulfuric acid H2SO4 + H2O → HSO4 + H3O+
nitric acid HNO3 + H2O → NO3 + H3O+
perchloric acid HClO4 + H2O → ClO4 + H3O+
Solutions of these acids in water are really solutions of the ionic species shown in heavy type on the right. This being the case, it follows that what we call a 1 M solution of "hydrochloric acid" in water, for example, does not really contain a significant concentration of HCl at all; the only real a acid present in such a solution is H3O+!
These considerations give rise to two important rules:
1. H3O+ is the strongest acid that can exist in water;
2. All strong acids appear to be equally strong in water.
The Leveling Effect
The second of these statements is called the leveling effect. It means that although the inherent proton-donor strengths of the strong acids differ, they are all completely dissociated in water. Chemists say that their strengths are "leveled" by the solvent water.
A comparable effect would be seen if one attempted to judge the strengths of several adults by conducting a series of tug-of-war contests with a young child. One would expect the adults to win overwhelmingly on each trial; their strengths would have been "leveled" by that of the child.
Weak acids
Most acids, however, are able to hold on to their protons more tightly, so only a small fraction of the acid is dissociated. Thus hydrocyanic acid, HCN, is a weak acid in water because the proton is able to share the lone pair electrons of the cyanide ion CN more effectively than it can with those of H2O, so the reaction
$HCN + H_2O \rightarrow H_3O^+ + CN^–$
proceeds to only a very small extent. Since a strong acid binds its proton only weakly, while a weak acid binds it tightly, we can say that
Strong acids are "weak" and weak acids are "strong." If you are able to explain this apparent paradox, you understand one of the most important ideas in acid-base chemistry!
Examples of proton donor-acceptor reactions
reaction
acid
base
conjugate acid
conjugate base
autoionization of water H2O H2O H2O H3O+ OH
ionization of hydrocyanic acid HCN HCN H2O H3O+ CN
ionization of ammonia NH3 in water NH3 H2O NH4+ OH
hydrolysis of ammonium chloride NH4Cl NH4+ H2O H3O+ NH3
hydrolysis of sodium acetate CH3COO- Na+ H2O CH3COO CH3COOH OH
neutralization of HCl by NaOH HCl OH H2O Cl
neutralization of NH3 by acetic acid CH3COOH NH3 NH4+ CH3COO
dissolution of BiOCl (bismuth oxychloride) by HCl 2 H3O+ BiOCl Bi(H2O)3+ H2O, Cl
decomposition of Ag(NH3)2+ by HNO3 2 H3O+ Ag(NH3)2+ NH4+ H2O
displacement of HCN by CH3COOH CH3COOH CN HCN CH3COO
Strong acids have weak conjugate bases
This is just a re-statement of what is implicit in what has been said above about the distinction between strong acids and weak acids. The fact that HCl is a strong acid implies that its conjugate base Cl is too weak a base to hold onto the proton in competition with either H2O or H3O+. Similarly, the CN ion binds strongly to a proton, making HCN a weak acid.
Salts of weak acids give alkaline solutions
The fact that HCN is a weak acid implies that the cyanide ion CN reacts readily with protons, and is thus is a relatively good base. As evidence of this, a salt such as KCN, when dissolved in water, yields a slightly alkaline solution:
CN + H2O → HCN + OH
This reaction is still sometimes referred to by its old name hydrolysis ("water splitting"), which is literally correct but tends to obscure its identity as just another acid-base reaction. Reactions of this type take place only to a small extent; a 0.1M solution of KCN is still, for all practical purposes, 0.1M in cyanide ion.
In general, the weaker the acid, the more alkaline will be a solution of its salt. However, it would be going to far to say that "ordinary weak acids have strong conjugate bases." The only really strong base is hydroxide ion, OH, so the above statement would be true only for the very weak acid H2O.
Strong bases and weak bases
The only really strong bases you are likely to encounter in day-to-day chemistry are alkali-metal hydroxides such as NaOH and KOH, which are essentially solutions of the hydroxide ion. Most other compounds containing hydroxide ions such as Fe(OH)3 and Ca(OH)2 are not sufficiently soluble in water to give highly alkaline solutions, so they are not usually thought of as strong bases.
There are actually a number of bases that are stronger than the hydroxide ion — best known are the oxide ion O2– and the amide ion NH2, but these are so strong that they can rob water of a proton:
O2– + H2O → 2 OH
NH2 + H2O → NH3 + OH
This gives rise to the same kind of leveling effect we described for acids, with hydroxide ion as the strongest base in water.
Hydroxide ion is the strongest base that can exist in aqueous solution.
The most common example of this is ammonium chloride, NH4Cl, whose aqueous solutions are distinctly acidic:
NH4+ + H2O → NH3 + H3O+
Because this (and similar) reactions take place only to a small extent, a solution of ammonium chloride will only be slightly acidic.
Autoprotolysis
From some of the examples given above, we see that water can act as an acid
CN + H2O → HCN + OH
and as a base
NH4+ + H2O → NH3 + H3O+
If this is so, then there is no reason why "water-the-acid" cannot donate a proton to "water-the-base":
This reaction is known as the autoprotolysis of water.
Chemists still often refer to this reaction as the "dissociation" of water and use the Arrhenius-style equation H2O → H+ + OHas a kind of shorthand. As discussed in the previous lesson, this process occurs to only a tiny extent. It does mean, however, that hydronium and hydroxide ions are present in any aqueous solution.
Ammonia and Sulfuric acid
Other liquids also exhibit autoprotolysis with the most well-known example is liquid ammonia:
2 NH3 → NH4+ + NH2
Even pure liquid sulfuric acid can play the game:
2 H2SO4→ H3SO4+ + HSO4
Each of these solvents can be the basis of its own acid-base "system", parallel to the familiar "water system".
Ampholytes
Water, which can act as either an acid or a base, is said to be amphiprotic: it can "swing both ways". A substance such as water that is amphiprotic is called an ampholyte.
As indicated here, the hydroxide ion can also be an ampholyte, but not in aqueous solution in which the oxide ion cannot exist.
It is of course the amphiprotic nature of water that allows it to play its special role in ordinary aquatic acid-base chemistry. But many other amphiprotic substances can also exist in aqueous solutions. Any such substance will always have a conjugate acid and a conjugate base, so if you can recognize these two conjugates of a substance, you will know it is amphiprotic.
The carbonate system
For example, the triplet set {carbonic acid, bicarbonate ion, carbonate ion} constitutes an amphiprotric series in which the bicarbonate ion is the ampholyte, differing from either of its neighbors by the addition or removal of one proton:
If the bicarbonate ion is both an acid and a base, it should be able to exchange a proton with itself in an autoprotolysis reaction:
$HCO_3^– + HCO_3^– \rightarrow H_2CO_3 + CO_3^{2-}$
Carbonic Acid
Your very life depends on the above reaction! CO2, a metabolic by-product of every cell in your body, reacts with water to form carbonic acid H2CO3 which, if it were allowed to accumulate, would make your blood fatally acidic. However, the blood also contains carbonate ion, which reacts according to the reverse of the above equation to produce bicarbonate which can be safely carried by the blood to the lungs. At this location the autoprotolysis reaction runs in the forward direction, producing H2CO3 which loses water to form CO2 which gets expelled in the breath. The carbonate ion is recycled back into the blood to eventually pick up another CO2 molecule.
If you can write an autoprotolysis reaction for a substance, then that substance is amphiprotic. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.03%3A_Acid-base_reactions_a_la_Brnsted.txt |
Learning Objectives
It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Sketch out a proton-energy diagram for a strong acid, a weak acid or base, and for a strong base.
• Describe how the pH affects the relative concentrations of a conjugate acid-base pair.
Will this acid react with that base? And if so, to what extent? These questions can be answered quantitatively by carrying out the detailed equilibrium calculations you will learn about in another lesson. However, modern acid-base chemistry offers a few simple principles that can enable you to make a qualitative decision at a glance. More importantly, the ideas which we develop in this section are guaranteed to give you a far better conceptual understanding of proton-based acid-base reactions in general.
Which base gets the proton?
Will acid HA react with base B? We stated above that the outcome of any acid-base reaction depends on how well two different bases can compete in the tug-of-war for the proton
$A^– \leftarrow H^+\rightarrow B^– \label{9.4.1}$
The proton will always go to the stronger base. Some insight into this can be had by thinking of the proton as having different potential energies when it is bound to different acceptors. We can draw a useful analogy with the electrons in an atom, which, you will recall, will always fall into the lowest-potential energy orbitals available, filling them from the bottom up. In a similar way, protons will "fall" into the lowest-energy empty spots (bases) they can find.
Consider the scheme shown here, which depicts two hypothetical acid-base conjugate pairs. Take careful note of the labeling of this diagram: the acids HA and HB are proton sources and the conjugate bases A and B are proton sinks. This "source-sink" terminology is synonymous with the "donor-acceptor" language that Brønsted taught us, but it also carries an implication about the relative energies of the proton as it exists in the two molecules HA and HB. If, as is indicated here, the proton is at a higher "potential energy" when it is in the form of HA than in HB, the reaction HA + B → HB + A will be favored compared to the reverse process HB + A → HA + B, which would require elevating the proton up to the A level. In this example, HA is the stronger acid because its proton can fall to a lower potential energy when it joins with B to form HB.
We will refer to diagrams such as the one in Figure $1$ as "proton-energy diagrams", which is not quite correct, but we do not want to get into thermodynamics at this point. (If you already know something about chemical thermodynamics, we are really referring to Gibbs energy.)
It follows, then, that if we can arrange all the common acid-base conjugate pairs on this kind of a scale, we can predict the direction of any simple acid-base reaction without resorting to numbers. This will be illustrated further on, but in order to keep things simple, let's look at a few proton-energy diagrams that illustrate some of the acid-base chemistry that we discussed in the preceding section.
Strong acids and weak acids
The hydronium ion is the dividing line; a strong acid, you will recall, is one whose conjugate base A loses out to the "stronger" base H2O in the competition for the proton:
$A^– \leftarrow H^+ \rightarrow H_2O \label{9.4.2}$
An acid that is a stronger proton donor than hydronium ion is a "strong" acid; if it is a weaker proton donor than H3O+, it is by definition "weak".This is seen most clearly in the diagram here, which contrasts the strong acid HA with the weak acid HB. HB "dissociates" to only a tiny extent because it is energetically unfavorable to promote its proton up to the H2O-H3O+ level (process 3 in the diagram).
Strong acids and leveling
A strong acid, you will recall, is one whose conjugate base A loses out to the "stronger" base H2O in the competition for the proton:
$A^- ← H^+→ H_2O \label{9.4.3}$
Because the reaction
$HA + H_2O \rightarrow A^–+ H_3O^+ \label{9.4.4}$
for any strong acid HA is virtually complete, all strong acids appear to be equally strong in water (the leveling effect.)
From the proton-energy standpoint, a strong acid is one in which the energy of the proton is substantially greater when attached to the anion Athan when it is attached to H2O. Adding a strong acid HA to water will put it in contact with a huge proton sink that drains off the protons from any such acid, leaving the conjugate base A along with hydronium ion, the strongest acid that can exist in water.
Weak bases
Conjugate bases of weak acids tend to accept protons from water, leaving a small excess of OH ions and thus an alkaline solution. As you can see in the diagram, the weak base ammonia accepts a proton from water:
$NH_3 + H_2O \rightarrow NH_4^+ + OH^– \label{9.4.6}$
The "weakness" of such a base is a consequence of the energetically unfavorable process (1) in which a proton must be raised up from the low-lying H2O-OH level. From the standpoint of the "proton sources" column on the left, you can think of this as similar to the situation for weak acids that we discussed above; it can be considered a special case in which the weak acid is H2O.
The weakest acid and the strongest base
For a very long time, chemists had regarded methane, CH4, as the weakest acid, making the methide ion CH3 (which is also the simplest carbanion) the strongest base. Methane still holds its position as the weakest acid, but in 2008, the ion LiOwas found to be an even stronger base than CH4. Because both of these bases are observable only in the gas phase, these facts have little obvious import on aqueous-solution chemistry.
Autoprotolysis
Because water is amphiprotic, one H2O molecule can donate a proton to another, as explained above. In this case the proton has to acquire considerable energy to make the jump (1) from the H2O-OH level to the H3O+-H2O level, so the reaction
$2 H_2O \rightarrow H_3O^++ OH^– \label{9.4.7}$
occurs only to a minute extent. Think of this as the special case of the "weakest" acid H2O reacting with the "weakest" base H2O.
Strong bases
Finally, what is a strong base? Just as a strong acid lies above the H3O+-H2O level, so does a strong base lie below the H2O-OH level. And for the same reason that H3O+ is the strongest acid that can exist in water, OH is the strongest base that can exist in water. The example of the oxide ion O2– is shown here. Sodium oxide Na2O is a white powder that dissolves in water to give oxide ions which immediately decompose into hydroxide ions
$O^{2–} + H_2O \rightarrow 2 OH^– \label{9.4.8}$
Putting it all together, and the meaning of pH
This table combines common examples covering the entire range of acid-base strengths, from the strong to the very weak. The energy scale at the left gives you some idea of the relative proton-energy levels for each conjugate pair; notice that the zero is arbitrarily set to that of the H3O+-H2O pair.
Of more importance is the pH scale on the right. The pH that corresponds to any conjugate pair is the pH at which equal concentrations of that pair are in their acid and base forms. For example, acetic acid CH3COOH is "half ionized" at a pH of 4.7. If another strong acid such as HCl is added so as to reduce the pH, the proportion of acetate ion decreases, while if sodium hydroxide is added to force the pH higher, a larger fraction of the acetic acid will be "dissociated".
This illustrates another aspect of pH: at its most fundamental level, pH is an inverse measure of the "proton intensity" in the solution. The lower the pH, the higher the proton intensity, and the greater will be the fraction of higher-energy proton levels populated— which translates to higher acid-to-conjugate base concentration ratios. It is easy to see why acids such as H2SO4 and bases such as the amide ion NH2 cannot exist in aqueous solution; the pH would have to be at the impossible level of –6 for the former and +23 for the latter!
Why acids are titrated with hydroxide ion
When you titrate an acid with a base, you want virtually every molecule of the acid to react with the base. In the case of a weak acid such as hypochlorous, the reaction would be
$HOCl + OH^– \rightarrow OCl^– + H_2O \label{9.4.9}$
Because the proton level in HOCl is considerably above that in H2O, titration with NaOH solution will ensure that every last proton is eaten up by the hydroxide ion. If, instead, you used ammonia NH3 as a titrant, the closeness of the two proton levels would cause the reaction to be incomplete, yielding a less distinct equivalence point. And, of course, titration with a base that is weaker then hypochlorite ion (such as sodium bicarbonate) would be hopeless.
As a practical matter, you can usually estimate that when the pH differs by more than about two units from the value that corresponds to the conjugate-pair for a monoprotic acid, the concentration of the non-favored species will be down by a factor of around 1000. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.04%3A_Acid-Base_Reactions.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Write the equation for the proton transfer reaction involving a Brønsted-Lowry acid or base, and show how it can be interpreted as an electron-pair transfer reaction, clearly identifying the donor and acceptor.
• Give an example of a Lewis acid-base reaction that does not involve protons.
• Write equations illustrating the behavior of a given non-aqueous acid-base system.
The Brønsted-Lowry proton donor-acceptor concept has been one of the most successful theories of Chemistry. But as with any such theory, it is fair to ask if this is not just a special case of a more general theory that could encompass an even broader range of chemical science. In 1916, G.N. Lewis of the University of California proposed that the electron pair is the dominant actor in acid-base chemistry. The Lewis theory did not become very well known until about 1923 (the same year that Brønsted and Lowry published their work), but since then it has been recognized as a very powerful tool for describing chemical reactions of widely different kinds and is widely used in organic and inorganic chemistry. According to Lewis,
• An acid is a substance that accepts a pair of electrons, and in doing so, forms a covalent bond with the entity that supplies the electrons.
• A base is a substance that donates an unshared pair of electrons to a recipient species with which the electrons can be shared.
In modern chemistry, electron donors are often referred to as nucleophiles, while acceptors are electrophiles.
Proton-Transfer Reactions Involve Electron-Pair Transfer
Just as any Arrhenius acid is also a Brønsted acid, any Brønsted acid is also a Lewis acid, so the various acid-base concepts are all "upward compatible". Although we do not really need to think about electron-pair transfers when we deal with ordinary aqueous-solution acid-base reactions, it is important to understand that it is the opportunity for electron-pair sharing that enables proton transfer to take place.
This equation for a simple acid-base neutralization shows how the Brønsted and Lewis definitions are really just different views of the same process. Take special note of the following points:
• The arrow shows the movement of a proton from the hydronium ion to the hydroxide ion.
• Note carefully that the electron-pairs themselves do not move; they remain attached to their central atoms. The electron pair on the base is "donated" to the acceptor (the proton) only in the sense that it ends up being shared with the acceptor, rather than being the exclusive property of the oxygen atom in the hydroxide ion.
• Although the hydronium ion is the nominal Lewis acid here, it does not itself accept an electron pair, but acts merely as the source of the proton that coordinates with the Lewis base.
The point about the electron-pair remaining on the donor species is especially important to bear in mind. For one thing, it distinguishes a Lewis acid-base reaction from an oxidation-reduction reaction, in which a physical transfer of one or more electrons from donor to acceptor does occur. The product of a Lewis acid-base reaction is known formally as an "adduct" or "complex", although we do not ordinarily use these terms for simple proton-transfer reactions such as the one in the above example. Here, the proton combines with the hydroxide ion to form the "adduct" H2O. The following examples illustrate these points for some other proton-transfer reactions that you should already be familiar with.
Another example, showing the autoprotolysis of water. Note that the conjugate base is also the adduct.
Ammonia is both a Brønsted and a Lewis base, owing to the unshared electron pair on the nitrogen. The reverse of this reaction represents the hydrolysis of the ammonium ion.
Because $\ce{HF}$ is a weak acid, fluoride salts behave as bases in aqueous solution. As a Lewis base, F accepts a proton from water, which is transformed into a hydroxide ion.
The bisulfite ion is amphiprotic and can act as an electron donor or acceptor.
Acid-base Reactions without Transferring Protons
The major utility of the Lewis definition is that it extends the concept of acids and bases beyond the realm of proton transfer reactions. The classic example is the reaction of boron trifluoride with ammonia to form an adduct:
$\ce{BF_3 + NH_3 \rightarrow F_3B-NH_3}$
One of the most commonly-encountered kinds of Lewis acid-base reactions occurs when electron-donating ligands form coordination complexes with transition-metal ions.
Exercise $1$
Here are several more examples of Lewis acid-base reactions that cannot be accommodated within the Brønsted or Arrhenius models. Identify the Lewis acid and Lewis base in each reaction.
1. $\ce{Al(OH)_3 + OH^{–} \rightarrow Al(OH)_4^–}$
2. $\ce{SnS_2 + S^{2–} \rightarrow SnS_3^{2–}}$
3. $\ce{Cd(CN)_2 + 2 CN^– \rightarrow Cd(CN)_4^{2+}}$
4. $\ce{AgCl + 2 NH_3 \rightarrow Ag(NH_3)_2^+ + Cl^–}$
5. $\ce{Fe^{2+} + NO \rightarrow Fe(NO)^{2+}}$
6. $\ce{Ni^{2+} + 6 NH_3 \rightarrow Ni(NH_3)_5^{2+}}$
Applications to organic reaction mechanisms
Although organic chemistry is beyond the scope of these lessons, it is instructive to see how electron donors and acceptors play a role in chemical reactions. The following two diagrams show the mechanisms of two common types of reactions initiated by simple inorganic Lewis acids:
In each case, the species labeled "Complex" is an intermediate that decomposes into the products, which are conjugates of the original acid and base pairs. The electric charges indicated in the complexes are formal charges, but those in the products are "real".
In reaction 1, the incomplete octet of the aluminum atom in $\ce{AlCl3}$ serves as a better electron acceptor to the chlorine atom than does the isobutyl part of the base. In reaction 2, the pair of non-bonding electrons on the dimethyl ether coordinates with the electron-deficient boron atom, leading to a complex that breaks down by releasing a bromide ion.
Non-aqueous Protonic Acid-Base Systems
We ordinarily think of Brønsted-Lowry acid-base reactions as taking place in aqueous solutions, but this need not always be the case. A more general view encompasses a variety of acid-base solvent systems, of which the water system is only one (Table $1$). Each of these has as its basis an amphiprotic solvent (one capable of undergoing autoprotolysis), in parallel with the familiar case of water.
The ammonia system is one of the most common non-aqueous system in Chemistry. Liquid ammonia boils at –33° C, and can conveniently be maintained as a liquid by cooling with dry ice (–77° C). It is a good solvent for substances that also dissolve in water, such as ionic salts and organic compounds since it is capable of forming hydrogen bonds. However, many other familiar substances can also serve as the basis of protonic solvent systems as Table $1$ indicates:
Table $1$: Popular Solvent systems
solvent
autoprotolysis reaction
pKap
water 2 H2O → H3O+ + OH 14
ammonia 2 NH3 → NH4+ + NH2 33
acetic acid 2 CH3COOH → CH3COOH2+ + CH3COO 13
ethanol 2 C2H5OH → C2H5OH2+ + C2H5O 19
hydrogen peroxide 2 HO-OH → HO-OH2+ + HO-O 13
hydrofluoric acid 2 HF → H2F+ + F 10
sulfuric acid 2 H2SO4 → H3SO4+ + HSO4 3.5
One use of nonaqueous acid-base systems is to examine the relative strengths of the strong acids and bases, whose strengths are "leveled" by the fact that they are all totally converted into H3O+ or OH ions in water. By studying them in appropriate non-aqueous solvents which are poorer acceptors or donors of protons, their relative strengths can be determined. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.05%3A_Lewis_Acids_and_Bases.txt |
Learning Objectives
• Given the formula of a binary compound of hydrogen with an element of Z<18, predict whether its aqueous solution be acidic or basic, and write an appropriate equation.
• Do the same for an oxygen- or hydroxy compound of a similar element.
• Define an oxyacid, and explain why many of these are very strong acids.
• Write an equation describing the amphoteric nature of zinc or aluminum hydroxide.
• Define an acid anhydride, and write an equation describing its behavior.
• Explain how metal cations can give acidic solutions.
• Write equations showing why aqueous solutions of some salts are acidic, while others are alkaline.
• Write the formulas of an organic acid and an organic base, and write an equation showing why the latter gives an alkaline solution in water.
You will already have noticed that not every compound that contains hydrogen atoms is acidic; ammonia NH3, for example, gives an alkaline aqueous solution. Similarly, some compounds containing the group -OH are basic, but others are acidic. An important part of understanding chemistry is being able to recognize what substances will exhibit acidic and basic properties in aqueous solution. Fortunately, most of the common acids and bases fall into a small number of fairly well-defined groups, so this is not particularly difficult.
Binary Hydrides
Strictly speaking, the term hydride refers to ionic compounds of hydrogen with the electropositive metals of Groups 1-2; these contain the hydride ion, H, and are often referred to as "true" hydrides. However, the term is often used in its more general sense to refer to any binary compound MHn in which M stands for any element. The hydride ion is such a strong base that it cannot exist in water, so salts such as sodium hydride react with water to yield hydrogen gas and an alkaline solution:
$NaH + H_2O \rightarrow Na^+ + OH^– + H_2$
The more electronegative elements form covalent hydrides which generally react as acids, a well-known example being hydrogen chloride, a gas which dissolves readily in water to give the solution we know as hydrochloric acid
$HCl_{(g)} + H_2O_{(l)} \rightarrow H_3O^+ + Cl^–$
Most of the covalent hydrogen compounds are weak acids— in some cases, such as methane, CH4, so weak that their acidic properties are rarely evident. Many, such as H2O and NH3, are amphiprotic. The latter compound, ammonia, is a weaker acid than H2O, so it exhibits basic properties in water
$NH_3 + H_2O \rightarrow NH_4^+ + OH^–$
but behaves as an acid in non-aqueous solvents such as liquid ammonia itself:
$NH_3 + NH_3 \rightarrow NH_4^+ + NH_2^–$
In general, the acidity of the non-metallic hydrides increases with the atomic number of the element to which it is connected. Thus as the element M moves from left to right across the periodic table or down within a group, the acids MH become stronger, as indicated by the acid dissociation constants shown at the right. Note that
• The formulas shown in red represent "strong" acids (that is, acids stronger than H3O+.)
• Hydrofluoric acid is the only weak member of the hydrohalogen acids.
• Acids weaker than water do not behave as acids in aqueous solution. Thus for most practical purposes, methane and ammonia are not commonly regarded as acids. H2O itself is treated as an acid only in the narrow context of aqueous solution chemistry.
Attempts to explain these trends in terms of a single parameter such as the electronegativity of M tend not to be very useful. The reason is that acid strengths depend on a number of factors such as the strength of the M-H bond and the energy released when the resultant ions become hydrated in solution. This last factor plays a major role in making HF something of an anomaly amongst the strong acids of Group 17.
Ammonia is such a weak acid that its conjugate base, amide ion NH2, cannot exist in water. In aqueous solution, NH3 acts as a weak base, accepting a proton from water and leaving a OH ion. An aqueous solution of NH3 is sometimes called “ammonium hydroxide”. This misnomer reflects the pre-Brønsted view that all bases contain –OH units that yield hydroxide ions on dissociation according to the Arrhenius scheme
$NH_4OH \rightleftharpoons NH_4^+ + OH^–$
A solution of ammonia in water is more correctly referred to as "aqueous ammonia" and represented by the formula NH3(aq). There is no physical evidence for the existence of NH4OH, but the name seems to remain forever etched on reagent bottles in chemical laboratories and in the vocabularies of chemists.
Hydroxy Compounds
Compounds containing the hydroxyl group –OH constitute the largest category of acids, especially if the organic acids (discussed separately farther on) are included. M–OH compounds also include many of the most common bases.
Whether a compound of the general type M–O–H will act as an acid or a base depends is influenced by the relative tendencies of the M–O and the O–H bonds to break apart in water. If the M–O bond cleaves more readily, then the –OH part will tend to retain its individuality and with its negative charge will become a hydroxide ion. If the O–H bond breaks, the MO-part of the molecule will remain intact as an oxyanion MO and release of the proton will cause the MOH compound to act as an acid.
This is not solely a matter of the relative strengths of the two bonds; the energy change that occurs when the resulting ions interact with water molecules is also an important factor.
In general, if M is a metallic element, the metal hydroxide compound $\ce{MOH}$ will be basic. The case of the highly electropositive elements of Groups 1 and 2 is somewhat special in that their solid MOH compounds exist as interpenetrating lattices of metal cations and OH ions, so those that can dissolve readily in water form strongly alkaline solutions; KOH and NaOH are well known examples of strong bases. From the Brønsted standpoint, these different “bases” are really just different sources for the single strong base OH. As one moves into Group 2 of the periodic table the M-OH compounds become less soluble; thus a saturated solution of Ca(OH)2 (commonly known as limewater) is only weakly alkaline. Hydroxides of the metallic elements of the p-block and of the transition metals are so insoluble that their solutions are not alkaline at all. Nevertheless these solids dissolve readily in acidic solutions to yield a salt plus water, so they are formally bases.
The acidic character of hydroxy compounds of the nonmetals, known collectively as oxyacids, is attributed to the displacement of negative charge from the hydroxylic oxygen atom by the electronegative central atom. The net effect is to make the oxygen slightly more positive, thus easing the departure of the hydrogen as H+ . The presence of other electron-attracting groups on the central atom has a marked effect on the strength of an oxyacid. Of special importance is the doubly-bonded oxygen atom. With the exception of the halogen halides, all of the common strong acids contain one or more of these oxygens, as in sulfuric acid SO2(OH), nitric acid NO2(OH) and phosphoric acid PO(OH)3. In general the strengths of these acids depends more on the number of oxygens than on any other factor, so periodic trends are not so important.
Most of the halogen elements form more than one oxyacid. Fluorine is an exception; being more electronegative than oxygen, no oxyacids of this element are known. Chlorine is the only halogen for which all four oxyacids are known, and the Ka values for this series show how powerfully the Cl–O oxygen atoms affect the acid strength.
Oxygen compounds
Binary oxides that contain no hydrogen atoms can exhibit acid-base behavior when they react with water. The division between acidic and basic oxygen oxides largely parallels that between the hydroxy compounds. The oxygen compounds of the highly electropositive metals of Groups 1-2 actually contain the oxide ion O. This ion is another case of a proton acceptor that is stronger than OH, and thus cannot exist in aqueous solution. Ionic oxides therefore tend to give strongly alkaline solutions:
$\ce{O^{-} + H2O -> 2OH^{-} (aq)}$
In some cases, such as that of MgO, the solid is so insoluble that little change in pH is noticed when it is placed in water. CaO, however, which is known as quicklime, is sufficiently soluble to form a strongly alkaline solution with the evolution of considerable heat; the result is the slightly-soluble slaked lime, Ca(OH)2. Oxygen compounds of the transition metals are generally insoluble solids having rather complex extended structures. Although some will dissolve in acids, they display no acidic properties in water.
Amphoteric oxides and hydroxides
The oxides and hydroxides of the metals of Group 3 and higher tend to be only weakly basic, and most display an amphoteric nature. Most of these compounds are so slightly soluble in water that their acidic or basic character is only obvious in their reactions with strong acids or bases. In general, these compounds tend to be more basic than acidic; thus the oxides and hydroxides of aluminum, iron, and zinc all dissolve in mildly acidic solutions, whereas they require treatment with concentrated hydroxide ion solutions to react as acids.
Al(OH)3 +3 H+→ Al3+(aq) +3H2O Al(OH)3(s) +OH → Al(OH)3 3 (aq)
Zn(OH)2 +3 H+ → Zn2+(aq) +2H2O Zn(OH)2(s) +2 OH → Zn(OH)34–(aq)
FeO(OH) + 3 H+ → Fe3+(aq) +3H2O Fe2O3(s) + 3 OH → 2 FeO2+(aq) +3 H2O
The product ions in the second column are known as aluminate, zincate, and ferrate. Other products, in which only some of the –OH groups of the parent hydroxides are deprotonated, are also formed, so there are actually whole series of these oxyanions for most metals.
Amphiprotic vs. amphoteric: what's the difference?
An amphoteric substance is one that can act as either an acid or a base. An amphiprotic substance can act as either a proton donor or a proton acceptor. So all amphiprotic compounds are also amphoteric. An example of an amphoteric compound that is not amphiprotic is ZnO, which can act as an acid even though it has no protons to donate:
$\ce{ZnO(s) + 4 OH^{-} (aq) -> Zn(OH)^{2-}4 (aq)}$
As a base, it "accepts" protons but does not retain them:
$\ce{ZnO(s) + 2H^{+} <=> Zn^{2+} + H_2O}$
The same remarks can be made about the other compounds shown in the table above. For most practical purposes, the distinction between amphiprotic and amphoteric is not worth worrying about.
Acid anhydrides
The binary oxygen compounds of the non-metallic elements tend to produce acidic solutions when they are added to water. Such compounds are sometimes referred to as acid anhydrides (“acids without water”.)
$CO_2 + H_2O \rightarrow H_2CO_3$
$SO_2 + H_2O \rightarrow [H_2SO_3]$
$SO_3 + H_2O \rightarrow H_2SO_4$
$P_4O_{10} + 6 H_2O \rightarrow 4 H_3PO_4$
In some cases, the reaction involves more than simply incorporating the elements of water. Thus nitrogen dioxide, used in the commercial preparation of nitric acid, is not an anhydride in the strict sense:
$3 NO_2 + H_2O \rightarrow 2 HNO+3 + NO$
Metal cations as acids
When sodium chloride is dissolved in pure water, the pH remains unchanged because neither ion reacts with water. However, a solution of magnesium chloride will be faintly acidic, and a solution of iron(III) chloride FeCl3 will be distinctly so. How can this be? Since none of these cations contains hydrogen, we can only conclude that the protons come from the water.
The water molecules in question are those that find themselves close to any cation in aqueous solution; the positive field of the metal ion interacts with the polar H2O molecule through ion-dipole attraction, and at the same time increases the acidity of these loosely-bound waters by making facilitating the departure H+ ion. In general, the smaller and more highly charged the cation, the more acidic will it be; the acidity of the alkali metals and of ions like Ag+(aq) is negligible, but for more highly-charged ions such as Mg2+, Pb2+ and Al3+, the effect is quite noticeable.
Most of the transition-metal cations form organized coordination complexes in which four or six H2O molecules are chemically bound to the metal ion where they are well within the influence of the coulombic field of the cation, and thus subject to losing a proton. Thus an aqueous solution of "Fe3+" is really a solution of the ion hexaaquo iron III, whose first stage of "dissociation" can be represented as
$Fe(H_2O)_6^{3+} + H_2O \rightarrow Fe(H_2O)_5(OH)^{2+} + H_3O^+$
As a consequence of this reaction, a solution of FeCl3 turns out to be a stronger acid than an equimolar solution of acetic acid. A solution of FeCl2, however, will be a much weaker acid; the +2 charge is considerably less effective in easing the loss of the proton.
It should be possible for a hydrated cation to lose more than one proton. For example, an Al(H2O)63+ ion should form, successively, the following species:
AlOH(H2O)52+ → Al(OH)2(H2O)4+→ Al(OH)3(H2O)30 → Al(OH)4(H2O)2→ Al(OH)5(H2O)2–→ Al(OH)63–
However, removal of protons becomes progressively more difficult as the charge decreases from a high positive value to a negative one; the last three species have not been detected in solution. In dilute solutions of aluminum chloride the principal species are actually Al(H2O)63+ (commonly represented simply as Al3+) and AlOH(H2O)52+ ("AlOH2+").
Salts
When salts dissolve in water, they yield solutions of anions and cations, so their effects on the pH of the solution will depend on the properties of the particular pair of ions. For a salt such as sodium chloride, the solution wil remain neutral because sodium ions have no acidic properties and chloride ions, being conjugate to the strong acid HCl have negligible proton-accepting tendencies. Ions of this kind are often referred to as "strong" ions (that is, derived from a strong acid and a strong base— HCl and NaOH in the case of NaCl.) The possible outcomes for the other three possibilities are shown below.
salt derived from example pH reaction
weak acid + strong base $\ce{NaF}$ >7 F + H2O → HF + OH
strong acid + weak base $\ce{NH4Cl}$ <7 NH4+ + H2O → NH3 + H3O+
weak acid + weak base $\ce{NH4F}$ ? depends on competition between above two reactions; need to do calculation
The reactions that cause salt solutions to have non-neutral pH values are sometimes still referred to by the older term hydrolysis (“water splitting”)— a reminder of times before the concept of proton transfer acid-base reactions had developed.
Organic Acids and Bases
The carboxyl group –CO(OH) is the characteristic functional group of the organic acids. The acidity of the carboxylic hydrogen atom is due almost entirely to electron-withdrawal by the non-hydroxylic oxygen atom; if it were not present, we would have an alcohol –COH whose acidity is smaller even than that of H2O. This partial electron withdrawal from one atom can affect not only a neighboring atom, but that atom’s neighbor as well. Thus the strength of a carboxylic acid will be affected by the bonding environment of the carbon atom to which it is connected. This propagation of partial electron withdrawal through several adjacent atoms is known as the inductive effect and is extremely important in organic chemistry.
A very good example of the inductive effect produced by chlorine (another highly electronegative atom) is seen by comparing the strengths of acetic acid and of the successively more highly substituted chloroacetic acids:
CH3–COOH
acetic acid
1.8 × 10–5
ClCH2–COOH
monochloroacetic acid
0.0014
Cl2CH–COOH
dichloroacetic acid
0.055
Cl3C–COOH
trichloroacetic acid
0.63
Phenols
The acidic character of the carboxyl group is really a consequence of the enhanced acidity of the –OH group as influenced by the second oxygen atom that makes up the –COOH group. The benzene ring has a similar although weaker electron-withdrawing effect, so hydroxyl groups that are attached to benzene rings also act as acids. The most well known example of such an acid is phenol, C6H5OH, also known as carbolic acid. Compared to carboxylic acids, phenolic acids are quite very weak, as indicated by the acid dissociation constants listed below:
CH3–COOH
acetic acid
1.8 × 10–5
C6H5–OH
phenol
1.1× 10–10
C6H5–COOH
benzoic acid
6.3 × 10–5
Amines and organic bases
We have already discussed organic acids, so perhaps a word about organic bases would be in order. The –OH group, when bonded to carbon, is acidic rather than basic, so alcohols are not the analogs of the inorganic hydroxy compounds. The amines, consisting of the –NH2 group bonded to a carbon atom, are the most common class of organic bases. Amines give weakly alkaline solutions in water:
$CH_3NH_2 + H_2O \rightarrow CH_3NH_3^+ + OH^–$
Amines are end products of the bacterial degradation of nitrogenous organic substances such as proteins. They tend to have rather unpleasant “rotten fish” odors. This is no coincidence, since seafood contains especially large amounts of nitrogen-containing compounds which begin to break down very quickly. Methylamine CH3NH2, being a gas at room temperature, is especially apt to make itself known to us. Addition of lemon juice or some other acidic substance to fish will convert the methylamine to the methylaminium ion CH3NH3+ . Because ions are not volatile they have no odor. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.06%3A_Types_of_Acids_and_Bases.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Describe some of the special properties of sulfuric acid that make it especially important both in the laboratory and in industry.
• Name the major acids and bases that are important to the fertilizer industry
• Name the natural sources of any three of the major organic acids.
• What are fatty acids? In what major way do the physical properties of saturated and unsaturated fatty acids differ?
• Describe the general structure of an amino acid, and state why they are important.
Acids and bases are of interest not only to the chemically inclined; they play a major role in our modern industrial society — so anyone who participates in it, or who is interested in its history and development, needs to know something about them. Five of the major acids and bases fall into the "Top 20" industrial chemicals manufactured in the world. The following table shows year-2000 figures for the U.S:
cal and rank
Sulfuric acid - 1
Lime (CaO)
3
Phosphoric acid - 4
Ammonia
5
Sodium hydroxide - 9
Nitric acid - 11
production in 109kg 40 20 16 15 11 8
major use chemicals cement fertilizers fertilizers chemicals chemicals
The mineral acids
This term refers to any inorganic acid, but its common use is usually limited to the major strong acids plus phosphoric acid. The major mineral acids— sulfuric, nitric, and hydrochloric— have been known since medieval times. Their discovery is usually credited to the Persian alchemist Abu Musa Jabir ibn Hayyan, known in the West by his Latinized name Geber. Jabir also invented aqua regia, the mixture of nitric and hydrochloric acids that has the unique ability to dissolve gold.
Sulfuric acid
More sulfuric acid is manufactured than any other industrial chemical, and it is the cheapest industrial acid worldwide. It has been continuously manufactured in the U.S. since 1793 and in Europe for much longer.
What you should know about it
• Pure anhydrous H2SO4 is a dense, viscous liquid which melts at 10.4°C and boils at about 300°C, decomposing back into its constituents, H2O and SO3.
• The acid undergoes autoprotolysis
2 H2SO4 → H3SO4+ + HSO4
H2SO4 + 2 NaCl → 2 Na+ + SO42– + HCl(g)
C12H22O11(s) → 12 C(s) + 11 H2O
H2SO4 + H2SO4 → HS2O7 + H3O+
• Its high boiling point makes the acid ideal for making other acids, such as nitric and hydrochloric, which are more volatile; removal of the gaseous product drives the reaction to the right as predicted by the Le Chatelier Principle:
• Sulfuric acid has a voracious appetite for water, and thus is an excellent dehydrating agent. This is seen most spectacularly if some concentrated acid is poured onto a small pile of table sugar; after a short time, a vigorous reaction ensues resulting in a pile of porous, steaming carbon
• Sulfuric acid can even dehydrate itself!
• , some of which have been detected on the surface of Jupiter's moon Europa.
• Owing to the autoprotolysis and self-dehydration reactions described above, "pure" sulfuric acid contains at least six minority species in addition to H2SO4.
How it is made
Sulfur trioxide, the anhydride of sulfuric acid, is the immediate precursor. Gaseous SO3 reacts vigorously with water, liberating much heat in the process:
$SO_{3(g)} + H_2O_{(l)} → H_2SO_{4(l)}$
Industrial manufacture of the acid starts with sulfur dioxide, prepared from burning elemental sulfur or obtained as a byproduct from roasting sulfide ores. The oxidation of SO2 to SO3 looks simple
$SO_{2(g)} + ½ O_{2(g)} → SO_{3(g)}$
but there are several complications:
• All chemical reactions take place more rapidly at higher temperatures, but because this reaction is highly exothermic, raising the temperature decreases the yield.
• Because of the decrease in volume (1.5 moles of gases to 1 mole), raising the pressure will increase the yield, so the reaction is carried out at a temperature below 600°C but at very high pressure.
• Specialized catalysts are used to speed up the reaction at these lower temperatures.
• Dissolving the SO3 directly in water would release large amounts of heat, creating a mist of fine acid droplets that would escape into the atmosphere. The SO3 is instead dissolved in sulfuric acid to form pyrosulfuric acid or oleum, sometimes known as fuming sufuric acid:
$H_2SO_{4(l)} + SO_{3(g)} → H_2S_2O_{7(l)}$
$H_2S_2O_{7(l)} + H_2O_{(l)} → 2 H_2SO_{4(l)}$
• The oleum is then treated with water to form industrial grade (96-98%) sulfuric acid:
What it is used for
Sulfuric acid has a broad spectrum of industrial uses, and the annual tonnage follows the economic cycle quite closely.
• Sixty percent of worldwide product goes into the manufacture of phosphoric acid H3PO4 which is used to make phosphate fertilizers and phosphate-based household detergents.
• An important nitrogen fertilizer, ammonium sulfate (NH4)2SO4, is made by reacting sulfuric acid with ammonia, the latter often obtained from the thermal decomposition of coal.
• Sulfuric acid is the major component of pickling acid that is used to remove surface oxide scale from steel before it is fabricated into steel product for the automobile and other industries.
• Aluminum sulfate (made from bauxite Al2O3 with H2SO4), is widely used in the papermaking industry to coagulate the cellulose fibers, producing a smooth, hard paper surface. Another major use is to make aluminum hydroxide which is used to filter out particulate matter in water treatment facilities.
• The familiar lead-acid storage battery employs sulfuric acid as an electrolyte. As the battery discharges, the concentraton of sulfuric acid in the electrolyte decreases as sulfate ions are taken up as PbSO4. Owing to the high density of the acid, the state of charge of the battery can be measured by means of a hydrometer.
Sulfuric acid in the environment
Acid Rain - Combustion of fossil fuels which contain organic sulfur compounds releases SO2 into the atmosphere. Photochemical oxidation of this compound to SO3, which rapidly takes up moisture, leads to the formation of H2SO4, a major component of acid rain.
Acid mine drainage results when sediments of the very common iron pyrite, FeS2, are exposed to air and are oxidized:
FeS2(s) + 7/2 O2 + H2O → Fe2+ 2 SO42– + 2 H+
further oxidation of the iron to Fe3+ results in additional reactions. The resulting drainage liquid is often orange-brown in color and can a have a pH of below zero.
Nitric acid
Anhydrous HNO3 is a colorless liquid boiling at 82.6°C, but "pure" HNO3 only exists as the solid which melts at –41.6°C. In its liquid and gaseous states, the acid is always partially decomposed into nitrogen dioxide:
2 HNO3 → 2 NO2 + ½ O2 + H2O
This reaction, which is catalyzed by light, accounts for the brownish color of HNO3 solutions.
What you should know about it
• HNO2 undergoes autoprotolysis to a greater extent than any other liquid. Further reactions of the conjugate acid H2NO3+ with HNO3 lead to a complicated mixture of species in the liquid.
• Dilute nitric acid can be concentrated by distillation up to a maximum of 68%, at which point it forms a constant-boiling (azeotropic) mixture with water. Higher concentrations require dehydration with sulfuric acid; the result is fuming nitric acid.
• "Concentrated nitric acid" is sold as a 70% solution in water, corresponding to a concentration of about 16M.
• Nitric acid is a very strong oxidizing agent, which adds to its corrosive behavior with organic materials including, of course, skin, which it turns yellow owing to a reaction with the protein keratin. Reactions with many organic compounds are highly exothermic and often violent. The well-known reaction of nitric acid with metallic copper produces copious amounts of brown nitrogen dioxide gas.
How it is made
The simplest method, which was used industrially before 1900, was by treatment of sodium nitrate ("Chile saltpeter", NaNO3) with sulfuric acid. The direct synthesis of the acid from atmospheric nitrogen and oxygen is thermodynamically favorable
½ N2 + 5/4 O2 + ½ H2O → HNO3
but is kinetically hindered by an extremely high activation energy, a fact for which we can be most thankful (see sidebar.) The first industrial nitrogen fixation process, developed in 1903, used this reaction to produce nitric acid, but it required the use of an electric arc to supply the activation energy and was therefore too energy-intensive to be economical.
Note
If it were not for the high activation energy required to sustain this reaction, all of the oxygen in the atmosphere would be consumed and the oceans would be a dilute solution of nitric acid.
The modern Ostwald process involves the catalytic oxidation of ammonia to nitric oxide NO, which is oxidized in a further step to NO2; reaction of the latter with water yields HNO3. This route, first developed in 1901, did not become practical until the large-scale production of ammonia by the Haber-Bosch process in 1910.
What it is used for
The major industrial uses of nitric acid are for the production of ammonium nitrate fertilizer, and in the manufacture of explosives. On a much small scale, the acid is used in metal pickling, etching semiconductors and electronic circuit boards, and in the manufacture of acrylic fibers.
In the laboratory, the acid finds use in a wide variety of roles.
In the environment
High-temperature combustion processes (in internal combustion engines, power plants, and incinerators) can oxidize atmospheric nitrogen to nitric oxide (NO) and other oxides ("NOx"); the NO is then photooxidized to NO2, which reacts with water to form HNO3 which is a major component of acid rain. NO2is the major precursor of photochemical smog.
Hydrochloric acid
Unlike the other major acids, there is no such substance as "pure" hydrochloric acid; what we call "hydrochloric acid" is just an aqueous solution of hydrogen chloride gas (bp –84°C). But in a sense it is more "pure" than the acids discussed above, since there is no autoprotolysis; hydronium and chloride ions are the only significant species in the solution. Hydrochloric acid is usually sold as a 32-38% (12M) solution of HCl in water; concentrations greater than this are known as fuming hydrochloric acid.
Note
Hydrochloric acid is still sometimes sold under its older name muriatic acid for cleaning bricks and other household purposes. The name comes from the same root as marine, reflecting its preparation from salt.
The acid has been known to chemists (and alchemists), and used for industrial purposes since the middle ages. Its composition HCl was demonstrated by Humphrey Davy in 1816.
What you should know about it
• Hydrochloric acid is the least hazardous of the strong mineral acids to work with because unlike the other ones, it is not an oxidizing agent. It is usually the acid of choice for titrations and other operations in which the main requirement is simply a strong source of hydronium ions.
• The concentrated acid boils at 48°C. As boiling continues, it loses HCl and the boiling point rises to 109°C, at which point a constant-boiling (azeotropic) solution remains, consisting of 20.2% HCl.
What it is used for
The uses of hydrochloric acid are far too many to enumerate individually, but the following stand out:
• A major industrial use is to remove surface scale from iron or steel ("pickling") before it is processed into sheets or other forms, or galvanized or coated.
• Production of chlorinated organic chemicals, particularly vinyl chloride, polyurethanes, and other construction polymers, consumes huge amounts of HCl.
• The acid is widely used for pH control of water, including neutralization of wastwater streams, and for regenerating ion-exchange water softeners.
How it is made
The ancient method of treating salt with sulfuric acid to release HCl has long since been supplanted by more efficient processes, including direct synthesis by "burning" hydrogen gas in chlorine:
$H_{2(g)} + Cl_{2(g)} \rightarrow 2 HCl_{(g)}$
Most hydrochloric acid production now comes from reclaiming byproduct hydrogen chloride gas from other processes, especially those associated with the production of industrial organic compounds.
2 The Alkali Metals
The term alkali usually means a basic salt of a Group 1 or 2 ("alkali" or "alkaline earth") metal. All alkalies are of course bases, but the latter term is much more general, whether defined according to the Arrhenius, Brønsted-Lowry, or Lewis concepts. The word alkali comes from the Arabic al-qali, which refers to the ashes from which sodium and potassium hydroxides (potash, "ashes remaining in the pot", and the origin of the element name potassium) were extracted as a step in the making of soap.
Sodium hydroxide
Pure sodium hydroxide is a white solid consisting of Na+ and OH ions in a crystal lattice. Although it is widely thought of as an ionic solid, van der Waals forces make a substantial contribution to its stability.
What you should know about it
• Owing to its deliquescence (ability to take up moisture) and its tendency to react with carbon dioxide, the solid must be stored in a closed container.
• In industry, sodium hydroxide is commonly known as caustic soda or simply as caustic; NaOH sold for household purposes is usually known as lye.
• Sodium hydroxide slowly attacks glass to form sodium silicate. Glass vessels used to store concentrated solutions gradually develop a cloudy coating on the inside.
• Some metals, notably aluminum, zinc, and titanium, react with strongly alkaline solutions, but iron and copper are immune to this kind of attack.
• Highly alkaline solutions also soften and dissolve skin, accounting for the slippery feeling associated with strong bases. Sodium hydroxide was once used to dispose of animal carcasses, digesting them into an easily disposable liquid form.
How it is made
Sodium hydroxide is now manufactured by the electrolysis of brine solutions, and along with chlorine, is one of the two major products of the chloralkali industry.
Electrolysis of aqueous NaCl produces Cl2 at the anode, but because H2O can be reduced more readily than Na+, the water is decomposed to H2 and OH at the cathode, leaving a solution of NaOH. An older mercury cell process reduces the Na+ to Na within a mercury amalgam (alloy), and the metallic sodium is then combined with water to produce NaOH and hydrogen. The net reaction for the reduction step is the same for both methods:
$2 Na^+ + 2 H_2O + 2e^– \rightarrow H_{2(g)] + 2 NaOH$
The resulting solution is usually evaporated to such a high concentration that it solidifies at ordinary temperatures. It is commonly shipped in rail cars or barges which can be heated with steam to liquefy the mixture for removal. (It is obviously uneconomical to ship large quantities of water across the country!)
What it is used for
• Sodium hydroxide is one of the most diverse industrial chemicals in terms of its applications. Most householders know it as the active ingredient of drain cleaning agents.
• Huge quantities are consumed by the pulp and paper industry, which is probably its single largest specific industrial application. It is used to remove the lignin component of wood pulp from the cellulose so that the latter can be processed into paper.
• About half of the NaOH output goes into the production of a wide variety of other industrial chemicals, and in degreasing steel drums and other industrial surfaces.
• "Lye" plays a role in the processing of many types of foods, including chocolate, olives, pretzels, and the "hominy" and "grits" corn products used in the Southern U.S. It also acts as a chemical peeling agent for fruits and vegetables.
The economic push and pull of caustic and chlorine
In contrast to the extremely diverse applications of sodium hydroxide which makes the demand for this commodity relatively immune to the ups and downs of the economic cycle, the consumption of chlorine is directly dependent on the economy as reflected in the demand for polyvinyl chloride products that are now widely used in the the construction and home furnishings industries. Because chlorine, being a gas, is expensive to store, the output of the chloralkali industry is governed largely by the demand for this commodity. When times are good this presents no problem; caustic is then largely a by-product and can easily be stockpiled if supply exceeds demand. But during an economic downturn, the demand for chlorine declines, limiting its production along with that of caustic. But because the demand for caustic tends to decline much less, it becomes scarce and its price rises, thus tending to drive the industrial economy into even deeper trouble.
Sodium carbonate
This compound is known industrially as soda ash, and domestically as washing soda. The common form is the heptahydrate, Na2CO3·7 H2O. The white crystals of this substance spontaneously lose water (effloresce) when exposed to the air, forming the monohydrate.
What it is used for
• Although carbonates are much weaker bases than hydroxides, a solution of sodium carbonate can still have a pH of 11 or so, sufficiently high to allow it to substitute for sodium hydroxide in many applications— especially when the price of caustic is high.
• The single most important use of soda ash is in the manufacture of glass, where it serves to lower the melting point of the principal component, SiO2.
• Another emerging major use is to neutralize the SO2 emissions of fossil fuel-burning power plants.
• The older use of sodium carbonate as a cleaning agent (hence the name washing soda) was based partly on the ability of its alkaline solutions to emulsify grease, but mainly as a means of precipitating the insoluble carbonates of calcium and magnesium before these ions (commonly present in hard water) could form undesirable preciptates with soaps. The use of modern detergents has largely eliminated this once important market.
How it is made
Most of the world's sodium carbonate is made by the ammonia-soda "Solvay" process developed in 1861 by the Belgian chemist Ernest Solvay (1838-1922) whose patents made him into a major industrialist and a rich philanthropist. This process involves a set of simple reactions that essentially converts limestone (CaCO3), ammonia NH3 and brine (NaCl) into sodium bicarbonate NaHCO3 and eventually Na2CO3, recycling several of the intermediate products in an ingenious way.
A minor source of soda ash (but quite significant in some countries, such as the U.S.) is the mining of natural evaporites (the remains of ancient lakes), such as the trona found in Southern California.
Ammonia
Ammonia NH3 is of course not a true alkali, but it is conveniently included in this section for discussion purposes. Most people are familiar with the pungent odor of this gas, which can be detected at concentrations as low as 20-50 ppm.
Tradition dies slowly: a non-existent chemical available in bottles!
What you should know about it
• More moles of ammonia are manufactured than of any other industrial chemical.
• Ammonia is extremely soluble in water. An aqueous solution of ammonia is still sometimes referred to in commerce as "ammonium hydroxide", but this term is no longer favored by chemists because no such compound as NH4OH has ever been shown to exist. At neutral pH, about 99% of the ammonia in water exists as NH4+ ions.
• Ammonia is an end product of nitrogen metabolism in most organisms. One source that may be familiar to parents of infants is the bacterial decomposition of the contents of diapers.
• Liquid ammonia (bp –33°C) is often used as an ionizing laboratory solvent.
What it is used for
• The major use of ammonia (about 80%) is as a fertilizer, most commonly as anhydrous ammonia (the gas is injected directly into the soil) or after conversion to (NH4)2SO4, NH4NO3, or urea O–C(NH3)2.
• Ammonia is used in the production of numerous polymers, including nylons and polyurethanes.
• Explosives manufacture accounts for about 5% of ammonia production.
• Beyond these, there are hundreds of minor uses, including as a household cleaning agent (aqua ammonia). a refrigerant, and as a laboratory reagent.
How it is made
Ammonia is made by direct synthesis from the elements:
$N_{2(g)} + 3 H_{2(g)} \rightarrow 2 NH_{3(g)}$
... a simple-looking reaction, but one that required some very creative work to implement; the Haber-Bosch process is considered to be the most important chemical synthesis of the 20th Century.
3 Some important organic acids
Most acids are organic— there are millions of them. The acidic function is usually a hydroxyl group connected to a carbon that is bonded to an electron-withdrawing oxygen atom; the combination is the well-known carboxyl group, –COOH. Here are a few that are part of everyone's life.
Acetic acid
This is next to formic acid in being the simplest of the organic acids, and in the form of vinegar (a 5-8% solution in water) its characteristic odor is known to everyone. The pure acid is a colorless liquid above 16.7°C; below this temperature it forms a crystalline solid, hence the term "glacial acetic acid" that is commonly applied to the pure substance. The name of the acid comes from acetum, the Latin word for vinegar.
What you should know about it
• The pure acid, although quite weak in the proton-donor sense, is quite corrosive and its vapors are very irritating.
• A 1.0M solution of the acid has a pH of about 2.4, corresponding to only four out of every thousand CH3COOH molecules being dissociated.
What it is used for
Slightly less than half of the world production of acetic acid goes into the production of polymers. The end product visible to most people would be the flexible plastic bottles in which drinking water is sold. Other uses are related mostly to the production of other chemicals, mainly acetic anhydride, but also including aspirin.
How it is made
Bacterial fermentation of sugars has been the source of vinegar since ancient times, and is still accounts for most food-grade acetic acid and vinegar, but it now amounts to only about 10% of total acetic acid production:
C6H12O6 → 3 CH3COOH
There are several important synthetic routes to acetic acid production, but the major one is by treating methanol with carbon monoxide:
CH3OH + CO → CH3COOH
A brief look at some other interesting organic acids
Formic acid
(Methanoic acid) HCOOH
mp 8.4°C.
This, the simplest of the carboxylic acids, is the chemical weapon that Nature has given ants and bees (the Latin word for ant is formicus.) Known since the 15th Century, it was first distilled from ants, but is now made synthetically. Its main uses are as a preservative and antibacterial agent in livestock feed. Chemists find it a useful source of carbon monoxide (just add sulfuric acid.)
Oxalic acid
HOOC–COOH
Just two carboxyl groups joined together, this acid has a special ability to grab up divalent metal ions and bind them into five-membered MO2C2rings. Because many such metal ions (Ca2+, for example) are essential nutrients, this can be dangerous to your health. The acid occurs in many plants, notably rhubarb (it is what makes the leaves poisonous), parsley and spinach. If you have ever noticed a funny feeling in your mouth after drinking milk with a rhubarb desert, it is due to precipitation of calcium oxalate. This same solid is often a major component of kidney stones, and it contributes to the miseries of gout.
Lactic acid
Structurally, lactic acid is both an alcohol and a carboxylic acid— not all that uncommon. We know it as the acid found in sour milk, yogurt, and in tired muscles. When the blood cannot deliver enough oxygen to oxidize glucose all the way to CO2 and water, your muscles go into a much-less-energy-efficient "anaerobic" mode that generates lactate. In milk products, lactic acid is produced in about the same way by acidophilus bacteria that eat the lactose ("milk sugar") and who don't know how to utilize oxygen.
Citric acid
Known by Arabic alchemists in the 8th Century; first isolated by Scheele in 1784
Although it is one part alcohol and three parts acid, citric acid is quite weak, but nevertheless strong enough to make it unpleasant to suck on a lemon, of which it can comprise as much as 8% of the dry weight of this fruit. Biochemists know it as part of the citric acid cycle, a sequence of reactions involved in extracting energy from the oxidative metabolism of foods. Its major industrial use is as a food flavoring and preservative agent; large quantities are used to make soft drink beverages. The acid also finds use in cleaning agents.
Ascorbic acid
No carboxyl groups here, but the –OH group one carbon away from the double bond is still fairly acidic. One of its geometric isomers, L-ascorbic acid, is more widely known as Vitamin C; discovery of the essential role of this substance in preventing the disease scurvy (from which its name is derived) yielded two Nobel prizes in 1937. Most animals are capable of synthesizing their own Vitamin C, but primates (along with guinea pigs) seem to have lost the required gene somewhere along the way, so we must depend on fruits and veggies for our supply. (Because ascorbic acid is soluble in water, it tends to be leached out of vegetables when they are boiled, so it is much healthier to steam them.)
The major industrial use of ascorbic acid is as an antioxidant, so it is often added to foods and other materials as a preservative.
Salicylic acid
2-hydroxybenzoic acid is found in willow trees
Does chemistry give you a headache? If so, pull a small shoot off of a willow tree and chew on it for a while. This has been a folk-remedy for pains and fever since ancient times. Willow bark (salix in Latin) contains, and lent its name to the active principle, salicylic acid.
The acid itself turned out to be a bit too much of an irritant to the stomach's lining, so the German firm Bayer began marketing a tamed version, acetylsalicylic acid (ASA) under the name Aspirin in 1899, and it has been going strong ever since. Interestingly, the detailed chemistry of its pain-and-fever reliving action was not discovered until 1970.
Fatty acids
This generic term refers to carboxylic acids built from a chain of 4 to 22 carbon atoms. Fatty acids, as the name implies, are derived from fats, in which they are bound to glycerol in the form of triglycerides. Fats, which occur in all animal tissues, are the most efficient means of storing metabolic energy. Vegetable oils are another source.
There are two general categories of fatty acids:
• Saturated fatty acids consist of straight chains of carbon atoms with a methyl group on one end and a carboxyl (acidic) group on the other, with methylene groups -CH2- in between. Thus stearic (octadecanoic) acid is CH3(CH2)16COOH. The linear shape of these molecules allows them to pack together very efficiently, contributing to their ability to store energy in a small space; the higher saturated acids tend to be waxy solids.
• Unsaturated fatty acids contain one or more carbon-carbon double bonds, which introduces a complication: because free rotation around the axis of a double bond is not possible, the neighboring hydrogens can be either on the same side of the bond (cis) or on opposite sites (trans)— thus the cis- and trans-fatty acids. These double bonds introduce kinks into the carbon chain, especially in the case of the cis-acids. The bent and curved carbon chains that result cannot pack together compactly enough to interact very strongly with each other, so unsaturated fatty acids tend to be liquids.
The human body is able to synthesize most of the fatty acids it needs, but two classes of unsaturated acids, known as the essential fatty acids, can be obtained only from foods. The compounds are known as ω-3 and ω-6 fatty acids; the ω (omega)-n notation means that the double bond is located n carbons away from the terminal CH3group of the molecule.
Note
Most of the unsaturated fatty acids found in nature have cis configurations around their double bonds. The trans fatty acids in our foods that we hear much about are made from vegetable oils that have had some of their double bonds changed to single bonds by a chemical process called hydrogenation. The object of this is to make the original liquid fatty acids into solid forms that are more suitable for use in foods (as in margarine) and particularly as shortenings in baking. Unfortunately, the hydrogenation process also changes the remaining cis double bonds into their trans forms which are believed to be implicated in cardiac disease.
Amino acids
Amino acids are the building blocks of proteins, which is what you are largely made of. There are twenty of them, of which about half can be manufactured by the body; we must depend on our food to supply the remainder, which are known as the essential amino acids. Each one comes in two mirror-image forms; only the "left" forms are found in most proteins.
All amino acids have the basic structure shown above; they differ in the nature of the group of atoms designated by "R" in the diagram.
Although we call them "acids", the amino acids are really chemical hermaphrodites; you will recall that amines are weak bases. The balance between their acidic and basic properties can be shifted simply by changing the pH.
The carboxylic acid part of one amino acid can react with the amine part of another to form a peptide bond (an amide linkage) shown here.
The polymeric chains that result are known as peptides if they are fairly short (2 to about 20-50 amino acid units); longer chains, or aggregate made up of multiple peptide units, are proteins. Owing to their very large size (500 amino acid residues is quite common, but some have as many as 1500), proteins are able to fold in various ways, so the amino acid sequence alone is not sufficient to determine their properties. | textbooks/chem/General_Chemistry/Chem1_(Lower)/10%3A_Fundamentals_of_Acids_and_Bases/10.07%3A_Acid-Base_Gallery.txt |
The laws of chemical equilibrium define the direction in which a chemical reaction will proceed, as well as the quantities of reactants and products that will remain after the reaction comes to an end. An understanding of chemical equilibrium and how it can be manipulated is essential for anyone involved in Chemistry and its applications.
• 11.1: Introduction to Chemical Equilibrium
Chemical change is one of the two central concepts of chemical science, the other being structure. The very origins of Chemistry itself are rooted in the observations of transformations such as the combustion of wood, the freezing of water, and the winning of metals from their ores that have always been a part of human experience. It was the quest for some kind of constancy underlying change that led the Greek thinkers of around 200 BCE to the idea of elements and later to that of the atom.
• 11.2: Le Chatelier's Principle
The previous Module emphasized the dynamic character of equilibrium as expressed by the Law of Mass Action. This law serves as a model explaining how the composition of the equilibrium state is affected by the "active masses" (concentrations) of reactants and products. In this lesson, we develop the consequences of this law to answer the very practical question of how an existing equilibrium composition is affected by the addition or withdrawal of one of the components.
• 11.3: Reaction Quotient
Consider a simple reaction such as the gas-phase synthesis of hydrogen iodide from its elements: $H_2 + I_2 \rightarrow 2 HI$ Suppose you combine arbitrary quantities of $H_2$, $I_2$ and $HI$. Will the reaction create more HI, or will some of the HI be consumed as the system moves toward its equilibrium state? The concept of the reaction quotient, which is the focus of this short lesson, makes it easy to predict what will happen.
• 11.4: Equilibrium Expressions
You know that an equilibrium constant expression looks something like K = [products] / [reactants]. But how do you translate this into a format that relates to the actual chemical system you are interested in? This lesson will show you how to write the equilibrium constant expressions that you will need to use when dealing with the equilibrium calculation problems in the chapter that follows this one.
• 11.5: Equilibrium Calculations
This page presents examples that cover most of the kinds of equilibrium problems you are likely to encounter in a first-year university course. Reading this page will not teach you how to work equilibrium problems! The only one who can teach you how to interpret, understand, and solve problems is yourself.
• 11.6: Phase Distribution Equilibria
If two immiscible liquid phases are in contact and one contains a solute, how will the solute tend to distribute itself between the two phases? One’s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases. This, however, does not take into the account the differing solubilities the solute might have in the two solvents; the solute will preferentially migrate into the phase in which it is more soluble.
11: Chemical Equilibrium
Learning Objectives
• Define "the equilibrium state of a chemical reaction system". What is its practical significance?
• State the meaning and significance of the following terms:
• reversible reaction
• quantitative reaction
• kinetically inhibited reaction
• Explain the meaning of the statement "equilibrium is macroscopically static, but microscopically dynamic". Very important!
• Explain how the relatve magnitudes of the forward and reverse reaction rate constants in the Mass Action expression affect the equilibrium composition of a reaction system.
• Describe several things you might look for during an experiment that would help determine if a reaction system is in its equilibrium state.
Chemical change is one of the two central concepts of chemical science, the other being structure. The very origins of Chemistry itself are rooted in the observations of transformations such as the combustion of wood, the freezing of water, and the winning of metals from their ores that have always been a part of human experience. It was, after all, the quest for some kind of constancy underlying change that led the Greek thinkers of around 200 BCE to the idea of elements and later to that of the atom. It would take almost 2000 years for the scientific study of matter to pick up these concepts and incorporate them into what would emerge, in the latter part of the 19th century, as a modern view of chemical change.
Chemical change: how far, how fast?
Chemical change occurs when the atoms that make up one or more substances rearrange themselves in such a way that new substances are formed. These substances are the components of the chemical reaction system; those components which decrease in quantity are called reactants, while those that increase are products. A given chemical reaction system is defined by a balanced net chemical equation which is conventionally written as
$\text{reactants} \rightarrow \text{products}$
The first thing we need to know about a chemical reaction represented by a balanced equation is whether it can actually take place. If the reactants and products are all substances capable of an independent existence, then in principle, the answer is always "yes". This answer must be qualified, however, by the following considerations:
How complete is the reaction?
That is, what fraction of the reactants are converted into products? Some reactions convert essentially 100% of reactants to products, while for others the quantity of products may be undetectable. Many are somewhere in between, meaning that significant quantities of all components remain at the end. Later on, in another part of the course, you will learn that the tendency of a reaction to occur can be predicted entirely from the properties of the reactants and products through the laws of thermodynamics.
How fast does the reaction occur?
Some reactions are over in microseconds; others take years. The speed of any one reaction can vary over a huge range depending on the temperature, the state of matter (gas, liquid, solid) and the presence of a catalyst. Unlike the question of completeness, there is no simple way of predicting reaction speed.
What is the mechanism of the reaction?
What happens, at the atomic or molecular level, when reactants are transformed into products? What intermediate species (those that are produced but later consumed so that they do not appear in the net reaction equation) are involved? This is the microscopic, or kinetic view of chemical change, and cannot be predicted by theory as it is presently developed and must be inferred from the results of experiments.
A reaction that is thermodynamically possible but for which no reasonably rapid mechanism is available is said to be kinetically limited. Conversely, one that occurs rapidly but only to a small extent is thermodynamically limited. As you will see later, there are often ways of getting around both kinds of limitations, and their discovery and practical applications constitute an important area of industrial chemistry.
What is equilibrium?
Basically, the term refers to what we might call a "balance of forces". In the case of mechanical equilibrium, this is its literal definition. A book sitting on a table top remains at rest because the downward force exerted by the earth's gravity acting on the book's mass (this is what is meant by the "weight" of the book) is exactly balanced by the repulsive force between atoms that prevents two objects from simultaneously occupying the same space, acting in this case between the table surface and the book. If you pick up the book and raise it above the table top, the additional upward force exerted by your arm destroys the state of equilibrium as the book moves upward. If you wish to hold the book at rest above the table, you adjust the upward force to exactly balance the weight of the book, thus restoring equilibrium.
An object is in a state of mechanical equilibrium when it is either static (motionless) or in a state of unchanging motion. From the relation f = ma , it is apparent that if the net force on the object is zero, its acceleration must also be zero, so if we can see that an object is not undergoing a change in its motion, we know that it is in mechanical equilibrium.
Thermal equilibrium
Another kind of equilibrium we all experience is thermal equilibrium. When two objects are brought into contact, heat will flow from the warmer object to the cooler one until their temperatures become identical. Thermal equilibrium arises from the tendency of thermal energy to become as dispersed or "diluted" as possible.
A metallic object at room temperature will feel cool to your hand when you first pick it up because the thermal sensors in your skin detect a flow of heat from your hand into the metal, but as the metal approaches the temperature of your hand, this sensation diminishes. The time it takes to achieve thermal equilibrium depends on how readily heat is conducted within and between the objects; thus a wooden object will feel warmer than a metallic object even if both are at room temperature because wood is a relatively poor thermal conductor and will therefore remove heat from your hand more slowly.
Thermal equilibrium is something we often want to avoid, or at least postpone; this is why we insulate buildings, perspire in the summer and wear heavier clothing in the winter.
Chemical equilibrium
When a chemical reaction takes place in a container which prevents the entry or escape of any of the substances involved in the reaction, the quantities of these components change as some are consumed and others are formed. Eventually this change will come to an end, after which the composition will remain unchanged as long as the system remains undisturbed. The system is then said to be in its equilibrium state, or more simply, "at equilibrium".
Why reactions go toward equilibrium
What is the nature of the "balance of forces" that drives a reaction toward chemical equilibrium? It is essentially the balance struck between the tendency of energy to reside within the chemical bonds of stable molecules, and its tendency to become dispersed and diluted. Exothermic reactions are particularly effective in this, because the heat released gets dispersed in the infinitely wider world of the surroundings.
In the reaction represented here, this balance point occurs when about 60% of the reactants have been converted to products. Once this equilibrium state has been reached, no further net change will occur. The only spontaneous changes that are allowed follow the arrows pointing toward maximum dispersal of energy.
Equilibrium is death!
Chemical equilibrium is something you definitely want to avoid for yourself as long as possible. The myriad chemical reactions in living organisms are constantly moving toward equilibrium, but are prevented from getting there by input of reactants and removal of products. So rather than being in equilibrium, we try to maintain a "steady-state" condition which physiologists call homeostasis — maintenance of a constant internal environment. Equilibrium is death!
For the time being, it's very important that you know this definition:
A chemical reaction is in equilibrium when there is no tendency for the quantities of reactants and products to change.
The direction in which a chemical reaction is written (and thus which components are considered reactants and which are products) is arbitrary. Consider the following two reactions:
$\underset{\text{synthesis of hydrogen iodide}}{H_2 + I_2 \rightarrow 2 HI} \label{10.1}$
$\underset{\text{dissociation of hydrogen iodide}}{2 HI \rightarrow H_2 + I_2} \label{10.2}$
Equations $\ref{10.1}$ and $\ref{10.2}$ represent the same chemical reaction system in which the roles of the components are reversed, and both yield the same mixture of components when the change is completed. This is central to the concept of chemical equilibrium. It makes no difference whether we start with two moles of HI or one mole each of H2 and I2; once the reaction has run to completion, the quantities of these two components will be the same. In general, then, we can say that the composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition. Once this equilibrium composition has been attained, no further change in the quantities of the components will occur as long as the system remains undisturbed.
The composition of a chemical reaction system will tend to change in a direction that brings it closer to its equilibrium composition.
The two diagrams below show how the concentrations of the three components of this chemical reaction change with time. Examine the two sets of plots carefully, noting which substances have zero initial concentrations, and are thus "products" of the reaction equations shown. Satisfy yourself that these two sets represent the same chemical reaction system, but with the reactions occurring in opposite directions. Most importantly, note how the final (equilibrium) concentrations of the components are the same in the two cases.
Whether we start with an equimolar mixture of H2 and I2 (left) or a pure sample of hydrogen iodide (shown on the right, using twice the initial concentration of HI to keep the number of atoms the same), the composition after equilibrium is attained (shaded regions on the right) will be the same.
The equilibrium composition is independent of the direction from which it is approached (i.e., the initial conditions).
A chemical equation of the form A → B represents the transformation of A into B, but it does not imply that all of the reactants will be converted into products, or that the reverse reaction B → A cannot also occur. In general, both processes (forward and reverse) can be expected to occur, resulting in an equilibrium mixture containing finite amounts of all of the components of the reaction system. (We use the word components when we do not wish to distinguish between reactants and products.)
If the equilibrium state is one in which significant quantities of both reactants and products are present (as in the hydrogen iodide example given above), then the reaction is said to incomplete or reversible. The latter term is preferable because it avoids confusion with "complete" in its other sense of being completed or finished, implying that the reaction has run its course and is now at equilibrium.
Note that there is no fundamental difference between the meanings of A → B and A B. Some older textbooks just use A = B.
• If it is desired to emphasize the reversibility of a reaction, the single arrow in the equation is replaced with a pair of hooked lines pointing in opposite directions, as in A B.
• A reaction is said to be complete or quantitative when the equilibrium composition contains no significant amount of the reactants. However, a reaction that is complete when written in one direction is said "not to occur" when written in the reverse direction.
In principle, all chemical reactions are reversible, but this reversibility may not be observable if the fraction of products in the equilibrium mixture is very small, or if the reverse reaction is very slow (the chemist's term is "kinetically inhibited")
How did Napoleon Bonaparte help discover reversible reactions?
We can thank Napoleon for bringing the concept of reaction reversibility to Chemistry. Napoleon recruited the eminent French chemist Claude Louis Berthollet (1748-1822) to accompany him as scientific advisor on the most far-flung of his campaigns, the expedition in Egypt in 1798. Once in Egypt, Berthollet noticed deposits of sodium carbonate around the edges of some the salt lakes found there. He was already familiar with the reaction
$Na_2CO_3 + CaCl_2 \rightarrow CaCO_3 + 2 NaCl \label{10.3}$
which was known to proceed to completion in the laboratory. He immediately realized that the Na2CO3 must have been formed by the reverse of this process brought about by the very high concentration of salt in the slowly-evaporating waters. This led Berthollet to question the belief of the time that a reaction could only proceed in a single direction. His famous textbook Essai de statique chimique(1803) presented his speculations on chemical affinity and his discovery that an excess of the product of a reaction could drive it in the reverse direction.
Unfortunately, Berthollet got a bit carried away by the idea that a reaction could be influenced by the amounts of substances present, and maintained that the same should be true for the compositions of individual compounds. This brought him into conflict with the recently accepted Law of Definite Proportions (that a compound is made up of fixed numbers of its constituent atoms), so his ideas (the good along with the bad) were promptly discredited and remained largely forgotten for 50 years. (Ironically, it is now known that certain classes of compounds do in fact exhibit variable composition of the kind that Berthollet envisioned.)
What is the law of mass action?
Berthollet's ideas about reversible reactions were finally vindicated by experiments carried out by others, most notably the Norwegian chemists (and brothers-in-law) Cato Guldberg and Peter Waage. During the period 1864-1879 they showed that an equilibrium can be approached from either direction (see the hydrogen iodide illustration above), implying that any reaction
$aA + bB \rightarrow cC + dD$
is really a competition between a "forward" and a "reverse" reaction. When a reaction is at equilibrium, the rates of these two reactions are identical, so no net (macroscopic) change is observed, although individual components are actively being transformed at the microscopic level.
Chemical Equilibrium is dynamic!
The "active masses" are essentially the concentrations of the reactants and products that combine directly in the manner represented by the reaction equation. The meaning of the terms on the right sides of the equations is simply that the rate is proportional to the concentrations of the components.
Guldberg and Waage showed that for a reaction
$aA + bB \rightarrow cC + dD$
the rate (speed) of the reaction in either direction is proportional to what they called the "active masses" of the various components:
• rate of forward reaction = kf [A]a [B]b
• rate of reverse reaction = kr [C]c [D]d
in which the proportionality constants k are called rate constants and the quantities in square brackets represent concentrations. If we combine the two reactants A and B, the forward reaction starts immediately; then, as the products C and D begin to build up, the reverse process gets underway. As the reaction proceeds, the rate of the forward reaction diminishes while that of the reverse reaction increases. Eventually the two processes are proceeding at the same rate, and the reaction is at equilibrium:
rate of forward reaction = rate of reverse reaction
$k_f [A]^a [B]^b = k_r [C]^c [D]^d$
It is very important that you understand the significance of this relation. The equilibrium state is one in which there is no net change in the quantities of reactants and products. But do not confuse this with a state of "no change"; at equilibrium, the forward and reverse reactions continue, but at identical rates, essentially cancelling each other out.
Equilibrium is macroscopically static, but is microscopically dynamic! To further illustrate the dynamic character of chemical equilibrium, suppose that we now change the composition of the system previously at equilibrium by adding some C or withdrawing some A (thus changing their "active masses"). The reverse rate will temporarily exceed the forward rate and a change in composition ("a shift in the equilibrium") will occur until a new equilibrium composition is achieved.
Composition of the Equilibrium State Depends on the ratio of the forward- and reverse rate constants.
Be sure you understand the difference between the rate of a reaction and a rate constant. The latter, usually designated by k, relates the reaction rate to the concentration of one or more of the reaction components — for example, rate = k [A]. At equilibrium the rates of the forward and reverse processes are identical, but the rate constants are generally different. To see how this works, consider the simplified reaction A → B in the following three scenarios.
kf >> kr
If the rate constants are greatly different (by many orders of magnitude), then this requires that the equilibrium concentrations of products exceed those of the reactants by the same ratio. Thus the equilibrium composition will lie strongly on the "right"; the reaction can be said to be "complete" or "quantitative".
kf << kr
The rates can only be identical (equilibrium achieved) if the concentrations of the products are very small. We describe the resulting equilibrium as strongly favoring the left; very little product is formed. In the most extreme cases, we might even say that "the reaction does not take place".
kfkr
If kf and kr have comparable values (within, say, several orders of magnitude), then significant concentrations of products and reactants are present at equilibrium; we say the the reaction is "incomplete" and "reversible".
The images shown below offer yet another way of looking at these three cases. The plots show how the relative concentrations of the reactant and product change during the course of the reaction. The plots differ in the assumptions we make about the ratio of kf to kr. The equilibrium composition of the system is illustrated by the proportions of A and B in the horizontal parts of each plot where the composition remains unchanged. In each case, the two rate constants are sufficiently close in magnitude that each reaction can be considered "incomplete".
• In plot (i) the forward rate constant is twice as large as the reverse rate constant, so product (B) is favored, but there is sufficient reverse reaction to maintain a significant quantity of A.
• In (ii), the forward and reverse rate constants have identical magnitudes. Not surprisingly, so are the equilibrium values of [A] and [B].
• In (iii), the reverse rate constant exceeds the forward rate constant, so the equilibrium composition is definitely "on the left".
The Law of Mass Action is thus essentially the statement that the equilibrium composition of a reaction mixture can vary according to the quantities of components that are present. This of course is just what Berthollet observed in his Egyptian salt ponds, but we now understand it to be a consequence of the dynamic nature of chemical equilibrium.
When is a Reaction at Equilibrium?
Clearly, if we observe some change taking place— a change in color, the release of gas bubbles, the appearance of a precipitate, or the release of heat, we know the reaction is not yet at equilibrium. However, the absence of any apparent change does not by itself establish that the reaction is at equilibrium. The equilibrium state is one in which not only no change in composition take place, but also one in which no energetic tendency for further change is present. Unfortunately, "tendency" is not a property that is directly observable! Consider, for example, the reaction representing the synthesis of water from its elements:
$2 H_{2(g)} + O_{2(g)} \rightarrow 2 H_2O_{(g)} \label{10.5}$
You can store the two gaseous reactants in the same container indefinitely without any observable change occurring. But if you create an electrical spark in the container or introduce a flame, bang! After you pick yourself up off the floor and remove the shrapnel from what's left of your body, you will know very well that the system was not initially at equilibrium! It happens that this particular reaction has a tremendous tendency to take place, but for reasons that we will discuss in a later chapter, nothing can happen until we "set it off" in some way— in this case by exposing the mixture to a flame or spark, or (in a more gentle way) by introducing a platinum wire, which acts as a catalyst.
A reaction of this kind is said to be highly favored thermodynamically, but inhibited kinetically. The similar reaction of hydrogen and iodine
$H_{2(g)} + I_{2(g)} \rightarrow 2 HI_{(g)} \label{10.6}$
by contrast is only moderately favored thermodynamically (and is thus incomplete), but its kinetics are both unspectacular and reasonably facile.
Some Simple Tests for the Equilibrium State
• As we explained above in the context of the law of mass action, addition or removal of one component of the reaction will affect the amounts of all the others. For example, if we add more of a reactant, we would expect to see the concentration of a product change. If this does not happen, then it is likely that the reaction is kinetically inhibited and that the system is unable to attain equilibrium.
• It is almost always the case, however, that once a reaction actually starts, it will continue on its own until it reaches equilibrium, so if we can observe the change as it occurs and see it slow down and stop, we can be reasonably certain that the system is in equilibrium. This is by far the chemist's most common criterion.
• There is one other experimental test for equilibrium in a chemical reaction, although it is really only applicable to the kind of reactions we described above as being reversible. As we shall see later, the equilibrium state of a system is always sensitive to the temperature, and often to the pressure, so any changes in these variables, however, small, will temporarily disrupt the equilibrium, resulting in an observable change in the composition of the system as it moves toward its new equilibrium state.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above.
• Any reaction that can be represented by a balanced chemical equation can take place, at least in principle. However, there are two important qualifications:
• The tendency for the change to occur may be so small that the quantity of products formed may be very low, and perhaps negligible.
A reaction of this kind is said to be thermodynamically inhibited. The tendency for chemical change is governed solely by the properties of the reactants and products, and can be predicted by applying the laws of thermodynamics.
• The rate at which the reaction proceeds may be very small, or even zero, in which case we say the reaction is kinetically inhibited. Reaction rates depend on the mechanism of the reaction— that is, on what actually happens to the atoms as reactants are transformed into products. Reaction mechanisms cannot generally be predicted, and must be worked out experimentally. Also, the same reaction may have different mechansims under different conditions.
• As a chemical change proceeds, the quantities of the components on one side of the reaction equation will decrease, and those on the other side will increase. Eventually the reaction slows down and the composition of the system stops changing. At this point the reaction is in its equilibrium state, and no further change in composition will occur as long as the system is left undisturbed.
• For many reactions, the equilibrium state is one in which components on both sides of the equation (that is, both reactants and products) are present in significant amounts. Such a reaction is said to be incomplete or reversible.
• The equilibrium composition is independent of the direction from which it is approached; the labeling of substances as "reactants" or "products" is entirely a matter of convenience. (See the hydrogen iodide reaction plots above.)
• The law of mass action states that any chemical change is a competition between a forward reaction (left-to-right in the chemical equation) and a reverse reaction. The rate of each of these processes is governed by the concentrations of the substances reacting; as the reaction proceeds, these rates approach each other and at equilibrium they become identical.
• From the above, it follows that equilibrium is a dynamic process in which microscopic change (the forward and reverse reactions) continues to occur, but macroscopic change (changes in the quantities of substances) is absent.
• When a chemical reaction is at equilibrium, any disturbance of the system, such as a change in temperature, or addition or removal of one of the reaction components, will "shift" the composition to a new equilibrium state. This is the only unambiguous way of verifying that a reaction is at equilibrium. The fact that the composition remains static does not in itself prove that a reaction is at equilibrium, because the change may be kinetically inhibited. | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.01%3A_Introduction_to_Chemical_Equilibrium.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• A system in its equilibrium state will remain in that state indefinitely as long as it is undisturbed. If the equilibrium is destroyed by subjecting the system to a change of pressure, temperature, or the number of moles of a substance, then a net reaction will tend to take place that moves the system to a new equilibrium state. Le Chatelier's principle says that this net reaction will occur in a direction that partially offsets the change.
• The Le Chatelier Principle has practical effect only for reactions in which signficant quantities of both reactants and products are present at equilibrium— that is, for reactions that are thermodynamically reversible.
• Addition of more product substances to an equilibrium mixture will shift the equilibrium to the left; addition of more reactant substances will shift it to the right. These effects are easily explained in terms of competing forward- and reverse reactions— that is, by the law of mass action.
• If a reaction is exothermic (releases heat), an increase in the temperature will force the equilibrium to the left, causing the system to absorb heat and thus partially ofsetting the rise in temperature. The opposite effect occurs for endothermic reactions, which are shifted to the right by rising temperature.
• The effect of pressure on an equilibrium is significant only for reactions which involve different numbers of moles of gases on the two sides of the equation. If the number of moles of gases increases, than an increase in the total pressure will tend to initiate a reverse reaction that consumes some the products, partially reducing the effect of the pressure increase.
• The classic example of the practical use of the Le Chatelier principle is the Haber-Bosch process for the synthesis of ammonia, in which a balance between low temperature and high pressure must be found.
The previous Module emphasized the dynamic character of equilibrium as expressed by the Law of Mass Action. This law serves as a model explaining how the composition of the equilibrium state is affected by the "active masses" (concentrations) of reactants and products. In this lesson, we develop the consequences of this law to answer the very practical question of how an existing equilibrium composition is affected by the addition or withdrawal of one of the components.
Le Chatelier's Principle
If a reaction is at equilibrium and we alter the conditions so as to create a new equilibrium state, then the composition of the system will tend to change until that new equilibrium state is attained. (We say "tend to change" because if the reaction is kinetically inhibited, the change may be too slow to observe or it may never take place.) In 1884, the French chemical engineer and teacher Henri Le Chatelier (1850-1936) showed that in every such case, the new equilibrium state is one that partially reduces the effect of the change that brought it about.
This law is known to every Chemistry student as the Le Chatelier principle. His original formulation was somewhat complicated, but a reasonably useful paraphrase of it reads as follows:
Le Chatelier principle: If a system at equilibrium is subjected to a change of pressure, temperature, or the number of moles of a component, there will be a tendency for a net reaction in the direction that reduces the effect of this change.
To see how this works (and you must do so, as this is of such fundamental importance that you simply cannot do any meaningful chemistry without a thorough working understanding of this principle), look again at the hydrogen iodide dissociation reaction
$2 HI \rightarrow H_2 + I_2$
Consider an arbitrary mixture of these three components at equilibrium, and assume that we inject more hydrogen gas into the container. Because the H2 concentration now exceeds its new equilibrium value, the system is no longer in its equilibrium state, so a net reaction now ensues as the system moves to the new state.
The Le Chatelier principle states that the net reaction will be in a direction that tends to reduce the effect of the added H2. This can occur if some of the H2 is consumed by reacting with I2 to form more HI; in other words, a net reaction occurs in the reverse direction. Chemists usually simply say that "the equilibrium shifts to the left".
To get a better idea of how this works, carefully examine the diagram below which follows the concentrations of the three components of this reaction as they might change in time (the time scale here will typically be about an hour):
Disruption and restoration of equilibrium. At the left, the concentrations of the three components do not change with time because the system is at equilibrium. We then add more hydrogen to the system, disrupting the equilibrium. A net reaction then ensues that moves the system to a new equilibrium state (right) in which the quantity of hydrogen iodide has increased; in the process, some of the I2 and H2 are consumed. Notice that the new equilibrium state contains more hydrogen than did the initial state, but not as much as was added; as the Le Chatelier principle predicts, the change we made (addition of H2) has been partially counteracted by the "shift to the right". Table $1$ Contains several examples showing how changing the quantity of a reaction component can shift an established equilibrium.
system
change
result
Table $1$: Example of Le Chatelier's principle
CO2 + H2 → H2O(g) + CO a drying agent is added to absorb H2O Shift to the right. Continuous removal of a product will force any reaction to the right
H2(g) + I2(g) → 2HI(g) Some nitrogen gas is added No change; N2 is not a component of this reaction system.
NaCl(s) + H2SO4(l) → Na2SO4(s)+ HCl(g) reaction is carried out in an open container Because HCl is a gas that can escape from the system, the reaction is forced to the right. This is the basis for the commercial production of hydrochloric acid.
H2O(l) → H2O(g) water evaporates from an open container Continuous removal of water vapor forces the reaction to the right, so equilibrium is never achieved.
HCN(aq) → H+(aq) + CN(aq) the solution is diluted Shift to right; the product [H+][CN] diminishes more rapidly than does [HCN].
AgCl(s) → Ag+(aq) + Cl(aq) some NaCl is added to the solution Shift to left due to increase in Cl concentration. This is known as the common ion effect on solubility.
N2 + 3 H2 → 2 NH3 a catalyst is added to speed up this reaction No change. Catalysts affect only the rate of a reaction; the have no effect at all on the composition of the equilibrium state.
How do changes in temperature affect Equilibria?
Virtually all chemical reactions are accompanied by the liberation or uptake of heat. If we regard heat as a "reactant" or "product" in an endothermic or exothermic reaction respectively, we can use the Le Chatelier principle to predict the direction in which an increase or decrease in temperature will shift the equilibrium state. Thus for the oxidation of nitrogen, an endothermic process, we can write
$\text{[heat]} + N_2 + O_2 \rightleftharpoons 2 NO$
Suppose this reaction is at equilibrium at some temperature $T_1$ and we raise the temperature to $T_2$. The Le Chatelier principle tells us that a net reaction will occur in the direction that will partially counteract this change. Since the reaction is endothermic, a shift of the equilibrium to the right will take place.
Nitric oxide, the product of this reaction, is a major air pollutant which initiates a sequence of steps leading to the formation of atmospheric smog. Its formation is an unwanted side reaction which occurs when the air (which is introduced into the combustion chamber of an engine to supply oxygen) gets heated to a high temperature. Designers of internal combustion engines now try, by various means, to limit the temperature in the combustion region, or to restrict its highest-temperature part to a small volume within the combustion chamber.
How do changes in pressure affect equilibria?
You will recall that if the pressure of a gas is reduced, its volume will increase; pressure and volume are inversely proportional. With this in mind, suppose that the reaction
$2 NO_{2(g)} \rightleftharpoons N_2O_{4(g)}$
is in equilibrium at some arbitrary temperature and pressure, and that we double the pressure, perhaps by compressing the mixture to a smaller volume. From the Le Chatelier principle we know that the equilibrium state will change to one that tends to counteract the increase in pressure. This can occur if some of the NO2 reacts to form more of the dinitrogen tetroxide, since two moles of gas are being removed from the system for every mole of N2O4 formed, thereby decreasing the total volume of the system. Thus increasing the pressure will shift this equilibrium to the right.
It is important to understand that changing the pressure will have a significant effect only on reactions in which there is a change in the number of moles of gas. For the above reaction, this change
$Δn_g = (n_{products} – n_{reactants}) = 1 – 2 = –1.$
In the case of the nitrogen oxidation reaction
$N_2 + O_2 \rightleftharpoons 2 NO$
Δng = 0 and changing pressure will have no effect on the equilibrium.
The volumes of solids and liquids are hardly affected by the pressure at all, so for reactions that do not involve gaseous substances, the effects of pressure changes are ordinarily negligible. Exceptions arise under conditions of very high pressure such as exist in the interior of the Earth or near the bottom of the ocean. A good example is the dissolution of calcium carbonate
$CaCO_{3(s)} \rightleftharpoons Ca^{2+} + CO_3^{2–}$
There is a slight decrease in the volume when this reaction takes place, so an increase in the pressure will shift the equilibrium to the right, with the results that calcium carbonate becomes more soluble at higher pressures.
The skeletons of several varieties of microscopic organisms that inhabit the top of the ocean are made of CaCO3, so there is a continual rain of this substance toward the bottom of the ocean as these organisms die. As a consequence, the floor of the Atlantic ocean is covered with a blanket of calcium carbonate. This is not true for the Pacific ocean, which is deeper; once the skeletons fall below a certain depth, the higher pressure causes them to dissolve. Some of the seamounts (undersea mountains) in the Pacific extend above the solubility boundary so that their upper parts are covered with CaCO3 sediments.
The effect of pressure on a reaction involving substances whose boiling points fall within the range of commonly encountered temperature will be sensitive to the states of these substances at the temperature of interest. For reactions involving gases, only changes in the partial pressures of those gases directly involved in the reaction are important; the presence of other gases has no effect.
Example $1$: steam reforming of methane
The commercial production of hydrogen is carried out by treating natural gas with steam at high temperatures and in the presence of a catalyst (“steam reforming of methane”):
$CH_4 + H_2O \rightleftharpoons CH_3OH + H_2 \nonumber$
Given the following boiling points: CH4 (methane) = –161°C, H2O = 100°C, CH3OH = 65°, H2 = –253°C, predict the effects of an increase in the total pressure on this equilibrium at 50°, 75° and 120°C.
Solution
Calculate the change in the moles of gas for each process:
temp
equation
$Δn_g$
shift
50°
CH4(g) + H2O(l) → CH3OH(l) + H2(g) 0 none
75° CH4(g) + H2O(l) → CH3OH(g) + H2(g) +1 to left
120° CH4(g) + H2O(g) → CH3OH(g) + H2(g) 0 none
The Haber Process and why is it important
The Haber process for the synthesis of ammonia is based on the exothermic reaction
N2(g) + 3 H2(g) → 2 NH3(g) ΔH = –92 kJ/mol
The Le Chatelier principle tells us that in order to maximize the amount of product in the reaction mixture, it should be carried out at high pressure and low temperature. However, the lower the temperature, the slower the reaction (this is true of virtually all chemical reactions.) As long as the choice had to be made between a low yield of ammonia quickly or a high yield over a long period of time, this reaction was infeasible economically.
Nitrogen is available for free, being the major component of air, but the strong triple bond in N2 makes it extremely difficult to incorporate this element into species such as NO3 and NH4+ which serve as the starting points for the wide variety of nitrogen-containing compounds that are essential for modern industry. This conversion is known as nitrogen fixation, and because nitrogen is an essential plant nutrient, modern intensive agriculture is utterly dependent on huge amounts of fixed nitrogen in the form of fertilizer. Until around 1900, the major source of fixed nitrogen was the NaNO3 found in extensive deposits in South America. Several chemical processes for obtaining nitrogen compounds were developed in the early 1900's, but they proved too inefficient to meet the increasing demand.
Although the direct synthesis of ammonia from its elements had been known for some time, the yield of product was found to be negligible. In 1905, Fritz Haber (1868-1934) began to study this reaction, employing the thinking initiated by Le Chatelier and others, and the newly-developing field of thermodynamics that served as the basis of these principles. From the Le Chatelier law alone, it is apparent that this exothermic reaction is favored by low temperature and high pressure. However, it was not as simple as that: the rate of any reaction increases with the temperature, so working with temperature alone, one has the choice between a high product yield achieved only very slowly, or a very low yield quickly. Further, the equipment and the high-strength alloy steels need to build it did not exist at the time. Haber solved the first problem by developing a catalyst that would greatly speed up the reaction at lower temperatures.
The second problem, and the development of an efficient way of producing hydrogen, would delay the practical implementation of the process until 1913, when the first plant based on the Haber-Bosch process (as it is more properly known, Carl Bosch being the person who solved the major engineering problems) came into operation. The timing could not have been better for Germany, since this country was about to enter the First World War, and the Allies had established a naval blockade of South America, cutting off the supply of nitrate for the the German munitions industry.
Bosch's plant operated the ammonia reactor at 200 atm and 550°C. Later, when stronger alloy steels had been developed, pressures of 800-1000 atm became common. The source of hydrogen in modern plants is usually natural gas, which is mostly methane:
CH4 + H2O → CO + 3 H2 formation of synthesis gas from methane
CO + H2O → CO2 + H2 shift reaction carried out in reformer
The Haber-Bosch process is considered the most important chemical synthesis developed in the 20th century. Besides its scientific importance as the first large-scale application of the laws of chemical equilibrium, it has had tremendous economic and social impact; without an inexpensive source of fixed nitrogen, the intensive crop production required to feed the world's growing population would have been impossible. Haber was awarded the 1918 Nobel Prize in Chemistry in recognition of his work. Carl Bosch, who improved the process, won the Nobel Prize in 1931.
The Le Chatelier Principle in Physiology
Many of the chemical reactions that occur in living organisms are regulated through the Le Chatelier principle.
Oxygen transport by the blood
Few of these are more important to warm-blooded organisms than those that relate to aerobic respiration, in which oxygen is transported to the cells where it is combined with glucose and metabolized to carbon dioxide, which then moves back to the lungs from which it is expelled.
hemoglobin + O2 oxyhemoglobin
The partial pressure of O2 in the air is 0.2 atm, sufficient to allow these molecules to be taken up by hemoglobin (the red pigment of blood) in which it becomes loosely bound in a complex known as oxyhemoglobin. At the ends of the capillaries which deliver the blood to the tissues, the O2 concentration is reduced by about 50% owing to its consumption by the cells. This shifts the equilibrium to the left, releasing the oxygen so it can diffuse into the cells.
Maintence of blood pH
Carbon dioxide reacts with water to form a weak acid H2CO3 which would cause the blood pH to fall to dangerous levels if it were not promptly removed as it is excreted by the cells. This is accomplished by combining it with carbonate ion through the reaction
$H_2CO_3 + CO_3^{2–} \rightleftharpoons 2 HCO_3^– \nonumber$
which is forced to the right by the high local CO2 concentration within the tissues. Once the hydrogen carbonate (bicarbonate) ions reach the lung tissues where the CO2 partial pressure is much smaller, the reaction reverses and the CO2 is expelled.
Carbon monoxide poisoning
Carbon monoxide, a product of incomplete combustion that is present in automotive exhaust and cigarette smoke, binds to hemoglobin 200 times more tightly than does O2. This blocks the uptake and transport of oxygen by setting up a competing equilibrium
O2-hemoglobin hemoglobin CO-hemoglobin
Air that contains as little as 0.1 percent carbon monoxide can tie up about half of the hemoglobin binding sites, reducing the amount of O2 reaching the tissues to fatal levels. Carbon monoxide poisoning is treated by administration of pure O2 which promotes the shift of the above equilibrium to the left. This can be made even more effective by placing the victim in a hyperbaric chamber in which the pressure of O2 can be made greater than 1 atm. | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.02%3A_Le_Chatelier%27s_Principle.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• When arbitrary quantities of the different components of a chemical reaction system are combined, the overall system composition will not likely correspond to the equilibrium composition. As a result, a net change in composition ("a shift to the right or left") will tend to take place until the equilibrium state is attained.
• The status of the reaction system in regard to its equilibrium state is characterized by the value of the equilibrium expression whose formulation is defined by the coefficients in the balanced reaction equation; it may be expressed in terms of concentrations, or in the case of gaseous components, as partial pressures.
• The various terms in the equilibrium expression can have any arbitrary value (including zero); the value of the equilibrium expression itself is called the reaction quotient Q.
• If the concentration or pressure terms in the equilibrium expression correspond to the equilibrium state of the system, then Q has the special value K, which we call the equilibrium constant.
• The ratio of Q/K (whether it is 1, >1 or <1) thus serves as an index of how far the system is from its equilibrium composition, and its value indicates the direction in which the net reaction must proceed in order to reach its equilibrium state.
• When Q = K, then the equilibrium state has been reached, and no further net change in composition will take place as long as the system remains undisturbed.
Consider a simple reaction such as the gas-phase synthesis of hydrogen iodide from its elements: $H_2 + I_2 \rightarrow 2 HI$ Suppose you combine arbitrary quantities of $H_2$, $I_2$ and $HI$. Will the reaction create more HI, or will some of the HI be consumed as the system moves toward its equilibrium state? The concept of the reaction quotient, which is the focus of this short lesson, makes it easy to predict what will happen.
What is the Equilibrium Quotient?
In the previous section we defined the equilibrium expression for the reaction
$a A + b B \rightleftharpoons c C + d D$
as
$K = \underbrace{\dfrac{a_C^c a_D^d}{a_A^a a_b^b}}_{\text{in terms} \ \text{of activities}} \approx \underbrace{\dfrac{[C]^c[D]^d}{[A]^a[B]^b}}_{\text{in terms} \ \text{of concetrations}}$
In the general case in which the concentrations can have any arbitrary values (including zero), this expression is called the reaction quotient (the term equilibrium quotient is also commonly used.) and its value is denoted by $Q$ (or $Q_c$ or $Q_p$ if we wish to emphasize that the terms represent molar concentrations or partial pressures.) If the terms correspond to equilibrium concentrations, then the above expression is called the equilibrium constant and its value is denoted by $K$ (or $K_c$ or $K_p$).
$K$ is thus the special value that $Q$ has when the reaction is at equilibrium
The value of Q in relation to K serves as an index how the composition of the reaction system compares to that of the equilibrium state, and thus it indicates the direction in which any net reaction must proceed. For example, if we combine the two reactants A and B at concentrations of 1 mol L–1 each, the value of Q will be 0÷1=0. The only possible change is the conversion of some of these reactants into products. If instead our mixture consists only of the two products C and D, Q will be indeterminately large (1÷0) and the only possible change will be in the reverse direction.
For example, if we combine the two reactants A and B at concentrations of 1 mol L–1 each, the value of Q will be 0÷1=0. The only possible change is the conversion of some of these reactants into products. If instead our mixture consists only of the two products C and D, Q will be indeterminately large (1÷0) and the only possible change will be in the reverse direction.
It is easy to see (by simple application of the Le Chatelier principle) that the ratio of Q/K immediately tells us whether, and in which direction, a net reaction will occur as the system moves toward its equilibrium state. A schematic view of this relationship is shown below:
It is very important that you be able to work out these relations for yourself, not by memorizing them, but from the definitions of $Q$ and $K$.
Table $1$: Comparison of Q and K
Condition Status of System
Q > K Product concentration too high for equilibrium; net reaction proceeds to left.
Q = K System is at equilibrium; no net change will occur.
Q < K Product concentration too low for equilibrium; net reaction proceeds to right.
It is very important that you be able to work out these relations for yourself, not by memorizing them, but from the definitions of $Q$ and $K$.
Example $1$
The equilibrium constant for the oxidation of sulfur dioxide is Kp = 0.14 at 900 K.
$\ce{2 SO_2(g) + O_2(g) \rightleftharpoons 2 SO_3(g)} \nonumber$
If a reaction vessel is filled with SO3 at a partial pressure of 0.10 atm and with O2 and SO2 each at a partial pressure of 0.20 atm, what can you conclude about whether, and in which direction, any net change in composition will take place?
Solution:
The value of the equilibrium quotient Q for the initial conditions is
$Q= \dfrac{p_{SO_3}^2}{p_{O_2}p_{SO_2}^2} = \dfrac{(0.10\; atm)^2}{(0.20 \;atm) (0.20 \; atm)^2} = 1.25\; atm^{-1} \nonumber$
Since Q > K, the reaction is not at equilibrium, so a net change will occur in a direction that decreases Q. This can only occur if some of the SO3 is converted back into products. In other words, the reaction will "shift to the left".
The formal definitions of Q and K are quite simple, but they are of limited usefulness unless you are able to relate them to real chemical situations. The following diagrams illustrate the relation between Q and K from various standpoints. Take some time to study each one carefully, making sure that you are able to relate the description to the illustration.
Example $2$: Dissociation of dinitrogen tetroxide
For the reaction
$N_2O_{4(g)} \rightleftharpoons 2 NO_{2(g)} \nonumber$
Kc = 0.0059 at 298 K.
This equilibrium condition is represented by the red curve that passes through all points on the graph that satisfy the requirement that
$Q = \dfrac{[NO_2]^2}{ [N_2O_4]} = 0.0059 \nonumber$
There are of course an infinite number of possible Q's of this system within the concentration boundaries shown on the plot. Only those points that fall on the red line correspond to equilibrium states of this system (those for which $Q = K_c$). The line itself is a plot of [NO2] that we obtain by rearranging the equilibrium expression
$[NO_2] = \sqrt{[N_2O_4]K_c} \nonumber$
If the system is initially in a non-equilibrium state, its composition will tend to change in a direction that moves it to one that is on the line. Two such non-equilibrium states are shown. The state indicated by has $Q > K$, so we would expect a net reaction that reduces Q by converting some of the NO2 into N2O4; in other words, the equilibrium "shifts to the left". Similarly, in state , Q < K, indicating that the forward reaction will occur.
The blue arrows in the above diagram indicate the successive values that Q assumes as the reaction moves closer to equilibrium. The slope of the line reflects the stoichiometry of the equation. In this case, one mole of reactant yields two moles of products, so the slopes have an absolute value of 2:1.
Example $3$: Phase-change equilibrium
One of the simplest equilibria we can write is that between a solid and its vapor. In this case, the equilibrium constant is just the vapor pressure of the solid. Thus for the process
$I_{2(s)} \rightleftharpoons I_{2(g)} \nonumber$
all possible equilibrium states of the system lie on the horizontal red line and is independent of the quantity of solid present (as long as there is at least enough to supply the relative tiny quantity of vapor.)
So adding various amounts of the solid to an empty closed vessel (states and ) causes a gradual buildup of iodine vapor. Because the equilibrium pressure of the vapor is so small, the amount of solid consumed in the process is negligible, so the arrows go straight up and all lead to the same equilibrium vapor pressure.
Example $4$: Heterogeneous chemical reaction
The decomposition of ammonium chloride is a common example of a heterogeneous (two-phase) equilibrium. Solid ammonium chloride has a substantial vapor pressure even at room temperature:
$NH_4Cl_{(s)} \rightleftharpoons NH_{3(g)} + HCl_{(g)}$
Arrow traces the states the system passes through when solid NH4Cl is placed in a closed container. Arrow represents the addition of ammonia to the equilibrium mixture; the system responds by following the path back to a new equilibrium state which, as the Le Chatelier principle predicts, contains a smaller quantity of ammonia than was added. The unit slopes of the paths and reflect the 1:1 stoichiometry of the gaseous products of the reaction. | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.03%3A_Reaction_Quotient.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• The equilibrium quotient Q is the value of the equilibrium expression of a reaction for any arbitrary set of concentrations or partial pressures of the reaction components.
• The equilibrium constant K is the value of Q when the reaction is at equilibrium. K has a unique value for a given reaction at a fixed temperature and pressure.
• Q and K can be expressed in terms of concentrations, partial pressures, or, when appropriate, in some combination of these.
• For a reaction in which all the components are gases, Qc and Kc will have different values except in the special case in which the total number of moles of gas does not change.
• Concentration terms for substances whose concentrations do not change in the reaction do not appear in equilibrium expressions. The most common examples are [H2O] when the reaction takes place in aqueous solution (so that [H2O] is effectively constant at 55.6 M), and in heterogeneous reactions involving solids, in which the concentration of the solid is determined by the density of the solid itself.
• A reaction whose equilibrium constant is in the range of about 0.01 to 100 is said to be incomplete or [thermodynamically] reversible.
• Q and K are conventionally treated as dimensionless quantities, and need not ordinarily have units associated with them.
• Heterogeneous reactions are those in which two or more phases are involved; homogeneous reactions take place in a single phase. A common type of heterogeneous reaction is the loss of water of crystallization by a solid hydrate such as CuSO4·5H2O.
• The equilibrium expression can be manipulated and combined in the following ways:
• If the reaction is written in reverse, Q becomes Q–1;
• If the coefficients of an equation are multiplied by n, Q becomes Qn;
• Q for the sum of two reactions (that is, for two reactions that take place in sequence) is the product (Q1)(Q2).
You know that an equilibrium constant expression looks something like K = [products] / [reactants]. But how do you translate this into a format that relates to the actual chemical system you are interested in? This lesson will show you how to write the equilibrium constant expressions that you will need to use when dealing with the equilibrium calculation problems in the chapter that follows this one.
Pressures can Express Concentrations
Although we commonly write equilibrium quotients and equilibrium constants in terms of molar concentrations, any concentration-like term can be used, including mole fraction and molality. Sometimes the symbols $K_c$ , $K_x$ , and $K_m$ are used to denote these forms of the equilibrium constant. Bear in mind that the numerical values of K’s and Q’s expressed in these different ways will not generally be the same.
Most of the equilibria we deal with in this course occur in liquid solutions and gaseous mixtures. We can express $K_c$ values in terms of moles per liter for both, but when dealing with gases it is often more convenient to use partial pressures. These two measures of concentration are of course directly proportional:
$c=\dfrac{n}{V}=\dfrac{\dfrac{PV}{RT}}{V}=\dfrac{P}{RT} \label{Eq1}$
so for a reaction $A_{(g)} \rightarrow B_{(g)}$ we can write the equilibrium constant as
$K_p =\dfrac{P_B}{P_A} \label{Eq2}$
Owing to interactions between molecules, especially when ions are involved, all of these forms of the equilibrium constant are only approximately correct, working best at low concentrations or pressures. The only equilibrium constant that is truly “constant” (except that it still varies with the temperature!) is expressed in terms of activities, which you can think of as “effective concentrations” that allow for interactions between molecules. In practice, this distinction only becomes important for equilibria involving gases at very high pressures (such as are often encountered in chemical engineering) and in ionic solutions more concentrated than about 0.001 M. We will not deal much with activities in this course.
For a reaction such as
$CO_{2\, (g)} + OH^–_{(aq)} \rightleftharpoons HCO_{3\, (aq)}^- \label{Eq3}$
that involves both gaseous and dissolved components, a “hybrid” equilibrium constant is commonly used:
$K =\dfrac{[HCO_3^-]}{P_{CO_2}[OH^-]} \label{Eq4}$
Clearly, it is essential to be sure of the units when you see an equilibrium constant represented simply by "$K$".
A note about pressure and concentration units
In this lesson (and in most of the others in this set,) we express concentrations in mol L–1 and pressures in atmospheres. Although this reflects common usage among chemists (older ones, especially!), these units are not part of the SI system which has been the international standard since the latter part of the 20th Century.
Molar concentrations are now more properly expressed in mol dm–3 and the "standard atmosphere" corresponds to a pressure of 101.325 kPa. Until 1990, 1 atm was the "standard pressure" employed in calculations involving the gas laws, and also in thermodynamics. Since that date, "standard pressure" has been 100.000 kPa, also expressed as 1 bar.
For most practical purposes, the differences between these values are so small that they can be neglected.
Converting between $K_p$ and $K_c$
It is sometimes necessary to convert between equilibrium constants expressed in different units. The most common case involves pressure- and concentration equilibrium constants. Note that when V is expressed in liters and P in atmospheres, R must have the value 0.08206 L-atm/mol K.). The ideal gas law relates the partial pressure of a gas to the number of moles and its volume:
$PV = nRT$
Concentrations are expressed in moles/unit volume n/V, so by rearranging the above equation we obtain the explicit relation of pressure to concentration:
$P = \left(\dfrac{n}{V} \right)RT \label{Eq5}$
Conversely,
$c = \dfrac{n}{V} = \dfrac{P}{RT}$
so a concentration of a gas [A] can be expressed as $\dfrac{P_A}{RT}$.
For a reaction of the form $A + 3 B\rightleftharpoons 2C$, we can write
Assuming that all of the components are gases, the difference
$(\text{moles of gas in products}) – (\text{moles of gas in reactants}) = \Delta{n_g} \nonumber$
is given by
$\color{red} {K_p = K_c (RT)^{\Delta{n_g}} \label{Eq6}}$
Do not show unchanging concentrations!
Substances whose concentrations undergo no significant change in a chemical reaction do not appear in equilibrium constant expressions. How can the concentration of a reactant or product not change when a reaction involving that substance takes place? There are two general cases to consider.
The substance is also the Solvent
This happens all the time in acid-base chemistry. Thus for the hydrolysis of the cyanide ion
$\ce{CN^{-} + H2O <=> HCN + OH^{–}} \nonumber$
we write
$K_c= \dfrac{[\ce{HCN}][\ce{OH^-}]}{[\ce{CN^-}]} \nonumber$
in which no $[H_2O]$ term appears. The justification for this omission is that water is both the solvent and reactant, but only the tiny portion that acts as a reactant would ordinarily go in the equilibrium expression. The amount of water consumed in the reaction is so minute (because $K$ is very small) that any change in the concentration of $H_2O$ from that of pure water (55.6 mol L–1) will be negligible.
Similarly, for the "autodissociation" of water
$H_2O = H^+ + OH^– \nonumber$
the equilibrium constant is expressed as the "ion product"
$K_w = [H^+][OH^–] \nonumber$
Be careful about throwing away H2O whenever you see it. In the esterification reaction
$\ce{CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O} \nonumber$
that we discussed in a previous section, a [H2O] term must be present in the equilibrium expression if the reaction is assumed to be between the two liquids acetic acid and ethanol. If, on the other hand, the reaction takes place between a dilute aqueous solution of the acid and the alcohol, then the [H2O] term would not be included.
The substance is a solid or a pure liquid phase
This is most frequently seen in solubility equilibria, but there are many other reactions in which solids are directly involved:
$CaF_{2\, (s)} \rightarrow Ca^{2+}_{(aq)} + 2F^-_{(aq)} \label{Eq11}$
$Fe_3O_4(s) + 4 H_{2\, (g)} \rightarrow 4 H_2O_{(g)} + 3Fe_{(s)} \label{Eq12}$
These are heterogeneous reactions (meaning reactions in which some components are in different phases), and the argument here is that concentration is only meaningful when applied to a substance within a single phase.
Thus the term $[CaF_2]$ would refer to the “concentration of calcium fluoride within the solid $CaF_2$", which is a constant depending on the molar mass of $CaF_2$ and the density of that solid. The concentrations of the two ions will be independent of the quantity of solid $CaF_2$ in contact with the water; in other words, the system can be in equilibrium as long as any $CaF_2$ at all is present. Throwing out the constant-concentration terms can lead to some rather sparse-looking equilibrium expressions. For example, the equilibrium expression for each of the processes shown in the following table consists solely of a single term involving the partial pressure of a gas:
Table $1$:
$CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$
$K_p = P_{CO_2}$
Thermal decomposition of limestone, a first step in the manufacture of cement.
$Na_2SO_4 \cdot 10 H_2O_{(s)} Na_2SO_{4(s)} + 10 H_2O_{(g)}$
$K_p = P_{H_2O}^{10}$
Sodium sulfate decahydrate is a solid in which H2O molecules (“waters of hydration") are incorporated into the crystal structure.)
$I_{2(s)} \rightleftharpoons I_{2(g)}$
$K_p = P_{I_2}$
sublimation of solid iodine; this is the source of the purple vapor you can see above solid iodine in a closed container.
$H_2O_{(l)} \rightleftharpoons H_2O_{(g)}$
$K_p = P_{H_2O}$
Vaporization of water. When the partial pressure of water vapor in the air is equal to K, the relative humidity is 100%.
The last two processes represent changes of state (phase changes) which can be treated exactly the same as chemical reactions. In each of the heterogeneous processes shown in Table $1$, the reactants and products can be in equilibrium (that is, permanently coexist) only when the partial pressure of the gaseous product has the value consistent with the indicated $K_p$. Bear in mind also that these $K_p$'s all increase with the temperature.
Example $1$
What are the values of $K_p$ for the equilibrium between liquid water and its vapor at 25°C, 100°C, and 120°C? The vapor pressure of water at these three temperatures is 23.8 torr, 760 torr (1 atm), and 1489 torr, respectively.
Comment: These vapor pressures are the partial pressures of water vapor in equilibrium with the liquid, so they are identical with the $K_p$'s when expressed in units of atmospheres.
Solution
25°C
100°C
120°C
$K_p = \dfrac{23.8\; torr}{ 760\; torr/atm} = 0.031\; atm$ $K_p = 1.00\; atm$ $K_p = \dfrac{1489\; torr}{ 760\; torr/atm} = 1.96\; atm$ The partial pressure of H2O above the surface of liquid water in a closed container at 25°C will build up to this value. If the cover is removed so that this pressure cannot be maintained, the system will cease to be at equilibrium and the water will evaporate.
This temperature corresponds, of course, to the boiling point of water. The normal boiling point of a liquid is the temperature at which the partial pressure of its vapor is 1 atm.
The only way to heat water above its normal boiling point is to do so in a closed container that can withstand the increased vapor pressure. Thus a pressure cooker that operates at 120°C must be designed to withstand an internal pressure of at least 2 atm.
Values of Equilibrium Constants
or “reversible”.
As an equilibrium constant approaches the limits of zero or infinity, the reaction can be increasingly characterized as a one-way process; we say it is “complete” or “irreversible”. The latter term must of course not be taken literally; the Le Chatelier principle still applies (especially insofar as temperature is concerned), but addition or removal of reactants or products will have less effect.
Kinetically Hindered Reactions
Although it is by no means a general rule, it frequently happens that reactions having very large equilibrium constants are kinetically hindered, often to the extent that the reaction essentially does not take place.
The examples in the following table are intended to show that numbers (values of K), no matter how dull they may look, do have practical consequences!
Table $2$: Examples of Reversible Reactions
Reaction
K
remarks
$N_{2(g)} + O_{2(g)} \rightleftharpoons 2 NO_{(g)}$ $5 \times 10^{–31}$ at 25°C,
0.0013 at 2100°C
These two very different values of K illustrate very nicely why reducing combustion-chamber temperatures in automobile engines is environmentally friendly.
$3 H_{2(g)} + N_{2(g)} \rightleftharpoons 2 NH_{3(g)}$ $7 \times 10^5$ at 25°C,
56 at 1300°C
See the discussion of this reaction in the section on the Haber process.
$H_{2(g)} \rightleftharpoons 2 H_{(g)}$ $10^{–36}$ at 25°C,
$6 \times 10^{–5}$ at 5000°
Dissociation of any stable molecule into its atoms is endothermic. This means that all molecules will decompose at sufficiently high temperatures.
$H_2O_{(g)} \rightleftharpoons H_{2(g)} + ½ O_{2(g)}$ $8 \times 10^{–41}$ at 25°C You won’t find water a very good source of oxygen gas at ordinary temperatures!
$CH_3COOH_{(l)} \rightleftharpoons 2 H_2O_{(l)} + 2 C_{(s)}$
$K_c = 10^{13}$ at 25°C This tells us that acetic acid has a great tendency to decompose to carbon, but nobody has ever found graphite (or diamonds!) forming in a bottle of vinegar. A good example of a super kinetically-hindered reaction!
Do Equilibrium Constants have Units?
The equilibrium expression for the synthesis of ammonia
$3 H_{2(g)} + N_{2(g)} \rightarrow 2 NH_{3(g)} \label{Eq13}$
can be expressed as
$K_p =\dfrac{P^2_{NH_3}}{P_{N_2}P^3_{H_2}} \label{Eq14}$
or
$K_c = \dfrac{[NH_3]^2}{[N_2] [H_2]^3} \label{Eq15}$
so $K_p$ for this process would appear to have units of atm–1, and $K_c$ would be expressed in mol–2 L2. And yet these quantities are often represented as being dimensionless. Which is correct? The answer is that both forms are acceptable. There are some situations (which you will encounter later) in which K’s must be considered dimensionless, but in simply quoting the value of an equilibrium constant it is permissible to include the units, and this may even be useful in order to remove any doubt about the units of the individual terms in equilibrium expressions containing both pressure and concentration terms. In carrying out your own calculations, however, there is rarely any real need to show the units.
Strictly speaking, equilibrium expressions do not have units because the concentration or pressure terms that go into them are really ratios having the forms (n mol L–1)/(1 mol L–1) or (n atm)/(1 atm) in which the unit quantity in the denominator refers to the standard state of the substance; thus the units always cancel out. (But first-year students are not expected to know this!)
For substances that are liquids or solids, the standard state is just the concentration of the substance within the liquid or solid, so for something like CaF(s), the term going into the equilibrium expression is [CaF2]/[CaF2] which cancels to unity; this is the reason we don’t need to include terms for solid or liquid phases in equilibrium expressions. The subject of standard states would take us beyond where we need to be at this point in the course, so we will simply say that the concept is made necessary by the fact that energy, which ultimately governs chemical change, is always relative to some arbitrarily defined zero value which, for chemical substances, is the standard state.
How the Reaction Equation affects K
It is important to remember that an equilibrium quotient or constant is always tied to a specific chemical equation, and if we write the equation in reverse or multiply its coefficients by a common factor, the value of $Q$ or $K$ will change. The rules are very simple:
• Writing the equation in reverse will invert the equilibrium expression;
• Multiplying the coefficients by a common factor will raise Q or K to the corresponding power.
Here are some of the possibilities for the reaction involving the equilibrium between gaseous water and its elements:
Example 1: $\ce{2 H2 + O2 <=> 2 H2O}$ with equilibrium expression $K_p = \dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}} \nonumber$
Example 2: $\ce{10 H2 + 5 O2 <=> 10 H2O}$ with equilibrium expression \begin{align*} K_p &= \dfrac{P_{H_2O}^{10}}{P_{H_2}^{10}P_{O_2}^5} \[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{5}\end{align*}
Example 3: $\ce{H2 + 1/2 O2 <=> H2O}$ with equilibrium expression \begin{align*} K_p &= \dfrac{P_{H_2O}}{P_{H_2}P_{O_2}^{1/2}} \[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{1/2} \end{align*}
Example 4: $\ce{H2O <=> H2 + 1/2 O2 }$ with equilibrium expression \begin{align*} K_p &= \dfrac{P_{H_2}P_{O_2}^{1/2}}{P_{H_2O}} \[4pt] &= \left(\dfrac{P_{H_2O}^2}{P_{H_2}^2P_{O_2}}\right)^{-1/2} \end{align*}
Many chemical changes can be regarded as the sum or difference of two or more other reactions. If we know the equilibrium constants of the individual processes, we can easily calculate that for the overall reaction according to the following rule.
Multi-step Equilibria Rule
The equilibrium constant for the sum of two or more reactions is the product of the equilibrium constants for each of the steps.
Example $2$
Calculate the value of $K$ for the reaction
$\ce{CaCO3(s) + H^{+}(aq) <=> Ca^{2+}(aq) + HCO^{–}3(aq)} \nonumber$
given the following equilibrium constants:
$CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}$
$K_1 = 10^{–6.3}$
$HCO^–_{3(aq)} \rightleftharpoons H^+_{(aq)} + CO^{2–}_{3(aq)}$
$K_2 = 10^{–10.3}$
Solution
The net reaction is the sum of reaction 1 and the reverse of reaction 2:
$CaCO_{3(s)} \rightleftharpoons Ca^{2+}_{(aq)} + CO^{2–}_{3(aq)}$
$K_1 = 10^{–6.3}$
$H^+_{(aq)} + CO^{2–}_{3(aq)} \rightleftharpoons HCO^–_{3(aq)}$
$K_{–2} = 10^{–(–10.3)}$
$CaCO_{3(s)} + H^+_{(aq)} \rightarrow Ca^{2+}_{(aq)} + HCO^–_{3(aq)}$
$K = \dfrac{K_1}{K_2} = 10^{(-6.4+10.3)} =10^{+3.9}$
Comment:
This net reaction describes the dissolution of limestone by acid; it is responsible for the eroding effect of acid rain on buildings and statues. This an example of a reaction that has practically no tendency to take place by itself (small K1) being "driven" by a second reaction having a large equilibrium constant (K–2). From the standpoint of the Le Chatelier principle, the first reaction is "pulled to the right" by the removal of carbonate by hydrogen ion. Coupled reactions of this type are widely encountered in all areas of chemistry, and especially in biochemistry, in which a dozen or so reactions may be linked.
Example $3$
The synthesis of $\ce{HBr}$ from hydrogen and liquid bromine has an equilibrium constant $K_p = 4.5 \times 10^{15}$ at 25°C. Given that the vapor pressure of liquid bromine is 0.28 atm, find $K_p$ for the homogeneous gas-phase reaction at the same temperature.
Solution
The net reaction we seek is the sum of the heterogeneous synthesis of $\ce{HBr}$ and the reverse of the vaporization of liquid bromine:
$H_{2(g)} + Br_{2(l)} \rightleftharpoons 2 HBr_{(g)}$ $K_p = 4.5\times 10^{15}$
$Br_{2(g)} \rightleftharpoons Br_{2(l)}$ $K_p = (0.28)^{–1}$
$H_{2(g)} + Br_{2(g)} \rightleftharpoons 2 HBr_{(g)}$ $K_p = 1.6 \times 10^{16}$
More on heterogeneous reactions
Heterogeneous reactions are those involving more than one phase. Some examples:
$Fe(s) + O_2(g) \rightleftharpoons FeO_2(s)$ air-oxidation of metallic iron (formation of rust)
$CaF_2(s) \rightleftharpoons Ca(aq) + F^+(aq)\_ dissolution of calcium fluoride in water \(H_2O(s) \rightleftharpoons H_2O(g)$ sublimation of ice (a phase change)
$NaHCO_3(s) + H^+(aq) \rightleftharpoons CO_2(g) + Na^+(aq) + H_2O(g)$
formation of carbon dioxide gas from sodium bicarbonate when water is added to baking powder (the hydrogen ions come from tartaric acid, the other component of baking powder.)
The vapor pressure of solid hydrates
A particularly interesting type of heterogeneous reaction is one in which a solid is in equilibrium with a gas. The sublimation of ice illustrated in the above table is a very common example. The equilibrium constant for this process is simply the partial pressure of water vapor in equilibrium with the solid— the vapor pressure of the ice.
Many common inorganic salts form solids which incorporate water molecules into their crystal structures. These water molecules are usually held rather loosely and can escape as water vapor. Copper(II) sulfate, for example forms a pentahydrate in which four of the water molecules are coordinated to the Cu2+ ion while the fifth is hydrogen-bonded to SO42–. This latter water is more tightly bound, so that the pentahydrate loses water in two stages on heating:
$\ce{CuSO4 \cdot 5H2O ->[140^oC] CuSO4 \cdot 5H2O ->[400^oC] CuSO4} \nonumber$
These dehydration steps are carried out at the temperatures indicated above, but at any temperature, some moisture can escape from a hydrate. For the complete dehydration of the pentahydrate we can define an equilibrium constant:
$\ce{CuSO4 \cdot 5H2O(s) <=> CuSO4(s) + 5 H2O(g)} \quad K_p = 1.14 \times 10^{10} \nonumber$
The vapor pressure of the hydrate (for this reaction) is the partial pressure of water vapor at which the two solids can coexist indefinitely; its value is $K_p$1/5 atm. If a hydrate is exposed to air in which the partial pressure of water vapor is less than its vapor pressure, the reaction will proceed to the right and the hydrate will lose moisture. Vapor pressures always increase with temperature, so any of these compounds can be dehydrated by heating.
Loss of water usually causes a breakdown in the structure of the crystal; this is commonly seen with sodium sulfate, whose vapor pressure is sufficiently large that it can exceed the partial pressure of water vapor in the air when the relative humidity is low. What one sees is that the well-formed crystals of the decahydrate undergo deterioration into a powdery form, a phenomenon known as efflorescence.
Table $3$:
name formula vapor pressure, torr
25°C 30°C
sodium sulfate decahydrate Na2SO4·10H2O 19.2 25.3
copper(II) sulfate pentahydrate CuSO4·5H2O 7.8 12.5
calcium chloride monohydrate CaCl2·H2O 3.1 5.1
(water) H2O 23.5 31.6
Example $4$
At what relative humidity will copper sulfate pentahydrate lose its waters of hydration when the air temperature is 30°C? What is $K_p$ for this process at this temperature?
Solution
From the Table $3$, we see that the vapor pressure of the hydrate is 12.5 torr, which corresponds to a relative humidity of 12.5/31.6 = 0.40 or 40%. This is the humidity that will be maintained if the hydrate is placed in a closed container of dry air
For this hydrate, $K_p = \sqrt{p_{H_2O)}}$, so the partial pressure of water vapor that will be in equilibrium with the hydrate and the dehydrated solid (remember that both solids must be present to have equilibrium!), expressed in atmospheres, will be
$\left(\dfrac{12.5}{760}\right)^5 = 1.20 \times 10^{-9}. \nonumber$
One of the first hydrates investigated in detail was calcium sulfate hemihydrate (CaSO4·½ H2O) which Le Chatelier (he of the “principle”) showed to be the hardened form of CaSO4 known as plaster of Paris. Anhydrous CaSO4 forms compact, powdery crystals, whereas the elongated crystals of the hemihydrate bind themselves into a cement-like mass that makes this material useful for making art objects, casts for immobilizing damaged limbs, and as a construction material (fireproofing, drywall). | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.04%3A_Equilibrium_Expressions.txt |
Learning Objectives
The five Examples presented above were carefully selected to span the range of problem types that students enrolled in first-year college chemistry courses are expected to be able to deal with. If you are able to reproduce these solutions on your own, you should be well prepared on this topic.
• The first step in the solution of all but the simplest equilibrium problems is to sketch out a table showing for each component the initial concentration or pressure, the change in this quantity (for example, +2x), and the equilibrium values (for example, .0036 + 2x). In doing so, the sequence of calculations required to get to the answer usually becomes apparent.
• Equilibrium calculations often involve quadratic- or higher-order equations. Because concentrations, pressures, and equilibrium constants are seldom known to a precision of more than a few significant figures, there is no need to seek exact solutions. Iterative approximations (as in Example $3$) or use of a graphical calculator (Example $4$) are adequate and convenient.
• Phase distribution equilibria play an important role in chemical separation processes on both laboratory and industrial scales. They are also involved in the movement of chemicals between different parts of the environment, and in the bioconcentration of pollutants in the food chain.
This page presents examples that cover most of the kinds of equilibrium problems you are likely to encounter in a first-year university course. Reading this page will not teach you how to work equilibrium problems! The only one who can teach you how to interpret, understand, and solve problems is yourself. So don't just "read" this and think you are finished. You need to find and solve similar problems on your own. Look over the problems in your homework assignment or at the end of the appropriate chapter in a textbook, and see how they fit into the general types described below. When you can solve them without looking at the examples below, you will be well on your way!
Calculating Equilibrium Constants
Clearly, if the concentrations or pressures of all the components of a reaction are known, then the value of $K$ can be found by simple substitution. Observing individual concentrations or partial pressures directly may be not always be practical, however. If one of the components is colored, the extent to which it absorbs light of an appropriate wavelength may serve as an index of its concentration. Pressure measurements are ordinarily able to measure only the total pressure of a gaseous mixture, so if two or more gaseous products are present in the equilibrium mixture, the partial pressure of one may need to be inferred from that of the other, taking into account the stoichiometry of the reaction.
Example $1$
In an experiment carried out by Taylor and Krist (J. Am. Chem. Soc. 1941: 1377), hydrogen iodide was found to be 22.3% dissociated at 730.8 K. Calculate $K_c$ for
$\ce{ 2 HI(g) <=> H2(g) + I2} \nonumber$
Solution
No explicit molar concentrations are given, but we do know that for every $n$ moles of $\ce{HI}$, $0.223n$ moles of each product is formed and $(1–0.223)n = 0.777n$ moles of $\ce{HI}$ remains. For simplicity, we assume that $n=1$ and that the reaction is carried out in a 1.00-L vessel, so that we can substitute the required concentration terms directly into the equilibrium expression for $K_c$.
\begin{align*} K_c &= \dfrac{[\ce{H2}][\ce{I2}]}{[\ce{HI}]^2} \[4pt] &= \dfrac{(0.223)(0.223)}{(0.777)^2} \[4pt] &= 0.082 \end{align*}
Example $2$: Evaluating the Equilibrium Constant
Ordinary white phosphorus, $\ce{P4}$, forms a vapor which dissociates into diatomic molecules at high temperatures:
$\ce{P4(g) <=> 2 P2(g)} \nonumber$
A sample of white phosphorus, when heated to 1000°C, formed a vapor having a total pressure of 0.20 atm and a density of 0.152 g L–1. Use this information to evaluate the equilibrium constant $K_p$ for this reaction.
Solution
Before worrying about what the density of the gas mixture has to do with $K_p$, start out in the usual way by laying out the information required to express $K_p$ in terms of an unknown $x$
ICE Table $P_{4(g)}$ $P_{2(g)}$ comment
Initial (moles) 1 0 Since $K$ is independent of the number of moles, assume the simplest initial case.
Change (moles) $-x$ $2x$ x is the fraction of P4 that dissociates.
Equilibrium (moles) $1-x$ $2x$ The denominator is the total number of moles:
Mole Fraction $\chi_{P_4}=\dfrac{1-x}{1+x}$ $\chi_{P_2}=\dfrac{2x}{1+x}$ The denominator is the total number of moles:
Equilibrium
(pressures)
$p_{P_4} = \chi_{P_4} 0.2 = \left( \dfrac{1-x}{1+x}\right) 0.2$ $p_{P_2} = \chi_{P_2} 0.2 = \left( \dfrac{2x}{1+x}\right) 0.2$ Partial pressure is the mole fraction times the total pressure.
The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure:
$P_i = \chi_i P_{tot} \nonumber$
with the term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium:
$P_{tot} = 1-x + 2x = 1+x \nonumber$
Expressing the equilibrium constant in terms of $x$ gives
\begin{align*} K_p &= \dfrac{p^2_{P_2}}{p_{P_4}} \[4pt] &= \dfrac{\left(\dfrac{2x}{1+x}\right)^2 0.2^2} {\left(\dfrac{1-x}{1+x}\right) 0.2} \[4pt] &= \left(\dfrac{4x^2}{(1-x)(1+x)}\right) 0.2 \[4pt] &= \left(\dfrac{4x^2}{1+x^2}\right) 0.2 \end{align*}
Now we need to find the dissociation fraction $x$ of $P_4$, and at this point we hope you remember those gas laws that you were told you would be needing later in the course! The density of a gas is directly proportional to its molecular weight, so you need to calculate the densities of pure $P_4$ and pure $P_2$ vapors under the conditions of the experiment. One of these densities will be greater than 0.152 gL–1 and the other will be smaller; all you need to do is to find where the measured density falls in between the two limits, and you will have the dissociation fraction.
The molecular weight of phosphorus is 31.97, giving a molar mass of 127.9 g for $\ce{P4}$. This mass must be divided by the volume to find the density; assuming ideal gas behavior, the volume of 127.9 g (1 mole) of $\ce{P4}$ is given by RT/P, which works out to 522 L (remember to use the absolute temperature here.) The density of pure $\ce{P4}$ vapor under the conditions of the experiment is then
$\rho = \dfrac{m}{V} = (128\; g \;mol^{–1}) \times x = (522\; L mol^{–1}) = 0.245\; g\; L^{–1} \nonumber$
The density of pure $P_2$ would be half this, or 0.122 g L–1. The difference between these two limiting densities is 0.123 g L–1, and the difference between the density of pure $\ce{P4}$ and that of the equilibrium mixture is (0.245 – 0.152) g L–1 or 0.093 g L–1. The ratio 0.093/0.123 = 0.76 is therefore the fraction of $\ce{P4}$ that remains and its fractional dissociation is (1 – 0.76) = 0.24. Substituting into the equilibrium expression above gives $K_p = 1.2$.
Exercise $2$
Solve Example $2$ by using a different set of initial conditions to demonstrated that the initial conditions indeed have no effect on determining the Equilibrium state and $K_p$.
Calculating Equilibrium Concentrations
This is by far the most common kind of equilibrium problem you will encounter: starting with an arbitrary number of moles of each component, how many moles of each will be present when the system comes to equilibrium? The principal source of confusion and error for beginners relates to the need to determine the values of several unknowns (a concentration or pressure for each component) from a single equation, the equilibrium expression. The key to this is to make use of the stoichiometric relationships between the various components, which usually allow us to express the equilibrium composition in terms of a single variable. The easiest and most error-free way of doing this is adopt a systematic approach in which you create and fill in a small table as shown in the following problem example. You then substitute the equilibrium values into the equilibrium constant expression, and solve it for the unknown.
This very often involves solving a quadratic or higher-order equation. Quadratics can of course be solved by using the familiar quadratic formula, but it is often easier to use an algebraic or graphical approximation, and for higher-order equations this is the only practical approach. There is almost never any need to get an exact answer, since the equilibrium constants you start with are rarely known all that precisely anyway.
Example $3$
Phosgene ($\ce{COCl2}$) is a poisonous gas that dissociates at high temperature into two other poisonous gases, carbon monoxide and chlorine. The equilibrium constant Kp = 0.0041 at 600°K. Find the equilibrium composition of the system after 0.124 atm of $\ce{COCl2}$ is allowed to reach equilibrium at this temperature.
Solution
First we need a balanced chemical reaction
$\ce{COCl_2 <=> CO(g) + Cl2(g)} \nonumber$
Start by drawing up a table showing the relationships between the components:
ICE Table $COCl_2$ $CO_{(g)}$ $Cl_{2(g)}$
Initial (pressure) 0.124 atm 0 0
Change (pressure) -x +x +x
Equilibrium (pressure) 0.124 – x +x +x
Substitution of the equilibrium pressures into the equilibrium expression gives
$\dfrac{x^2}{0.124 - x} = 0.0041 \nonumber$
This expression can be rearranged into standard polynomial form
$x^2 +0.0041 x – 0.00054 = 0 \nonumber$
and solved by the quadratic formula, but we will simply obtain an approximate solution by iteration. Because the equilibrium constant is small, we know that x will be rather small compared to 0.124, so the above relation can be approximated by
$\dfrac{x^2}{0.124 - x} \approx \dfrac{x^2}{0.124}= 0.0041 \nonumber$
which gives x = 0.0225. To see how good this is, substitute this value of x into the denominator of the original equation and solve again:
$\dfrac{x^2}{0.124 - 0.0225} = \dfrac{x^2}{0.102}= 0.0041 \nonumber$
This time, solving for $x$ gives 0.0204. Iterating once more, we get
$\dfrac{x^2}{0.124 - 0.0204} = \dfrac{x^2}{0.104}= 0.0041 \nonumber$
and x = 0.0206 which is sufficiently close to the previous to be considered the final result. The final partial pressures are then 0.104 atm for COCl2, and 0.0206 atm each for CO and Cl2.
Comment: Using the quadratic formula to find the exact solution yields the two roots –0.0247 (which we ignore) and 0.0206, which show that our approximation is quite good.
Example $4$
The gas-phase dissociation of phosphorus pentachloride to the trichloride has $K_p = 3.60$ at 540°C:
$\ce{PCl5(g) <=> PCl3(g) + Cl2(g)} \nonumber$
What will be the partial pressures of all three components if 0.200 mole of $\ce{PCl5}$ and 3.00 moles of $\ce{PCl3}$ are combined and brought to equilibrium at this temperature and at a total pressure of 1.00 atm?
Solution
As always, set up a table showing what you know (first two rows) and then expressing the equilibrium quantities:
ICE Table $\ce{PC5(g)}$ $\ce{PCl3(g)}$ $\ce{Cl2(g)}$
Initial (moles) 0.200 3.00 0
Change (moles) –x +x +x
Equilibrium (moles) 0.200 – x 3.00 + x x
Equilibrium (partial pressures) $\dfrac{2.00 - x }{ 3.20 + x}$ $\dfrac{3.00 + x }{ 3.20 + x}$ $\dfrac{ x }{ 3.20 + x}$
The partial pressures in the bottom row were found by multiplying the mole fraction of each gas by the total pressure:
$P_i = \chi_i P_{tot} \nonumber$
with the term in the denominator of each mole fraction is the total number of moles of gas present at equilibrium:
$P_{tot} = (0.200 – x) + (3.00 + x) + x = 3.20 + x \nonumber$
Substituting the equilibrium partial pressures into the equilibrium expression, we have
$\dfrac{ (3.00 +x)(x)}{(0.200 -x)(3.20 +x)} = 3.60 \nonumber$
whose polynomial form is
$4.60x^2 + 13.80x – 2.304 = 0. \nonumber$
You can use the quadratic question to solve this or you can do it graphically (more useful for higher order equations). Plotting this on a graphical calculator yields $x = 0.159$ as the positive root:
Substitution of this root into the expressions for the equilibrium partial pressures in the table yields the following values:
• $P_{\ce{PCl5}}$ = 0.012 atm,
• $P_{\ce{PCl3}}$ = 0.94 atm,
• $P_{\ce{Cl2}}$ = 0.047 atm.
Effects of Dilution on Equilibrium
In the section that introduced the Le Chatelier principle, it was mentioned that diluting a weak acid such as acetic acid $\ce{CH3COOH}$ (“$\ce{HAc}$”) will shift the dissociation equilibrium to the right:
$\ce{HAc + H_2O \rightleftharpoons H_3O^{+} + Ac^{–}} \nonumber$
Thus a $0.10\,M$ solution of acetic acid is 1.3% ionized, while in a 0.01 M solution, 4.3% of the $\ce{HAc}$ molecules will be dissociated. This is because as the solution becomes more dilute, the product [H3O+][Ac] decreases more rapidly than does the $\ce{[HAc]}$ term. At the same time the concentration of $\ce{H2O}$ becomes greater, but because it is so large to start with (about 55.5 M), any effect this might have is negligible, which is why no $\ce{[H2O]}$ term appears in the equilibrium expression.
For a reaction such as
$\ce{CH_3COOH (l) + C_2H_5OH (l) \rightleftharpoons CH_3COOC_2H_5 (l) + H_2O (l) }$
(in which the water concentration does change), dilution will have no effect on the equilibrium; the situation is analogous to the way the pressure dependence of a gas-phase reaction depends on the number of moles of gaseous components on either side of the equation (i.e., $\Delta n_g$).
Example $5$
The biochemical formation of a disaccharide (double) sugar from two monosaccharides is exemplified by the reaction
$\text{fructose} (aq) + \text{glucose-6-phosphate} (aq) → \text{sucrose-6-phosphate} (aq) + \ce{H2O} (l)$
(Sucrose is ordinary table sugar.) To what volume should a solution containing 0.050 mol of each monosaccharide be diluted in order to bring 5% conversion to sucrose phosphate? The equilibrium constant for this reaction is $K_{c} = 7.1 \times 10^{-6}$ at room temperature.
Solution
The initial and final numbers of moles in this equation are as follows:
ICE Table fructose ($\text{fruc}$) glucose-6-P ($\text{gluc6P}$) sucrose-6-P ($\text{suc6P}$) water $\ce{H2O}$)
Initial (moles) 0.05 0.05 0 -
Change (moles) -x -x +x -
Equilibrium (moles) 0.05-x 0.05-x +x -
What is the value of $x$? That is when 5% of the reaction has proceeded or when 5% of the fructose (or glucose-6-P) is consumed:
$\dfrac{x}{0.05} = 0.05 \nonumber$
so $x = 0.0025 \nonumber$. The equilibrated concentrations are then
• $[\text{suc6P}]_{equil} = \dfrac{0.0025}{V}$
• $[\text{fruc}]_{equil} = \dfrac{0.0475}{V}$
• $[\text{glu6P}]_{equil} = \dfrac{0.0475}{V}$
Substituting into the values in for the expression of $K_c$ (in which the solution volume is the unknown), we have
\begin{align*} K_{c} &= \dfrac{[\text{suc6P}]_{equil} }{[\text{fruc}]_{equil} [\text{gluc6P}]_{equil} } \[4pt] &= \dfrac{\left(\dfrac{0.0025}{V}\right)}{\left(\dfrac{0.0475}{V}\right)^2} = 7.1 \times 10^{-6} \end{align*}
$V = (7.1 \times 10^{-6}) \dfrac{(0.0475)^2}{0.0025} \nonumber$
Solving for $V$ gives a final solution volume of $6.4 \times 10^{-4}\,L$ or $640 \mu L$. Why so small? The reaction is not favored and to push it forward, large concentrations of reactants are needed (Le Chatelier principle in action). | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.05%3A_Equilibrium_Calculations.txt |
Learning Objectives
• Phase distribution equilibria play an important role in chemical separation processes on both laboratory and industrial scales. They are also involved in the movement of chemicals between different parts of the environment, and in the bioconcentration of pollutants in the food chain.
It often happens that two immiscible liquid phases are in contact, one of which contains a solute. How will the solute tend to distribute itself between the two phases? One’s first thought might be that some of the solute will migrate from one phase into the other until it is distributed equally between the two phases, since this would correspond to the maximum dispersion (randomness) of the solute. This, however, does not take into the account the differing solubilities the solute might have in the two liquids; if such a difference does exist, the solute will preferentially migrate into the phase in which it is more soluble.
For a solute $S$ distributed between two phases a and b the process Sa = Sbis defined by the distribution law
$K_{a,b} = \dfrac{[S]_a}{[S]_b}$
in which
• $K_{a,b}$ is the distribution ratio (also called the distribution coefficient) and
• $[S]_i$ is the solubility of the solute in the phase.
biomagnification
The transport of substances between different phases is of immense importance in such diverse fields as pharmacology and environmental science. For example, if a drug is to pass from the aqueous phase with the stomach into the bloodstream, it must pass through the lipid (oil-like) phase of the epithelial cells that line the digestive tract. Similarly, a pollutant such as a pesticide residue that is more soluble in oil than in water will be preferentially taken up and retained by marine organism, especially fish, whose bodies contain more oil-like substances; this is basically the mechanism whereby such residues as DDT can undergo biomagnification as they become more concentrated at higher levels within the food chain. For this reason, environmental regulations now require that oil-water distribution ratios be established for any new chemical likely to find its way into natural waters. The standard “oil” phase that is almost universally used is octanol, C8H17OH.
In preparative chemistry it is frequently necessary to recover a desired product present in a reaction mixture by extracting it into another liquid in which it is more soluble than the unwanted substances. On the laboratory scale this operation is carried out in a separatory funnel as shown below. The two immiscible liquids are poured into the funnel through the opening at the top. The funnel is then shaken to bring the two phases into intimate contact, and then set aside to allow the two liquids to separate into layers, which are then separated by allowing the more dense liquid to exit through the stopcock at the bottom.
If the distribution ratio is too low to achieve efficient separation in a single step, it can be repeated; there are automated devices that can carry out hundreds of successive extractions, each yielding a product of higher purity. In these applications our goal is to exploit the Le Chatelier principle by repeatedly upsetting the phase distribution equilibrium that would result if two phases were to remain in permanent contact.
Video $1$: How to perform a liquid-liquid extraction using a separating funnel.
Example $1$
The distribution ratio for iodine between water and carbon disulfide is 650. Calculate the concentration of I2 remaining in the aqueous phase after 50.0 mL of 0.10M I2 in water is shaken with 10.0 mL of CS2.
Solution
The equilibrium constant is
$K_d = \dfrac{C_{CS_2}}{C_{H_2O}} = 650 \nonumber$
mmol, so m2 = (5.00 – m1) mmol and we now have only the single unknown m1. The equilibrium constant then becomes
$((5.00 – m_1) mmol / 10 mL) ÷ (m_1 mmol / 50 mL) = 650 \nonumber$
Simplifying and solving for m1 yields
$\dfrac{(0.50 – 0.1)m_1}{(0.02 m_1} = 650 \nonumber$
with m1 = 0.0382 mmol.
The concentration of solute in the water layer is (0.0382 mmol) / (50 mL) = 0.000763 M, showing that almost all of the iodine has moved into the CS2 layer. | textbooks/chem/General_Chemistry/Chem1_(Lower)/11%3A_Chemical_Equilibrium/11.06%3A_Phase_Distribution_Equilibria.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Discuss the roles of lattice- and hydration energy in determining the solubility of a salt in water.
• Explain what a qualitative analysis separation scheme is, and how it works.
• Write the solubility product expression for a salt, given its formula.
• Explain the distinction between an ion product and a solubility product.
• Given the formula of a salt and its Ks value, calculate the molar solubility.
• Explain the Le Chatelier principle leads to the common ion effect.
• Explain why a strong acid such as HCl will dissolve a sparingly soluble salt of a weak acid, but not a salt of a strong acid.
• Describe what happens (and why) when aqueous ammonia is slowly added to a solution of silver nitrate
Dissolution of a salt in water is a chemical process that is governed by the same laws of chemical equilibrium that apply to any other reaction. There are, however, a number of special aspects of of these equilibria that set them somewhat apart from the more general ones that are covered in the lesson set devoted specifically to chemical equilibrium. These include such topics as the common ion effect, the influence of pH on solubility, supersaturation, and some special characteristics of particularly important solubility systems.
Solubility: the dissolution of salts in water
Drop some ordinary table salt into a glass of water, and watch it "disappear". We refer to this as dissolution, and we explain it as a process in which the sodium and chlorine units break away from the crystal surface, get surrounded by H2O molecules, and become hydrated ions.
$NaCl_{(s)} \rightarrow Na^+_{(aq)}+ Cl^–_{(aq)}$
The designation (aq) means "aqueous" and comes from aqua, the Latin word for water. It is used whenever we want to emphasize that the ions are hydrated — that H2O molecules are attached to them.
Remember that solubility equilibrium and the calculations that relate to it are only meaningful when both sides (solids and dissolved ions) are simultaneously present. But if you keep adding salt, there will come a point at which it no longer seems to dissolve. If this condition persists, we say that the salt has reached its solubility limit, and the solution is saturated in NaCl. The situation is now described by
$NaCl_{(s)} \rightleftharpoons Na^+_{(aq)}+ Cl^–_{(aq)}$
in which the solid and its ions are in equilibrium.
Salt solutions that have reached or exceeded their solubility limits (usually 36-39 g per 100 mL of water) are responsible for prominent features of the earth's geochemistry. They typically form when NaCl leaches from soils into waters that flow into salt lakes in arid regions that have no natural outlets; subsequent evaporation of these brines force the above equilibrium to the left, forming natural salt deposits. These are often admixed with other salts, but in some cases are almost pure NaCl. Many parts of the world contain buried deposits of NaCl (known as halite) that formed from the evaporation of ancient seas, and which are now mined.
Solubilities are most fundamentally expressed in molar (mol L–1 of solution) or molal (mol kg–1 of water) units. But for practical use in preparing stock solutions, chemistry handbooks usually express solubilities in terms of grams-per-100 ml of water at a given temperature, frequently noting the latter in a superscript. Thus 6.9 20 means 6.9 g of solute will dissolve in 100 mL of water at 20° C. When quantitative data are lacking, the designations "soluble", "insoluble", "slightly soluble", and "highly soluble" are used. There is no agreed-on standard for these classifications, but a useful guideline might be that shown below.
The solubilities of salts in water span a remarkably large range of values, from almost completely insoluble to highly soluble. Moreover, there is no simple way of predicting these values, or even of explaining the trends that are observed for the solubilities of different anions within a given group of the periodic table.
Ultimately, the driving force for dissolution (and for all chemical processes) is determined by the Gibbs free energy change. But because many courses cover solubility before introducing free energy, we will not pursue this here. Dissolution of a salt is conceptually understood as a sequence of the two processes depicted above:
1. breakup of the ionic lattice of the solid,
2. followed by attachment of water molecules to the released ions.
The first step consumes a large quantity of energy, something that by itself would strongly discourage solubility. But the second step releases a large amount of energy and thus has the opposite effect. Thus the net energy change depends on the sum of two large energy terms (often approaching 1000 kJ/mol) having opposite signs. Each of these terms will to some extent be influenced by the size, charge, and polarizability of the particular ions involved, and on the lattice structure of the solid. This large number of variables makes it impossible to predict the solubility of a given salt. Nevertheless, there are some clear trends for how the solubilities of a series of salts of a given anion (such as hydroxides, sulfates, etc.) change with a periodic table group. And of course, there are a number of general solubility rules — for example, that all nitrates are soluble, while most sulfides are insoluble.
Solubility and temperature
Solubility usually increases with temperature - but not always. This is very apparent from the solubility-vs.-temperature plots shown here. (Some of the plots are colored differently in order to make it easier to distinguish them where they crowd together.) The temperature dependence of any process depends on its entropy change — that is, on the degree to which thermal kinetic energy can spread throughout the system. When a solid dissolves, its component molecules or ions diffuse into the much greater volume of the solution, carrying their thermal energy along with them. So we would normally expect the entropy to increase — something that makes any process take place to a greater extent at a higher temperature.
So why does the solubility of cerium sulfate (green plot) diminish with temperature? Dispersal of the Ce3+ and SO42– ions themselves is still associated with an entropy increase, but in this case the entropy of the water decreases even more owing to the ordering of the H2O molecules that attach to the Ce3+ ions as they become hydrated. It's difficult to predict these effects, or explain why they occur in individual cases — but they do happen.
The Importance of Sparingly Soluble Solids
All solids that dissociate into ions exhibit some limit to their solubilities, but those whose saturated solutions exceed about 0.01 mol L–1 cannot be treated by simple equilibrium constants owing to ion-pair formation that greatly complicates their behavior. For this reason, most of what follows in this lesson is limited to salts that fall into the "sparingly soluble" category. The importance of sparingly soluble solids arises from the fact that formation of such a product can effectively remove the corresponding ions from the solution, thus driving the reaction to the right. Consider, for example, what happens when we mix solutions of strontium nitrate and potassium chloride in a 1:2 mole ratio. Although we might represent this process by
$Sr(NO_3)_{2(aq)}+ 2 KCl_{(aq)}→ SrCl_{(aq)}+ 2 KNO_{3(aq)} \label{1}$
the net ionic equation
$Sr^{2+} + 2 NO_3^– + 2 K^+ + 2 Cl^– → Sr^{2+} + 2 NO_3^– + 2 K^+ + 2 Cl^–$
indicates that no net change at all has taken place! Of course if the solution were than evaporated to dryness, we would end up with a mixture of the four salts shown in Equation $\ref{1}$, so in this case we might say that the reaction is half-complete. Contrast this with what happens if we combine equimolar solutions of barium chloride and sodium sulfate:
$BaCl_{2(aq)}+ Na_2SO_{4(aq)}→ 2 NaCl_{(aq)}+ BaSO_{4(s)} \label{2}$
whose net ionic equation is
$Ba^{2+} + \cancel{ 2 Cl^–} + \cancel{2 Na^+} + SO_4^{2–} → \cancel{2 Na^+} + \cancel{2 Cl^–} + BaSO_{4(s)}$
which after canceling out like terms on both sides, becomes simply
$Ba^{2+} + SO_4^{2– }→ BaSO_{4(s)} \label{3}$
Because the formation of sparingly soluble solids is "complete" (that is, equilibria such as the one shown above for barium sulfate lie so far to the right), virtually all of one or both of the contributing ions are essentially removed from the solution. Such reactions are said to be quantitative, and they are especially important in analytical chemistry:
• Qualitative analysis: This most commonly refers to a procedural scheme, widely encountered in first-year laboratory courses, in which a mixture of cations (usually in the form of their dissolved nitrate salts) is systematically separated and identified on the basis of the solubilities of their various anion salts such as chlorides, carbonates, sulfates, and sulfides. Although this form of qualitative analysis is no longer employed by modern-day chemists (instrumental techniques such as atomic absorption spectroscopy are much faster and comprehensive), it is still valued as an educational tool for familiarizing students with some of the major classes of inorganic salts, and for developing basic skills relating to observing, organizing, and interpreting results in the laboratory.
• Quantitative gravimetric analysis: In this classical form of chemical analysis, an insoluble salt of a cation is prepared by precipitating it by addition of a suitable anion. The precipitate is then collected, dried, and weighed ("gravimetry") in order to determine the concentration of the cation in the sample. For example, a gravimetric procedure for determining the quantity of barium in a sample might involve precipitating the metal as the sulfate according to Equation $\ref{3}$ above, using an excess of sulfate ion to ensure complete removal of the barium. This method of quantitative analysis became extremely important in the latter half of the nineteenth century, by which time reasonably accurate atomic weights had become available, and sensitive analytical balances had been developed. It was not until the 1960's that it became largely supplanted by instrumental techniques which were much quicker and accurate. Gravimetric analysis is still usually included as a part of more advanced laboratory instruction, largely as a means of developing careful laboratory technique.
Figure $1$: The standard qualitative analysis separation scheme. Steps in a Typical Qualitative Analysis Scheme for a Solution That Contains Several Metal Ions
Solubility Products and Equilibria
Some salts and similar compounds (such as some metal hydroxides) dissociate completely when they dissolve, but the extent to which they dissolve is so limited that the resulting solutions exhibit only very weak conductivities. In these salts, which otherwise act as strong electrolytes, we can treat the dissolution-dissociation process as a true equilibrium. Although this seems almost trivial now, this discovery, made in 1900 by Walther Nernst who applied the Law of Mass Action to the dissociation scheme of Arrhenius, is considered one of the major steps in the development of our understanding of ionic solutions.
Using silver chromate as an example, we express its dissolution in water as
$Ag_2CrO_{4(s)} \rightarrow 2 Ag^+_{(aq)}+ CrO^{2–}_{4(aq)} \label{4a}$
When this process reaches equilibrium (which requires that some solid be present), we can write (leaving out the "(aq)s" for simplicity)
$Ag_2CrO_{4(s)} \rightleftharpoons 2 Ag^+ + CrO^{2–}_{4} \label{4b}$
The equilibrium constant is formally
$K = \dfrac{[Ag^+]^2[CrO_4^{2–}]}{[Ag_2CrO_{4(s)}]} = [Ag^+]^2[CrO_4^{2–}] \label{5a}$
But because solid substances do not normally appear in equilibrium expressions, the equilibrium constant for this process is
$[Ag^+]^2 [CrO_4^{2–}] = K_s = 2.76 \times 10^{–12} \label{5b}$
Because equilibrium constants of this kind are written as products, the resulting K's are commonly known as solubility products, denoted by $K_s$ or $K_{sp}$.
Strictly speaking, concentration units do not appear in equilibrium constant expressions. However, many instructors prefer that students show them anyway, especially when using solubility products to calculate concentrations. If this is done, $K_s$ in Equation $\ref{5b}$ would have units of mol3 L–3.
Equilibrium and non-equilibrium in solubility systems
An expression such as [Ag+]2 [CrO42–] in known generally as an ion product — this one being the ion product for silver chromate. An ion product can in principle have any positive value, depending on the concentrations of the ions involved. Only in the special case when its value is identical with Ks does it become the solubility product. A solution in which this is the case is said to be saturated. Thus when
$[Ag^+]^2 [CrO_4^{2–}] = 2.76 \times 10^{-12}$
at the temperature and pressure at which this value $K_s$ of applies, we say that the "solution is saturated in silver chromate".
A solution must be saturated to be in equilibrium with the solid. This is a necessary condition for solubility equilibrium, but it is not by itself sufficient. True chemical equilibrium can only occur when all components are simultaneously present. A solubility system can be in equilibrium only when some of the solid is in contact with a saturated solution of its ions. Failure to appreciate this is a very common cause of errors in solving solubility problems.
Undersaturated and supersaturated solutions
If the ion product is smaller than the solubility product, the system is not in equilibrium and no solid can be present. Such a solution is said to be undersaturated. A supersaturated solution is one in which the ion product exceeds the solubility product. A supersaturated solution is not at equilibrium, and no solid can ordinarily be present in such a solution. If some of the solid is added, the excess ions precipitate out and until solubility equilibrium is achieved.
How to know the saturation status of a solution
This is just a simple matter of comparing the ion product $Q_s$ with the solubility product $K_s$. So for the system
$Ag_2CrO_{4(s)} \rightleftharpoons 2 Ag^+ + CrO_4^{2–} \label{4ba}$
a solution in which $Q_s < K_s$ (i.e., $K_s /Q_s > 1$) is undersaturated (blue shading) and the no solid will be present. The combinations of [Ag+] and [CrO42–] that correspond to a saturated solution (and thus to equilibrium) are limited to those described by the curved line. The pink area to the right of this curve represents a supersaturated solution.
Example $1$
A sample of groundwater that has percolated through a layer of gypsum (CaSO4, Ks = 4.9E–5 = 10–4.3) is found to have be 8.4E–5 M in Ca2+ and 7.2E–5 M in SO42–. What is the equilibrium state of this solution with respect to gypsum?
Solution
The ion product
$Q_s = (8.4 \times 10^{–5})(7.2 \times 10^{-5}) = 6.0 \times 10^{–4}$
exceeds $K_s$, so the ratio Ks /Qs > 1 and the solution is supersaturated in CaSO4.
How are solubilities determined?
There are two principal methods, neither of which is all that reliable for sparingly soluble salts:
• Evaporate a saturated solution of the solid to dryness, and weigh what's left.
• Measure the electrical conductivity of the saturated solution, which will be proportional to the concentrations of the ions.
How Solubilities relate to solubility products
The solubility (by which we usually mean the molar solubility) of a solid is expressed as the concentration of the "dissolved solid" in a saturated solution. In the case of a simple 1:1 solid such as AgCl, this would just be the concentration of Ag+ or Cl in the saturated solution. However, for a more complicated stoichiometry such as as silver chromate, the solubility would be only one-half of the Ag+ concentration.
For example, let us denote the solubility of Ag2CrO4 as $S$ mol L–1. Then for a saturated solution, we have
• $[Ag^+] = 2S$
• $[CrO_4^{2–}] = S$
Substituting this into Equation $\ref{5b}$ above,
$(2S)^2 (S) = 4S^3 = 2.76 \times 10^{–12}$
$S= \left( \dfrac{K_s}{4} \right)^{1/3} = (6.9 \times 10^{-13})^{1/3} = 0.88 \times 10^{-4} \label{6a}$
thus the solubility is $8.8 \times 10^{–5}\; M$.
Note that the relation between the solubility and the solubility product constant depends on the stoichiometry of the dissolution reaction. For this reason it is meaningless to compare the solubilities of two salts having the formulas $A_2B$ and $AB_2$, on the basis of their $K_s$ values.
It is meaningless to compare the solubilities of two salts having different formulas on the basis of their $K_s$ values.
Example $2$
under these conditions.
Solution
moles of solute in 100 mL; S = 0.0016 g / 78.1 g/mol = 2.05E-5 mol
S = 2.05E–5 mol/0.100 L = 2.05E-4 M
Ks = [Ca2+][F]2 = (S)(2S)2 = 4 × (2.05E–4)3 = 3.44E–11
Example $3$
Estimate the solubility of La(IO3)3 and calculate the concentration of iodate in equilibrium with solid lanthanum iodate, for which Ks = 6.2 × 10–12.
Solution
The equation for the dissolution is
$La(IO_3)_3 \rightleftharpoons La^{3+ }+ 3 IO_3^–$
If the solubility is S, then the equilibrium concentrations of the ions will be
[La3+] = S and [IO3] = 3S. Then Ks = [La3+][IO3]3 = S(3S)3 = 27S4
27S4 = 6.2 × 10–12, S = ( ( 6.2 ÷ 27) × 10–12 )¼ = 6.92 × 10–4 M
[IO3] = 3S = 2.08 × 10–5 (M)
Example $4$ : Cadmium
Cadmium is a highly toxic environmental pollutant that enters wastewaters associated with zinc smelting (Cd and Zn commonly occur together in ZnS ores) and in some electroplating processes. One way of controlling cadmium in effluent streams is to add sodium hydroxide, which precipitates insoluble Cd(OH)2 (Ks = 2.5E–14). If 1000 L of a certain wastewater contains Cd2+ at a concentration of 1.6E–5 M, what concentration of Cd2+ would remain after addition of 10 L of 4 M NaOH solution?
Solution
As with most real-world problems, this is best approached as a series of smaller problems, making simplifying approximations as appropriate.
Volume of treated water: 1000 L + 10 L = 1010 L
Concentration of OH on addition to 1000 L of pure water:
(4 M) × (10 L)/(1010 L) = .040 M
Initial concentration of Cd2+ in 1010 L of water:
(1.6E–5 M) x (100/101) ≈ 1.6E–5 M
The easiest way to tackle this is to start by assuming that a stoichiometric quantity of Cd(OH)2 is formed — that is, all of the Cd2+ gets precipitated.
Concentrations [Cd2+], M [OH], M
initial 1.6E–5 0.04
change –1.6E–5 –3.2E–5
final: 0 0.04 – 3.2E–5 ≈ .04
Now "turn on the equilibrium" — find the concentration of Cd2+ that can exist in a 0.04M OH solution:
Concentrations [Cd2+], M [OH], M
initial o 0.04
change +x +2x
at equilibrium x 0.04 + 2x.04
Substitute these values into the solubility product expression:
$Cd(OH)_{2(s) } = [Cd^{2+}] [OH^–]^2 = 2.5 \times 10^{–14}$
$[Cd^{2+}] = \dfrac{2.5 \times 10^{–14}}{ 16 \times 10^{–4}} = 1.6 \times 10^{–13}\; M$
Note that the effluent will now be very alkaline:
$pH = 14 + \log 0.04 = 12.6$
so in order to meet environmental standards an equivalent quantity of strong acid must be added to neutralize the water before it is released.
All Just a Simplification of Reality
The simple relations between Ks and molar solubility outlined above, and the calculation examples given here, cannot be relied upon to give correct answers. Some of the reasons for this are explained in Part 2 of this lesson, and have mainly to do with incomplete dissociation of many salts and with complex formation in the presence of anions such as Cl and OH. The situation is nicely described in the article What Should We Teach Beginners about Solubility and Solubility Products? by Stephen Hawkes (J Chem Educ.1998 75(9) 1179-81). See also the earlier article by Meites, Pode and Thomas Are Solubilities and Solubility Products Related?(J Chem Educ. 1966 43(12) 667-72).
It turns out that solubility equilibria more often than not involve many competing processes and their rigorous treatment can be quite complicated. Nevertheless, it is important that students master these over-simplified examples. However, it is also important that they are not taken too seriously!
The Common Ion Effect
It has long been known that the solubility of a sparingly soluble ionic substance is markedly decreased in a solution of another ionic compound when the two substances have an ion in common. This is just what would be expected on the basis of the Le Chatelier Principle; whenever the process
$CaF_{2(s)} \rightleftharpoons Ca^{2+} + 2 F^– \label{7}$
is in equilibrium, addition of more fluoride ion (in the form of highly soluble NaF) will shift the composition to the left, reducing the concentration of Ca2+, and thus effectively reducing the solubility of the solid. We can express this quantitatively by noting that the solubility product expression
$[Ca^{2+}][F^–]^2 = 1.7 \times 10^{–10} \label{8}$
must always hold, even if some of the ionic species involved come from sources other than CaF2(s). For example, if some quantity x of fluoride ion is added to a solution initially in equilibrium with solid CaF2, we have
• $[Ca^{2+}] = S$
• $[F^–] = 2S + x$
so that
$K_s = [Ca^{2+}][ F^–]^2 = S (2S + x)^2 . \label{9a}$
University-level students should be able to derive these relations for ion-derived solids of any stoichiometry. In most practical cases x will be large compared to S so that the 2S term can be dropped and the relation becomes
$K_s ≈ S x^2$
or
$S ≈ \dfrac{K_s}{x^2} \label{9b}$
University-level students should be able to derive these relations for ion-derived solids of any stoichiometry.
The plots shown below illustrate the common ion effect for silver chromate as the chromate ion concentration is increased by addition of a soluble chromate such as $Na_2CrO_4$.
What's different about the plot on the right? If you look carefully at the scales, you will see that this one is plotted logarithmically (that is, in powers of 10.) Notice how a much wider a range of values can display on a logarithmic plot. The point of showing this pair of plots is to illustrate the great utility of log-concentration plots in equilibrium calculations in which simple approximations (such as that made in Equation $\ref{9b}$ can yield straight-lines within the range of values for which the approximation is valid.
Example $5$
Calculate the solubility of strontium sulfate (Ks = 2.8 × 10–7) in
1. pure water and
2. in a 0.10 mol L–1 solution of $Na_2SO_4$.
= [Sr2+][SO42–] = S2
$S = \sqrt{K_s} = \sqrt{ 2.8 \times 10^{–7} } = 5.3 \times 10^{–4}$
(b) In 0.10 mol L–1 Na2SO4, we have
= [Sr2+][SO42–] = S × (0.10 + S) = 2.8 × 10–7
Because S is negligible compared to 0.10 M, we make the approximation
= [Sr2+][SO42–] ≈ S × (0.10 M) = 2.8 × 10–7
so
This is roughly 100 times smaller than the result from (a).
Selective Precipitation and Separations
Differences in solubility are widely used to selectively remove one species from a solution containing several kinds of ions.
Example $6$
The solubility products of AgCl and Ag2CrO4 are 1.8E–10 and 2.0E–12, respectively. Suppose that a dilute solution of AgNO3 is added dropwise to a solution containing 0.001M Cl and 0.01M CrO42–.
1. Which solid, AgCl or Ag2CrO4, will precipitate first?
2. What fraction of the first anion will have been removed when the second just begins to precipitate? Neglect any volume changes.
Solution
The silver ion concentrations required to precipitate the two salts are found by substituting into the appropriate solubility product expressions:
• to precipitate AgCl: [Ag+] = 1.8E-10 / .001 = 1.8E-7 M
• to precipitate Ag2CrO4: [Ag+] = (2.0E-12 / .01)½ = 1.4E–5 M
The first solid to form as the concentration of Ag+ increases will be AgCl. Eventually the Ag+ concentration reaches 1.4E-5 M and Ag2CrO4 begins to precipitate. At this point the concentration of chloride ion in the solution will be 1.3E-5 M which is about 13% of the amount originally present.
The preceding example is the basis of the Mohr titration of chloride by Ag+, commonly done to determine the salinity of water samples. The equivalence point of this precipitation titration occurs when no more AgCl is formed, but there is no way of observing this directly in the presence of the white AgCl which is suspended in the container. Before the titration is begun, a small amount of K2CrO4 is added to the solution. Ag2CrO4 is red-orange in color, so its formation, which signals the approximate end of AgCl precipitation, can be detected visually.
Competing Equilibria involving solids
Solubility expression are probably the exception rather than the rule. Such equilibria are often in competition with other reactions with such species as H+or OH, complexing agents, oxidation-reduction, formation of other sparingly soluble species or, in the case of carbonates and sulfites, of gaseous products. The exact treatments of these systems can be extremely complicated, involving the solution of large sets of simultaneous equations. For most practical purposes it is sufficient to recognize the general trends, and to carry out approximate calculations.
Salts of weak acids are soluble in strong acids, but strong acids will not dissolve salts of strong acids
The solubility of a sparingly soluble salt of a weak acid or base will depend on the pH of the solution. To understand the reason for this, consider a hypothetical salt MA which dissolves to form a cation M+ and an anion Awhich is also the conjugate base of a weak acid HA. The fact that the acid is weak means that hydrogen ions (always present in aqueous solutions) and M+ cations will both be competing for the A:
The weaker the acid HA, the more readily will reaction take place, thus gobbling up A ions. If an excess of H+ is made available by addition of a strong acid, even more A ions will be consumed, eventually reversing reaction , causing the solid to dissolve.
In , for example, sulfate ions react with calcium ions to form insoluble CaSO4. Addition of a strong acid such as HCl (which is totally dissociated ) has no effect because CaCl2 is soluble. Although H+ can protonate some SO42– ions to form hydrogen sulfate ("bisulfate") HSO4, this ampholyte acid is too weak to reverse by drawing a significant fraction of sulfate ions out of CaSO4(s).
Example $7$: Aluminum Ion
Calculate the concentration of aluminum ion in a solution that is in equilibrium with aluminum hydroxide when the pH is held at 6.0.
The equilibria are
$Al(OH)_3 \rightleftharpoons Al^{3+} + 3 OH^–$
with
$K_s = 1.4 \times 10^{–34}$
and
$H_2O \rightleftharpoons H^+ + OH^–$
with
$K_w = 1 \times 10^{–14}$
Substituting the equilibrium expression for the second of these into that for the first, we obtain
$[OH^–]^3 = \left( \dfrac{K_w}{ [H^+]}\right)^3 = \dfrac{K_s}{[Al^{3+}]}$
(1.0 × 10–14) / (1.0 × 10–6)3 = (1.4 × 10–24) / [Al3+]
from which we find
$[Al^{3+}] = 1.4 \times 10^{–10}\; M$
Formation of a Competing Precipitate
If two different anions compete with a single cation to form two possible precipitates, the outcome depends not only on the solubilities of the two solids, but also on the concentrations of the relevant ions.
These kinds of competitions are especially important in groundwaters, which acquire solutes from various sources as they pass through sediment layers having different compositions. As the following example shows, competing equilibria of these kinds are very important for understanding geochemical processes involving the formation and transformation of mineral deposits.
Example $8$
Suppose that groundwater containing 0.001M F and 0.0018M CO32– percolates through a sediment containing calcite, CaCO3. Will the calcite be replaced by fluorite, CaF2?
The two solubility equilibria are
$\ce{CaCO3 <=> Ca^{2+} + CO3^{2–} \quad K_s = 10^{–8.1}$
$\ce{CaF2 <=> Ca^{2+} + 2 F^{–} \quad K_s = 10^{–10.4}$
Solution:
The equilibrium between the two solids and the two anions is
$CaCO_3 + 2 F^–\rightleftharpoons CaF_2 + CO_3^{2–}$
This is just the sum of the dissolution reaction for CaCO3 and the reverse of that for CaF2, so the equilibrium constant is
$K = \dfrac{[CO_3^{2–}]}{ [F^–]^2} = \dfrac{10^{–8.1}}{ 10^{–10.4}} = 200$
That is, the two solids can coexist only if the reaction quotient Q ≤ 200. Substituting the given ion concentrations we find that
$Q = \dfrac{0.0018}{0.0012} = 1800$
Since Q>K, we can conclude that the calcite will not change into fluorite.
Complex Ion Formation
Most transition metal ions possess empty d orbitals that are sufficiently low in energy to be able to accept electron pairs from electron donors from cations, resulting in the formation of a covalently-bound complex ion. Even neutral species that have a nonbonding electron pair can bind to ions in this way. Water is an active electron donor of this kind, so aqueous solutions of ions such as Fe3+(aq) and Cu2+(aq) exist as the octahedral complexes Fe(H2O)63+ and Cu(H6O)62+, respectively.
Many of the remarks made above about the relation between Ks and solubility also apply to calculations involving complex formation. See Stephen Hawkes' article Complexation Calculations are Worse Than Useless ("... to the point of absurdity...and should not be taught" in introductory courses.) (J Chem Educ. 1999 76(8) 1099-1100). However, it is very important that you understand the principles outlined in this section.
H2O is only one possible electron donor; NH3, CN and many other species (known collectively as ligands) possess lone pairs that can occupy vacantd orbitals on a metallic ion. Many of these bind much more tightly to the metal than does H2O, which will undergo displacement and substitution by one or more of these ligands if they are present in sufficiently high concentration.
If a sparingly soluble solid is placed in contact with a solution containing a ligand that can bind to the metal ion much more strongly than H2O, then formation of a complex ion will be favored and the solubility of the solid will be greater. Perhaps the most commonly seen example of this occurs when ammonia is added to a solution of copper(II) nitrate, in which the Cu2+(aq) ion is itself the complex hexaaquo complex ion shown at the left:
Because ammonia is a weak base, the first thing we observe is formation of a cloudy precipitate of Cu(OH)2 in the blue solution. As more ammonia is added , this precipitate dissolves, and the solution turns an intense deep blue, which is the color of hexamminecopper(II) and the various other related species such as Cu(H2O)5(NH3)2+, Cu(H2O)4(NH3)22+, etc.
In many cases, the complexing agent and the anion of the sparingly soluble salt are identical. This is particularly apt to happen with insoluble chlorides, and it means that addition of chloride to precipitate a metallic ion such as Ag+ will produce a precipitate at first, but after excess Cl has been added the precipitate will redissolve as complex ions are formed.
Some important solubility systems
In this section, we discuss solubility equilibria that relate to some very commonly-encountered anions of metallic salts. These are especially pertinent to the kinds of separations that most college students are required to carry out (and understand!) in their first-year laboratory courses.
Solubility of oxides and hydroxides
Metallic oxides and hydroxides both form solutions containing OH ions. For example, the solubilities of the [sparingly soluble] oxide and hydroxide of magnesium are represented by
$Mg(OH)_{2(s)} → Mg^{2+} + 2 OH^– \label{10}$
$MgO_{(S)} + H_2O → Mg^{2+} + 2 OH^– \label{11}$
If you write out the solubility product expressions for these two reactions, you will see that they are identical in form and value.
Recall that pH = –log10[H+], so that [H+] = 10–pH.
One might naïvely expect that the dissolution of an oxide such as MgO would yield as one of its products the oxide ion O2+. But the oxide ion is such a strong base that it grabs a proton from water, forming two hydroxide ions instead:
$O^{2+} + H_2O → 2 OH^–$
This is an example of the rule that the hydroxide ion is the strongest base that can exist in aqueous solution.2" is an equilibrium mixture of hydrated CO2molecules and carbonic acid. To keep things as simple as possible, we will not distinguish between them in what follows, and just use the formula H2CO3 to represent the two species collectively.
The other Group 2 metals, especially Mg, along with iron and several other transition elements are also found in carbonate sediments. When rain falls through the air, it absorbs atmospheric carbon dioxide, a small portion of which reacts with the water to form carbonic acid. Thus all pure water in contact with the air becomes acidic, eventually reaching a pH of 5.6.
As noted above, the equilibrium between bicarbonate and carbonate ions depends on the pH. Since the pH scale is logarithmic, it makes sense (and greatly simplifies the construction of the plot) to employ a log scale for the concentrations. The plot shown below corresponds to a total carbonate-system concentration of 10–3 M, which is representative of many ground waters. For river and lake waters, 10–5 M would be more typical; this would simply shift the curves downward without affecting their shapes.
Points 1 and 2 where adjacent curves overlap correspond to the two pK's. Recall that when the pH is the same as the pK, the concentrations of the two conjugate species are identical and half of the total system concentration. This places the crossover points at log 0.5 = –0.3 below the system concentration level.
A 10–3 M solution of sodium bicarbonate would have a pH denoted by point 3, with [H2CO3] and [CO32–] constituting only 1% (10–5 M) of the system. This corresponds to the equilibrium
$2 HCO_3^– \rightleftharpoons H_2CO_3 + CO_3^{2–}$
Carbonates act as bases and, as such, react with acids. Thus, the portion of the global water cycle that transports carbon from the air into natural waters constitutes a gigantic acid-base reaction that yields hydrogen carbonate ions, commonly referred to as bicarbonate. The natural waters that result have pH values between 6 and 10 and are essentially solutions of bicarbonates.
Limestone caves and sinkholes
When rainwater permeates into the soil, it can become even more acidic owing to the additional CO2 produced by soil organisms. Also, the deeper the water penetrates, the greater its hydrostatic pressure and the more CO2 it can hold, further increasing its acidity. If this water then works its way down through the fissures and cracks within a limestone layer, it will dissolve some of limestone, leaving void spaces which may eventually grow into limestone caves or form sinkholes that can swallow up cars or houses.
A well-known feature of limestone caves is the precipitated carbonate formations that decorate the ceilings and floors. These are known as stalactites and stalagmites, respectively. When water emerges from the ceiling of a cave that is open to the atmosphere, some of the excess CO2 it contains is released as it equilibrates with the air. This raises its pH and thus reduces the solubility of of the carbonates, which precipitate as stalactites. Some of the water remains supersaturated and does not precipitate until it drips to the cave floor, where it builds up the stalagmite formations.
Hard Water
This term refers to waters that, through contact with rocks and sediments in lakes, streams, and especially in soils (groundwaters), have acquired metallic cations such as Ca2+, Mg2+, Fe2+, Fe3+, Zn2+ Mn2+, etc. Owing to the ubiquity of carbonate sediments, the compensating negative charge is frequently supplied by the bicarbonate ion HCO3, but other anions such as SO42–, F, Cl, PO43– and SiO42– may also be significant.
Solid bicarbonates are formed only by Group 1 cations and all are readily soluble in water. But because HCO3 is amphiprotic, it can react with itself to yield carbonate:
$2 HCO_3^– → H_2O + CO_3^[2–} + CO_{2(g)}$
If bicarbonate-containing water is boiled, the CO2 is driven off, and the equilibrium shifts to the right, causing any Ca2+ or similar ions to form a cloudy precipitate. If this succeeds in removing the "hardness cations", the water has been "softened". Such water is said to possess carbonate hardness, sometimes known as "temporary hardness". Waters in which anions other than HCO3 predominate cannot be softened by boiling, and thus possess non-carbonate hardness or "permanent hardness".
Hard waters present several kinds of problems, both in domestic and industrial settings:
• Waters containing dissolved salts leave solid deposits when they evaporate. Residents of areas having hard water (about 85 percent of the U.S.) notice evaporative deposits on shower walls, in teakettles, and on newly-washed windows, glassware, and vehicles.
• Much more seriously from an economic standpoint, evaporation of water in boilers used for the production of industrial steam leaves coatings on the heat exchanger surfaces that impede the transfer of heat from the combustion chamber, reducing the thermal transfer efficiency. The resultant overheating of these surfaces can lead to their rupture, and in the case of high-pressure boilers, to disastrous explosions. In the case of calcium and magnesium carbonates, the process is exacerbated by the reduced solubility of these salts at high temperatures. Removal of boiler scales is difficult and expensive.
• Municipal water supplies in hard-water areas tend to be supersaturated in hardness ions. As this water flows through distribution pipes and the plumbing of buildings, these ions often tend to precipitate out on their interior surfaces. Eventually, this scale layer can become thick enough to restrict or even block the flow of water through the pipes. When scale deposits within appliances such as dishwashers and washing machines, it can severely degrade their performance.
• Cations of Group 2 and above react with soaps, which are sodium salts of fatty acids such as stearic acid, C17H35COOH. The sodium salts of such acids are soluble in water, which allows them to dissociate and act as surfactants:
$C_{17}H_{35}COONa → C_{17}H_{35}COO^– Na^+$
but the presence of polyvalent ions causes them to form precipitates
$2 C_{17}H_{35}COO^– + Ca^{2+} → (C_{17}H_{35}COO^–)_2Ca_{(s)}$
Calcium stearate is less dense than water, so it forms a scum that floats on top of the water surface; anyone who lives in a hard-water area is likely familiar with the unsightly "bathtub rings" it leaves around the high-water mark or the shower-wall stains.
Solubility Complications
All heterogeneous equilibria, on close examination, are beset with complications. But solubility equilibria are somewhat special in that there are more of them. Back in the days when the principal reason for teaching about solubility equilibria was to prepare chemists to separate ions in quantitative analysis procedures, these problems could be mostly ignored. But now that the chemistry of the environment has grown in importance — especially that relating to the ocean and natural waters — there is more reason for chemical scientists to at least know about the limitations of simple solubility products. This section will offer a quick survey of the most important of these complications, while leaving their detailed treatment to more advanced courses.
Tabulated Ks values are notoriously unreliable
Many of the $K_s$ values found in tables were determined prior to 1940 (some go back to the 1880s!) at a time before highly accurate methods became available. Especially suspect are many of those for highly insoluble salts which are more difficult to measure. A table showing the variations in $K_{sp}$ values for the same salts among ten textbooks was published by Clark and Bonikamp in J Chem Educ. 1998 75(9) 1183-85.A good An example that used a variety of modern techniques to measure the solubility of silver chromate was published by A.L. Jones et al in the Australian J. of Chemistry, 1971 24 2005-12.
Generations of chemistry students have amused themselves by comparing the disparate Ks values to be found in various textbooks and table. In some cases, they differ by orders of magnitude. There are several reasons for this in addition to the ones described in detail further on.
• The most direct methods of measuring solubilities tend to not be very accurate for sparingly soluble salts. Two-significant figure precision is about the best one can hope in a single measurement.
• Many insoluble salts can exist in more than one crystalline form (polymorphs), and in some cases also as amorphous solids. Precipitation under different conditions (in the presence of different ions, at different temperatures, etc.) can yield different or mixed polymorphs.
• Other ions present in the solution can often get incorporated into the crystalline solid, usually replacing an ion of similar size (substitutional solid solutions). When this happens, it is no longer valid to write the equilibrium condition as a simple "product". This is very common in mineral deposits, and an important consideration in geochemistry,
Most salts are not Completely Dissociated in Water
The dissolution of cadmium iodide is water is commonly represented as
$CdI_{2(s)} → Cd^{2+} + 2 I^–$
What you were likely taught about the dissociation of salts in water is wrong! To most students (and to most of their teachers!), this would imply that a 0.1M solution of this salt would contain 0.1M of Cd2+(aq) — and this would be seriously in error because it fails to take into account that the two ions react with each other.
Firstly, they combine to form neutral, largely-covalent molecular species:
$Cd^{2+}_{(aq)} + 2 I^–_{(aq)} → CdI_{2(aq)}$
This non-ionic form accounts for 78% of the Cd present in the solution! In addition, they form a molecular ion $CdI^–_{(aq)}$ according to the following scheme:
Table $2$: CdI2 species in solution
$CdI_{2(s)} \rightleftharpoons Cd^{2+} + 2 I^–$ $K_1 = 10^{–3.9}$
$Cd^{2+} + I^– \rightleftharpoons CdI^+$ $K_2= 10^{+2.3}$
$CdI2_{(s)} \rightleftharpoons CdI^++ I^–$ $K = 10^{–1.6} = 0.023$
The data shown Tables $1$ and $2$ are taken from the article Salts are Mostly NOT Ionized by Stephen Hawkes: 1996 J Chem Educ. 73(5) 421-423. This fact was stated by Arrhenius in 1887, but has been largely ignored and is rarely mentioned in standard textbooks.
As a consequence, the concentration of "free" Cd2+(aq) in an aqueous cadmium iodide solution is only about 2% of the value you would calculate by taking K1 as the solubility product. The principal component of such as solution is actually [covalently-bound] CdI2(aq). It turns out that many salts, especially those of metals beyond Group 2, are similarly only partially ionized in aqueous solution:
salt molarity % cation other species
Table $2$: Incomplete Dissociation of Select Salts
KCl 0.52 95 KCl(aq) 5%
MgSO4 0.04 58 MgSO4(aq) 42%
CaCl2 0.44 70 CaCl+(aq) 30%
CuSO4 0.045 56 CuSO4(aq) 44%
CdI2 0.50 2 CdI2(aq) 76%, CdI(aq) 22%
FeCl3 0.1 10 FeCl2+(aq) 42%, FeCl2(aq) 40%, FeOH2+(aq) 6%, Fe(OH)2+(aq) 2%
If you are enrolled in an introductory course and do not plan on taking more advanced courses in chemistry or biochemistry, you can probably be safe in ignoring this, since your instructor and textbook likely do so. However, if you expect to do more advanced work or teach, you really should take note of these points, since few textbooks mention them.
Formation of Hydrous Complexes
Transition metal ions form a large variety of complexes with H2O and OH, both of which have electron-pairs available to coordinate with the central ion. This gives rise to a large variety of soluble species that are in competition with an insoluble solid. Because of this, a single equilibrium constant (solubility product) cannot describe the behavior of a solid such as Fe(OH)3, which we summarize here as an example.
Aquo complexes: The electrostatic field of the positively-charged metal ion enhances the acidic nature of these H2O molecules, encouraging them to shed a proton and leaving OH groups in their place.
$Fe(H_2O)_6^{3+} → Fe(H_2O)_5(OH)^{2+}+H^+$
This is just the first of a series of similar reactions, each one having a successively smaller equilibrium constant:
$Fe(H_2O)_5(OH)^{2+}→ Fe(H_2O)_4(OH)_2^+→ Fe(H_2O)_3(OH)_3 → Fe(H_2O)_2(OH)_4^-$
Hydroxo complexes: But there's more: when the hydroxide ion acts as a ligand, it gives rise to a series of hydroxo complexes, of which the insoluble Fe(OH)3 can be considered a member:
Fe3+ + 3 OH → Fe(OH)3(s) 1/Ks = 1038
Fe3+ + H2O → Fe(OH)2+ + H+ K = 10–2.2
Fe3+ + 2H2O → FeOH+ + 2H+ K = 10–6.7
Fe3+ + 4H2O → Fe(OH)4 + 4H+ K = 10–23
2Fe3+ + 2H2O → Fe2(OH)24+ + 2H+ K = 10–2.8
Because five equilibria are involved in the hydroxy-complex behavior alone, a complete treatment of this system for $Fe^{3+}$ requires the solution of five simultaneous equations, which is not a lot of fun. Fortunately, it is possible to make some approximations which greatly simplify the construction of a log-concentration vs. pH plot as shown below.
The equilibria listed above all involve H+ and OH ions, and are therefore pH dependent, as illustrated by the straight lines in the plot, whose slopes reflect the pH dependence of the corresponding ionic species. At any given pH, the equilibrium with solid Fe(OH)3 is controlled by the ionic species having the highest concentration at any given pH. The corresponding lines in the plot therefore delineate the region (indicated by the orange shading) at which the solid can exist.
Ionic interactions: The "non-common ion effect"
A sparingly-soluble salt will be more soluble in a solution that contains non-participating ions. This is just the opposite of the common ion effect, and it might at first seem rather counter-intuitive: why would adding more ions of any kind make a salt more likely to dissolve?
Figure $2$: Solubility of thallium iodate in solutions containing dissolved salts
A clue to the answer can be found in another fact: the higher the charge of the foreign ion, the more pronounced is the effect. This tells us that inter-ionic (and thus electrostatic) interactions must play a role. The details are rather complicated, but the general idea is that all ions in solution, besides possessing tightly-held waters of hydration, tend to attract oppositely-charged ions ("counter-ions") around them. This "atmosphere" of counterions is always rather diffuse, but much less so (and more tightly bound) when one or both kinds of ions have greater charges. From a distance, these ion-counterion bodies appear to be almost electrically neutral, which keeps them from interacting with each other (as to form a precipitate).
The overall effect is to reduce the concentrations of the less-shielded ions that are available to combine to form a precipitate. We say that the thermodynamically-effective concentrations of these ions are less than their "analytical" concentrations. Chemists refer to these effective concentrations as ionic activities, and they denote them by curly brackets {Ag+} as opposed to square brackets [Ag+] which refer to the nominal or analytical concentrations.
Although the concentrations of ions in equilibrium with a sparingly soluble solid are so low that they are essentially the same as the activities, the presence of other ions at concentrations of about 0.001M or greater can materially reduce the activities of the dissolution products, permitting the solubilities to be greater than what simple equilibrium calculations would predict.
Measured solubilities are averages and depend on size
The heterogeneous nature of dissolution reactions leads to a number of peculiar effects relating to the nature of equilibria involving surfaces. These arise from the fact that the tendency of a crystalline solid to dissolve will depend on the particular face or location from which dissolution occurs. Since all crystals present a variety of faces to the solution, a measured Ks is really an average of values for these various faces.
And because many salts can exhibit different external shapes depending on the conditions under which they are formed, solubility products are similarly dependent on these conditions.
Very small crystals are more soluble than big ones
Molecules or ions that are situated on edges or corners are less strongly bound to the remainder of the solid than those on plane surfaces, and will consequently tend to dissolve more readily. Thus the leftmost face in the schematic lattice below will have more edge-bound molecular units than the other two, and this face (11) will be more soluble.
This means, among other things, that smaller crystals, in which the ratio of edges and corners is greater, will tend to have greater Ks values than larger ones. As a consequence, smaller crystals will tend to disappear in favor of larger ones. Practical use is sometimes made of this when the precipitate initially formed in a chemical analysis or separation is too fine to be removed by filtration. The suspension is held at a high temperature for several hours, during with time the crystallites grow in size. This procedure is sometimes referred to as digestion.
Formation of supersaturated solutions
Contrary to what you may have been taught, precipitates do not form when the ion concentration product reaches the solubility product of a salt in a solution that is pure and initially unsaturated; to form a precipitate from a homogeneous solution, a certain degree of supersaturation is required. The extent of supersaturation required to initiate precipitation can be surprisingly great. Thus formation of barium sulfate BaSO4 by combining the two kinds of ions does not occur until Qs exceeds Ks by a factor of 160 or more. In part, this reflects the fact that precipitation proceeds by a series of reactions beginning with formation of an ion-pair which eventually becomes an ion cluster:
Ba2+ + SO42– → (BaSO4)0 → (BaSO4)20 → (BaSO4)30 → etc.
Owing to their overall neutrality, these aggregates are not stabilized by hydration, so they are more likely to break up than not. But a few may eventually survive until they are large enough (but still submicroscopic in size) to serve as precipitation nuclei.
Many substances other than salts form supersaturated solutions, and some salts form them more readily than others. Supersaturated solutions are easily made by dissolving the solid to near its solubility limit in a warmed solvent, and then letting it cool.
Ks) and are inherently unstable; dropping a "seed" crystal of the solid into such a solution will usually initiate rapid precipitation. But as is explained below, even a tiny dust particle may be enough. An old chemist's trick is to use the tip of a glass stirring rod to scrape the inner surface of a container holding a supersaturated solution; the minute particles of glass that are released presumably serve as precipitation nuclei.
The nucleation problem: precipitation is [theoretically] impossible!
Any process in which a new phase forms within an existing homogeneous phase is beset by the nucleation problem: the smallest of these new phases — raindrops forming in air, tiny bubbles forming in a liquid at its boiling point — are inherently less stable than larger ones, and therefore tend to disappear. The same is true of precipitate formation: if smaller crystals are more soluble, then how can the tiniest, first crystal, form at all?
In any ionic solution, small clumps of oppositely-charged ions are continually forming by ordinary collisional processes. The smallest of these aggregates possess a higher free energy than the isolated solvated ions, and they rapidly dissociate. Occasionally, however, one of these proto-crystallites reaches a critical size whose stability allows it to remain intact long enough to serve as a surface (a "nucleus") onto which the deposition of additional ions can lead to still greater stability. At this point, the process passes from the nucleation to the growth stage.
Theoretical calculations predict that nucleation from a perfectly homogeneous solution is a rather unlikely process; tenfold supersaturation should produce only one nucleus per cm3 per year. Most nucleation is therefore believed to occur heterogeneously on the surface of some other particle, possibly a dust particle. The efficiency of this process is critically dependent on the nature and condition of the surface that gives rise to the nucleus. | textbooks/chem/General_Chemistry/Chem1_(Lower)/12%3A_Solubility_Equilibria.txt |
Acid-base chemistry can be extremely confusing, particularly when dealing with weak acids and bases. This set of lessons presents an updated view of the Brønsted-Lowry theory that makes it easy to understand answers to common questions: What's the fundamental difference between a strong acid and a weak acid? Can acid A neutralize base B? Why are some salts acidic and others alkaline? How do buffers work? What governs the shapes of titration curves?
• 13.1: Introduction to Acid/Base Equilibria
Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry.
• 13.2: Strong Monoprotic Acids and Bases
To a good approximation, strong acids, in the forms we encounter in the laboratory and in much of the industrial world, have no real existence; they are all really solutions of H3O+. So if you think about it, the labels on those reagent bottles you see in the lab are not strictly true! However, if the strong acid is highly diluted, the amount of H3O+ it contributes to the solution becomes comparable to that which derives from the autoprotolysis of water.
• 13.3: Finding the pH of weak Acids, Bases, and Salts
Most acids are weak; there are hundreds of thousands of them, whereas there are fewer than a dozen strong acids. We can treat weak acid solutions in much the same general way as we did for strong acids. The only difference is that we must now take into account the incomplete "dissociation"of the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which salts of the acid are present - these are known as buffer solutions.
• 13.4: Conjugate Pairs and Buffers
We often tend to regard the pH as a quantity that is dependent on other variables such as the concentration and strength of an acid, base or salt. But in much of chemistry (and especially in biochemistry), we find it more useful to treat pH as the "master" variable that controls the relative concentrations of the acid- and base-forms of one or more sets of conjugate acid-base systems. In this Module, we explore this approach in some detail, showing its application to buffer solutions.
• 13.5: Acid/Base Titration
The objective of an acid-base titration is to determine Ca, the nominal concentration of acid in the solution. In its simplest form, titration is carried out by measuring the volume of the solution of strong base required to complete the neutralization reaction . The point at which this reaction is just complete is known as the equivalence point, which is distinguished from the end point, which is the value we observe experimentally.
• 13.6: Applications of Acid-Base Equilibria
Acid-base reactions pervade every aspect of industrial-, physiological-, and environmental chemistry. In this unit we touch on a few highlights that anyone who studies or practices chemical science should be aware of.
• 13.7: Exact Calculations and Approximations
The methods for dealing with acid-base equilibria that we developed in the earlier units of this series are widely used in ordinary practice. Although many of these involve approximations of various kinds, the results are usually good enough for most purposes. Sometimes, however — for example, in problems involving very dilute solutions, the approximations break down, often because they ignore the small quantities of H+ and OH– ions always present in pure water.
13: Acid-Base Equilibria
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Write a chemical equation for the autoprotolysis of a species such as HCO3; write an equation for the equilibrium constant for this reaction.
• Write equations that express the chemical reactions and equilibrium constant expressions that define the strength of an acid or base in terms of proton transfer processes (Brønsted-Lowry concept).
• Write expressions defining the equilibrium constants Ka and Kb.
• Given the value of any two of Ka, Kb, and Kw, evaluate the third one. (This requires that you can show that Ka × Kb = Kw.)
• Explain the significance of the value of pKa for a given monoprotic acid-base system — that is, what it tells us about that particular acid and its conjugate base.
• Sketch out a simple proton free energy (PFE) diagrams illustrating the distinction between monoprotic strong and weak acids.
• State the meaning of the leveling effect, and construct a simple PFE diagram that illustrates its action.
• Describe, in your own words, the meaning of PFE, how it relates to pH, and in its significance in terms of the relative concentrations of a conjugate acid-base pair in a solution.
Acid-base reactions, in which protons are exchanged between donor molecules (acids) and acceptors (bases), form the basis of the most common kinds of equilibrium problems which you will encounter in almost any application of chemistry. In order to thoroughly understand the material in this unit, you are expected to be familiar with the following topics which were covered in the separate unit Introduction to Acid-Base Chemistry:
• The Arrhenius concept of acids and bases
• The Brønsted-Lowry concept, conjugate acids and bases
• Definition of pH and the pH scale
• Strong vs. weak acids and bases
You should also have some familiarity with the principles of chemical equilibrium, and know how to write and use equilibrium constants.
Proton transfer reactions in acid-base equilibria
As you should recall from your earlier introduction to acids and bases, the +1 electric charge of the tiny proton (a bare hydrogen nucleus) is contained in such a miniscule volume of space that the resulting charge density is far too large to enable its independent existence in solution; it will always attach to, and essentially bury itself in, the non-bonding orbitals of a solvent. Thus in aqueous solution, what we commonly represent as the "hydrogen ion" H+ is more accurately described as the hydronium ion H3O+.
Acid-base chemistry is a transactional process in which protons are exchanged between two chemical species. A molecule or ion that loses or "donates" a proton is acting as an acid; a species that receives or "accepts" that proton plays the role of a base.
Defining an acid as a substance that simply "dissociates" into its component parts
$HA \rightarrow H^+ + AB^–$
fails to capture the concept outlined in the above box. It is therefore important that you become comfortable with the Brønsted-Lowry model of acid-base behavior that represents the generalized acid-base reaction as a proton transfer process:
$\underbrace{HA}_{\text{acid}} + \underbrace{B}_{base} \rightarrow \underbrace{A^-}_{\text{conjugate base}} + \underbrace{HB^+}_{\text{conjugate acid}}$
The actual electric charges of the reaction products will of course depend on the particular nature of the species $A$, but the base will always have one more negative charge than the acid HA. Although anyone familiar with chemicals tends to regard certain substances as "acids" and others as "bases", it is important to bear in mind that these labels have meaning only in terms of the particular process being considered. For example, in the absence of water, ammonia, which we usually think of as a base, can lose a proton (and thus act as an acid) to a "stronger" base $B$ to form the amide ion:
$NH_3 + B → NH_2^– + HB^+$
Similarly, anhydrous sulfuric acid can accept a proton from a stronger acid AH:
$H_2SO_4 + AH \rightarrow H_3SO_4^+ + A^–$
However, this is rather exotic stuff that is beyond the scope of this introductory lesson. The acid-base processes we describe in this set of lessons take place in aqueous solution, so we will always assume that $H_2O$, which can act either as a proton donor or proton acceptor, plays an active role:
• A substance that acts as an acid in aqueous solution donates a proton to $H_2O$, yielding the hydronium ion $H_3O^+$.
$HA + H_2O → H_3O^+ + A^–$
• Similarly, a base will remove a proton from $H_2O$, yielding the hydroxide ion $OH^–$.
$B + H_2O → BH^+ + OH^–$
Autoprotolysis
A small number of acids are themselves amphiprotic, meaning that one molecule can transfer a proton to another molecule of the same kind. The best-known example of this process, known as autoprotolysis, is of course that of water:
$2 H_2O_{(l)} \rightleftharpoons OH^– + H_3O^+ \label{1-1}$
Pure liquid ammonia and sulfuric acid are other well-known examples:
$2 NH_{3(l)} \rightleftharpoons NH_2^– + NH_4^+$
$2 H_2SO_{4(l)} \rightleftharpoons HSO_4^– + H_3SO_4^+$
Autoprotolysis reactions such as these typically proceed to only a very small extent. The equilibrium constant for Equation $\ref{1-1}$ is commonly written as
$K_w = [H^+][OH^–] \label{1-2}$
and is usually known as the dissociation constant or ion product of water. For most purposes in elementary courses, we can use the value 1.0 × 10–14 for Kw. But you should understand that this value is correct only for pure water at 25° C and 1 atm pressure. Kw varies considerably with temperature, pressure, and the ionic content of water:
It should be obvious that anyone working in the fields of physiological or oceanographic chemistry cannot routinely use $1.0 \times 10^{–14}$ for Kw!
The equilibrium constants Ka and Kb
Acids and bases vary greatly in their "strength" — that is, their tendencies to donate or accept protons. As with any chemical reaction, we can define an equilibrium constant $K_c$ whose numerical value reflects the extent of the reaction — that is, the relative concentrations of the products and reactants when the reaction reaches equilibrium. The strength of an acid $HA$ in water can be defined by the equilibrium
$HA + H_2O \rightleftharpoons H_3O^+ + A^–$
$K_a =\dfrac{[H_3O^+][A^-]}{[HA]} \label{1-3a}$
Similarly, the strength of the base $A^−$ in water is defined by
$A^− + H_2O \rightleftharpoons HA + OH^–$
$K_b = \dfrac{'[HA][OH^-]}{[A^-]} \label{1-3b}$
Note carefully that reaction $\ref{1-3a}$ is not the reverse of $\ref{1-3b}$.
Ka and Kb are inversely proportional
How are Ka and Kb related? The answer can found by multiplying the above two expressions for Ka and Kb:
(1-3)
Alternatively, we can simply add the two corresponding reaction equations:
and — as you should recall from your earlier study of equilibrium, the equilibrium constant of the sum of two reactions is the product of the two K's. So either way you figure it, the $K_a$ for an acid and $K_b$ for its conjugate base are related in a very simple way:
$K_a K_b = K_w \label{1-4}$
Clearly, as the strength of a series of acids increases, the strengths of their conjugate bases will decrease, this inverse relation between Ka and Kb is implicit in Equation $\ref{1-4}$. You will recall that the pH scale serves as a convenient means of compressing a wide range of [H+]-values into a small range of numbers. Just as we defined the pH as the negative logarithm of the hydrogen ion concentration, we can define
$pK = −\log K \label{2-5}$
for any equilibrium constant. Acid and base strengths are very frequently expressed in terms of $pK_a$ and $pK_b$. From Equation $\ref{1-4}$, it should be apparent that
$pK_a + pK_b = pK_w \;(14.0\; at\; 25^o C) \label{1-5}$
pK values of some common acids and bases
In order to convert pKa into Ka, simply take the negative antilog of pKa. Thus, for acetic acid with
$pK_a = 4.7$
$K_a = 10^{–4.7} = 2.0 \times 10^{–5}$
The following table gives the pKa values for a number of commonly-encountered acid-base systems which are listed in order of decreasing acid strength. It's worth taking some time to study this table in some detail.
lease take special note of the following points:
• Acids (1-4) are the major strong acids. All dissociate in water to produce the hydronium ion (5), the strongest acid that can exist in water.
• (6) H2SO3 has never been detected; "sulfurous acid" and its formula refer to a solution of SO2 in water.
• The names bisulfate (3, 7), bisulfite (6, 16), and bicarbonate (13) are still commonly used by chemists, but the preferred names are now hydrogen sulfate, hydrogen sulfite, and hydrogen carbonate, respectively. Similarly, bisulfide (14, 26) is also known as hydrosulfide.
• The ion (9) as it exists in aqueous solution is often represented as Fe3+ and is known more commonly as the ferrous or iron(III) ion. The aqueous aluminum ion Al3+ (12) as well as many transition metal ions have similarly complex structures and give acidic solutions in water because the high positive charge increases the acidity of the attached H2O units. For more on these aquo complexes, see this brief discussion, or this much more detailed one by Jim Cook of the UK.
• The molecule H2CO3 (13) exists as a minority species in an aqueous solution of CO2 and has never been isolated. In most contexts, the formula H2CO3, the name carbonic acid and the pKa 6.3 refer to "total dissolved CO2" in the water. "True" H2CO3 is about a thousand times stronger than the pKa of its equilibrium mixture with CO2(aq).
• The sulfide, amide and oxide ions (26-28), being stronger bases than OH, cannot exist in water.
Common acids and bases
ref acid pKa conjugate base pKb
1 HClO4 perchloric acid ~ –7 ClO4 chlorate ~ 21
2 HCl hydrochloric acid ~ –3 Cl chloride ~ 17
3 H2SO4 sulfuric acid ~ –3 HSO4 bisulfate ~ 17
4 HNO3 nitric acid ~ –1 NO3 nitrate 15
5 H3O+ hydronium ion 0 H2O water 14.00
6 H2SO3 sulfurous acid 1.8 HSO3 bisulfite 12.2
7 HSO4 bisulfate ion 1.9 SO42– sulfate 12.1
8 H3PO4 phosphoric acid 2.12 H2PO4 dihydrogen phosphate 11.88
9 [Fe(H2O)6]3+ hexaaquoiron(III) 2.10 [Fe(H2O)5OH]2+ 11.90
10 HF hydrofluoric acid 3.2 F fluoride 10.8
11 CH3COOH acetic acid 4.76 CH3COO acetate 9.3
12 [Al(H2O)6]3+ hexaaquoaluminum(III) 4.9 [Al(H2O)5OH]2+ 9.1
13 H2CO3 carbonic acid 6.3 HCO3 bicarbonate 7.7
14 H2S hydrogen sulfide 7.04 HS bisulfide 6.96
15 H2PO4 dihydrogen phosphate 7.2 HPO4monohydrogen phosphate 6.8
16 HSO3 bisulfite ion 7.21 SO32– sulfite 6.79
17 HOCl hypochlorous acid 8.0 OCl hypochlorite 6.0
18 HCN hydrocyanic acid 9.2 CN cyanide 4.8
19 H3BO4 boric acid 9.30 B(OH)4 borate 4.79
20 NH4+ ammonium ion 9.25 NH3 ammonia 4.75
21 Si(OH)4 ortho-silicic acid 9.50 SiO(OH)3 silicate 4.50
22 HCO3 bicarbonate ion 10.33 CO32– carbonate 3.67
23 HPO42– monohydrogen phosphate ion 12.32 PO43– phosphate 1.67
24 SiO(OH)3 silicate ion 12.6 SiO2(OH)22– 1.4
25 H2O water 14.00 OH hydroxide 0
26 HS bisulfide ion ~ 19 S 2– sulfide ~ –5
27 NH3 ammonia ~ 23 NH2 amide ~ -9
28 OH hydroxide ion ~ 24 O 2– oxide ~ –10
The Fall of the proton
The concept of the "potential energy of the proton" (more properly, proton free energy), is not nearly as intimidating as it might sound; it is easy to grasp, requires no mathematics, and will enable you to achieve a far better understanding of acid-base chemistry.
Most courses (and their textbooks) do a good job of explaining the concept of proton transfer, but few of them provide you with a clear basis for understanding and making useful predictions about the direction and extent this transfer will take. For example, when the gas HCl is added to water, you probably know that the reaction proceeds strongly to the right:
$HCl + H_2O \rightleftharpoons H_3O^+ + Cl^– \label{2-1}$
...but what principle determines this, and what is to prevent the hydronium ion (also an acid) from pushing one of its own protons back onto a chloride ion, and thus reversing the reaction?
The concept of the proton free energy level was introduced in R.W. Gurney's classic Ionic Processes in Solution (1953). The best detailed exposition can be found in the first (1970) edition of Aquatic Chemistry by W. Stumm and J.L. Morgan.
The answer is that a proton is never really "pushed"; it will spontaneously seek out the lowest potential energy state (that is, the strongest base) it can find. Think how a macroscopic object such as a book will tend to fall to a location where its [gravitational] potential energy is as low as possible. Of course, the potential energy of a proton has nothing to do with gravity, but rather with how tightly it can bind to a base. This kind of potential energy is known as proton free energy (PFE) — but in the context of first-year chemistry, we can also just call it "proton energy".
You are not sure what free energy is? Don’t worry about it for the time being; just think of it as you would any other form of potential energy: something that falls when chemical reactions take place. See here for the lesson on free energy.
For the present, all you need to know is that a proton will tend to "fall" to the lowest potential energy state it can find. In keeping with this idea, you can think of an acid as a proton source, and a base as a proton sink. Using this terminology, we can depict the generalized process HA + B → AB + HB+ as in this plot, which depicts the proton on HA "falling" to the lower-PFE "sink" provided by the base B, leaving AB and HB+ as products.
We will be making a lot of use of schematics like this farther on, so take a moment to familiarize yourself with information it depicts.
This "source/sink" nomenclature recalls the tendency of water to flow down from a high elevation to a lower one; this tendency (which is related to the amount of energy that can be extracted in the form of electrical work if the water flows through a power station at the bottom of the dam) will be directly proportional to the difference in elevation (difference in potential energy) between the source (top of the dam) and the sink (bottom of the dam).
Thus you can regard proton free energy as entirely analogous to the gravitational potential energy of the water contained in a dam; in this case, we are dealing with chemical potential energy, more commonly known as free energy. Take particular note of the following:
• The ordinate represents the relative free energy of a mole of protons when attached to a given base; we label it simply "proton energy" for brevity.
• The gray vertical line at the center of the diagram separates the acidic form of each conjugate pair from its base form; the terms "proton sources" and "proton sinks" are there to remind us of the role these forms play in acid-base chemistry.
• The red diagonal arrow represents the actual process in which a mole of protons "falls" from the acid HA to the base B. The fact that the arrow slants downward indicates that this process is spontaneous, meaning that the reaction proceeds to completion. In other words, the equilibrium constant for this reaction is very large.
• The green horizontal arrows depict the changes (that is, the formation of the two new conjugate species) that result from the proton transfer as the protons "fall" (in free energy) from the acid HA to the base B.
Strong acids in water
When you were first introduced to the Brønsted-Lowry model, you learned that the proper way to represent the "dissociation" of a strong acid like hydrochloric is
$HCl + H_2O \rightarrow H_3O^+ + Cl^- \label{2-1a}$
Thus water is revealed as an active participant in the behavior of an acid, rather than as an inert spectator or solvent. Although analogous acid-base reactions can take place in other solvents such as liquid ammonia, virtually all of the acid-base chemistry we encounter in daily life is based on water. Since water is acting as the proton acceptor here, it must be fulfilling the role of the base B in the schematic given above. We can therefore construct a similar plot specifically for HCl (or, in fact, for any strong acid).
Recall that, for strong acids, this reaction is essentially complete. This means that a "solution of hydrochloric acid" is in reality a solution of hydronium and chloride ions; except in solutions much more concentrated than 1 M, the species HCl is only present in trace amounts.
Weak acids in water
When acetic acid CH3COOH is dissolved in water, the resulting solution displays only very mild acidic properties. For simplicity we will represent acetic acid and the acetate ion CH3COO by HAc and Ac, respectively:
$HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{2-2}$
But although this equation has the same form as (2-1) above, it hides an important qualitative difference: HAc is a weak acid, meaning that the proton transfer takes place to only a tiny extent; the equilibrium strongly favors the left side:
HAc + H2O H3O+ + Ac
The corresponding proton free energy diagram now shows the H3O+-H2O system at the top, with the proton-transfer arrow pointing upward, indicating that the protons get kicked up to a higher free-energy level in this process.
Where does this additional energy come from? Simply from the random thermal energy in the solution. As you might expect, this is not a thermodynamically favorable process, so it only happens to a very small extent. In contrast to the HCl example we described previously in which 99.99+ percent of the HCl molecules end up as hydronium ions, almost all of the acetic acid molecules remain unchanged, and only tiny amounts of H3O+ and Ac are produced.
Strong and weak acids both donate protons to water, yielding hydrogen ions and thus rendering the solution acidic.
What's different is that virtually all of the protons of the strong acid "fall" to the H2O level, whereas only a tiny fraction of the weak acid molecules acquire enough thermal energy to promote their protons up to the H2O level.
The fundamental difference between strong acids and weak acids is clearly evident when we combine the preceding two proton-free energy diagrams into one. Strong acids (strong "conjugate-pairs") are always above the PFE of the H3O+-H2O system, while the weak conjugate pairs are below it.
The other big difference with acetic acid is that water is acting as a base here, rather than as an acid as it did in the HCl example. So it's not surprising that water, the basis of almost all acid-base chemistry, should appear in the middle of these diagrams.
Water can swing both ways
Substances that can both donate and accept protons are said to be amphiprotic. Many oxides and hydroxides behave in this way, and since H2O falls into both of these classes, it is not surprising that water is amphiprotic:
• Water ("the-acid"): $H_2O + B \rightleftharpoons BH^+ + OH^–$
• Water ("the base"): $H_2O + AH \rightleftharpoons H_3O^+ + A^–$
("AH" above represents any acid; B stands for any base)
These two opposing tendencies can be represented by
In the case of water, the contest is a draw; not only does neither side win, but the fraction of
H2O molecules that break up in this way is miniscule: In pure water, only about one H2O molecule out of 109 is “dissociated” at any instant:
H2O H+ + OH (2-4)
The actual reaction, of course, is the proton transfer
(2-5)
for which the equilibrium constant
Kw = [H3O+][OH] (2-6)
is known as the ion product of water. The value of Kw at room temperature is 1.008 x 10–14. We can also express this in logarithmic terms: pKw = 14.0.
As with any equilibrium constant, the value of Kw is affected by the
• temperature: Kw undergoes a 10-fold increase between 0°C and 60°C
• pressure: Kw is about doubled at 1000 atm
• presence of ionic species in the solution.
Because most practical calculations involving Kw refer to ionic solutions rather than to pure water, the common practice of using 10–14 as if it were a universal constant is unwise; under the conditions commonly encountered in the laboratory, pKw can vary from about 11 to almost 15. In seawater, Kw is 6.3 × 10−12. (See Stephen J. Hawkes: “pKw is almost never 14.0”, J. Chem. Education 1995: 72(9) 799-802)
Notice that under conditions when Kw differs significantly from 1.0 × 10–14, the pH of a neutral solution will not be 7.0.
For example, at a pressure of 93 kbar and 527°C, Kw = 10−3.05, the pH of pure water would be 1.5. Similarly unusual conditions apply to deposits of water in geological formations and in undersea vents.
Example
At 60° C, the ion product of water is $9.6 \times 10^{-14}$. What is the pH of a neutral solution at this temperature?
Solution:
Under these conditions, [H+][OH] = 9.6E–14. If the solution is neutral, [H+] = [OH] = (9.6E–14)½ = 3.1E–7, corresponding to a pH of log 3.1E-7 = 6.5.
The leveling effect
The strengths of strong acids and bases are "leveled" in water. This means that all strong acids and strong bases are equally strong in water. To see this more clearly, notice that the free energies of protons in all of these acids (even nitric) are sufficiently above the H3O+/H2O level that their "dissociation" is essentially complete. Thus
The strongest acid that can exist in water is H3O+ ;
The strongest base that can exist in water is OH
Similarly, the hydroxide ion is the strongest base that can exist in water.
So what happens if you add a soluble oxide such as Na2O to water? Since O2 is a proton sink to H2O, it will react with the solvent, leaving OH as the strongest base present: Na2O + H2O → 2 OH + Na+ .
Thus all bases stronger than OH appear equally strong in water — simply because they are all converted to OH .
3 Why acids react with bases, but not always completely
PFE's are the key to understanding acid-base reactions
In the preceding section, we have employed simple PFE (proton free energy) diagrams to help you better understand the concepts of strong and weak acids, water as an acid or a base, and the leveling effect. But we can gain a more generalized and comprehensive view of acid-base chemistry with the aid of a PFE diagram that shows a whole series of acid-base systems, arranged in order of decreasing acid strengths.
How acid strengths relate to proton free energies
As we explained near the beginning of this lesson, we ordinarily express the strength of an acid by quoting its "dissociation constant" Ka, which is the equilibrium constant for transfer of a proton to the base H2O.
HA + H2O → H3O+ + A
We also introduced the concept of PFE as an alternative way of expressing the strength of an acid. If both PFE and Ka express the same thing, you may wonder how these two quantities are related.
If you have had an introduction to thermodynamics and specifically to free energy G, you might wish to look at the Appendix section of this lesson to review the relationship between the fall in free energy and the value of pKa.
A thorough answer to this question requires a bit of thermodynamics, which you may not have yet studied, but you don't really need any of the math here. It should suffice to show the results in the form of a plot, which yields a simple straight-line relationship when plotted on a semi-logarithmic scale.
Note that
• The extent of any chemical reaction is proportional to the fall in free energy when it takes place. So negative ΔG values correspond to larger equilibrium constants.
• For acid-base reactions specifically, Ka is inversely proportional to ΔG, and thus to the fall in PFE.
• This means that the proton on a strong acid (Ka > 1) has a large negative PFE; it has a "long way to fall" to the H2O level.
• Conversely, a proton on a weak acid (Ka <1) can only move up to the H2O level when supplied with thermal energy from the surroundings.
• A K value of 100 = 1 corresponds to a zero-free energy change, meaning that the reaction is at equilibrum when [HA] = [A].
• Proton-free energy changes exceeding –20 kJ mol–1 result in Ka values so large that the proton transfer can be considered "complete" for most practical purposes.
Understanding the PFE chart
On the PFE diagram shown below, each acid-base system is represented by a horizontal line connecting an acidic species with its conjugate base. Each line is placed at a height on the diagram that is proportional to its pKa, as measured on the logarithmic scale (labeled "pH") at the right. What we have, then, is a list of acid-base systems arranged in order of decreasing acid strength.
Think of this chart as an alternative view of the table of acid strengths displayed near the top of this page, in which some of the information that is "hidden" in the table is now revealed in an easily-grasped way.
Take particular note of the following points:
• The ΔG scale on the left measures the fall in potential energy when one mole of protons is transferred from a given acid to H2O. Negative values correspond to stronger acids.
• Any acid shown on the left side of the vertical line running down the center of the diagram can donate protons to any base (on the right side of the line) that appears below it.
The value of the equilibrium constant for such a reaction will exceed unity; for ΔG separations of more than about 10 kJ/mol, the reaction can be considered to be complete.
• The zero free energy assigned to the hydronium ion corresponds to the standard free energy of formation ΔGf° of this ion.
• Notice how the two conjugate pairs H3O+ / H2O and H2O/OH bracket the 80-kJ/mol region that comprises the weak acids. The weakness of these acids reflects the energetic cost of moving a proton "up-hill" to the H2O level.
• The strong acids at the top of the table can all be regarded as totally "dissociated" in water. This means that these acids cannot exist in water, as they are totally transformed into H3O+. Of course, they will also react with any bases below the H2O level that may be available. (See "leveling effect" below.)
• Similarly, the two strong bases NH2 and O2 shown at the bottom of the table cannot exist in water. Thus if solid sodium oxide Na2O is added to water, the oxide ions are such a low-energy sink that will totally fill up with protons from the H2O acid level which defines the top of the strong-base range.
• Notice the pH scale along the right side, and note the range of pH values possible in water. We will say more about the significance of pH in these diagrams farther on.
The PFE chart helps you visualize (and understand!) acid-base chemistry
The virtue of a PFE chart is that you can see, at a glance, the relations between conjugate acid-base pairs that enables you to make quick, qualitative determinations of what is going on even in complex mixtures of acids and bases. The speed and convenience this affords enables you to build a much deeper conceptual understanding of acid-base chemistry.
In order to do the same thing by consulting a simple list of acid Ka's or pKa's, you also need to be familiar with the principles of chemical equilibrium and thermodynamics — and then take the time to work through the arithmetic.
In the preceding sections we have shown how the PFE-chart approach can clarify the distinction between strong and weak acids. We will now go on to describe a variety of other principles that are easily grasped by consulting a PFE chart.
Significance of the pH scale
You will have noticed the scale labeled "pH" at the left side of the PFE plot.
First, please understand what this scale does NOT do; it does not tell you what the pH of the solution will be when you add an acid to pure water. This pH will depend on both the strength (Ka or pKa) of the acid, and on its concentration in the resulting solution; you will learn how to carry out pH calculations in later lessons.
All but the very weakest acids can drive the pH down to near the bottom of the scale if their solutions are sufficiently concentrated. The only thing you can be sure of is that the pH of a solution of a solution of an acid in pure water will never be greater than 7.
pH sets the average proton free energy
When pH was first introduced in 1909, it was defined in terms of the "hydrogen ion concentration" [H+]. This has since been amended to the hydrogen ion activity {H+}. The latter term refers to the "effective" concentration of these ions — that is to the "availability" of protons (regardless of whether they physically exist as H3O+ units or in other forms) to react with bases. But from the standpoint of thermodynamics, "availability" is just a loose term for "free energy". So we can also say that
pH is a measure of the average PFE in an aqueous solution
The more negative the pH, the higher the proton free energy in the solution. (The inverse relation is a consequence of the negative-logarithm definition of pH.)
The pH has the dual advantages of being both dimensionless and readily observable. More importantly, we can easily alter the pH of a solution by adding some strong acid or base. And when we do this, we are in effect using the pH as a tool for controlling the average PFE in the solution.
More on this in a moment; first, let us introduce an important relation that you may have seen in an earlier chemistry course, and which we will discuss more in a later lesson in this group.
pH can control the relative abundance of conjugate acid-base pairs
Recalling the equilibrium expression for a weak acid
$K_a = \dfrac{[H^+][A^-]}{[HA]} \label{3-1}$
We can solve this for $[H^+]$:
$H^+] = K_a \dfrac{[HA]}{[A^-]} \label{3-2}$
Re-writing this in terms of negative logarithms, this becomes
(3-3)
or, since pKa = – log Ka, we invert the ratio to preserve the positive sign:
We will make extensive use of this equation in a later lesson on acid-base buffering and titration.
(3-4)
This extremely important relation tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system. Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its pKa. According to the above equation, when pH = pKa, the log term becomes zero, so that the ratio
[AB] / [HA] = 100 = 1, meaning that [HA] = [AB]. In other words,
This pH adjustment can be easily made by adding appropriate amounts of a strong acid or strong base to the solution.
When the pH of a solution is set to the value of the pKa of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically:
This is the basis for the arrangement of the various acid-base systems depicted in the PFE diagram; each conjugate pair is placed at a location on the vertical pH scale that corresponds to the acid's pKa. It is instructive to consider how sensitively the concentration ratio [A] / [HA] depends on the pH.
Example $2$
Hypochlorous acid, HOCl, has a pKa of 8.0.
1. Estimate the ratio R = [OCl] / [HOCl] in a solution of sodium hypochlorite whose pH has been adjusted to 8.2.
2. If the initial concentration of the hypochlorite solution was 0.10M, what will be the concentrations of the acid and base forms at pH 8.2?
Solution
a) Substituting in Eq 3-4, the log term becomes (8.2 – 8.0) = 0.2,
so R = 100.2 = 1.6.
Note that it does not matter whether the initial solution consists of HOCl or NaOCl or some mixture thereof; adjusting the pH forces the system composition to the ratio calculated above.)
b) In the resulting solution, the base form OCl is the predominant species, the mole ratio R being 1.6/1. The mole fraction of OCl will therefore be 1.6/2.6 = 0.62, making that of HOCl (1.0 – .62) = 0.38. The actual concentrations will be 0.1 times these values, reflecting the 0.10M concentration of the solution.
Example $3$
To what pH must we adjust a solution of acetic acid (pKa 4.7) in order to convert 20 percent of the acid into its base form?
Solution:
In the final solution, the mole fractions will be [AB] = .20, [HA] = 0.80; the ratio [AB]/[HA] = 0.20/0.80 = 0.25. Substituting into Eq 2-9, we have
$pH = 4.7 + \log\' 0.25 = 4.7 – 0.6 = 4.1$
We can explore the relation between the pH and the distribution of conjugate species between the acid and base forms by carrying out calculations similar to those in Problem Example 2 for an acid-base system at different pH values.
The orange bars and blue numbers show the mole fractions of the acid and base forms at pH values of the pKa and the pKa ± 1. The green numbers give the percent dissociation of the acid at each pH.
The apparent symmetry between the two extreme pH values suggests a simple relation between the log of the ratio
[AB] / [HA] and the pH. This is seen very clearly when this ratio is plotted as a function of pH.
Solutions containing many different weak acid-base systems are very commonly encountered in nature, especially in biological fluids or natural waters, including the ocean. When the pH of such solutions is altered by the addition of strong acid or base (as can occur, for example, when wastes from industrial processes or mine drainage, or acidic rain, enter a lake or stream); the distribution of all conjugate acid-base systems will change.
It is important to note that the [AB] / [HA] ratios (calculated here from Eq 3-4) yield only a qualitative picture and do not take into account the concentrations of the different acid-base systems or proton transfers between them. The latter processes would cause the lowest proton-vacant levels (that is, the strongest bases) to be completely filled from the bottom up, depending on the number of protons available. An exact calculation would require solving a set of simultaneous equations and is beyond the scope of this lesson.
A. What do we mean by the "strength" of an acid ?
The Brønsted-Lowry model defines the strength of an acid HA as its tendency to donate protons to a base B. But bases vary in their abilities to accept protons, so the tendency for the complete transaction
HA + B → BH+ + A(A-1)
to occur depends on the two processes
HA → H+ + A K1 (A-2)
B + H+ → BH+ K2(A-3)
Ka's are always expressed on a scale relative to the Kb of the solvent
Because it is impossible to study either of these steps independently, what we might call the "absolute" strength of an acid (Eq A-2) cannot be measured. For this reason, we define a "standard" base, usually the solvent, and most commonly, water. Thus we replace Eq A-3 with
H2O + H+ → H3O+K3(A-4)
so that Eq A1 now expresses the strength of HA relative to the base H2O:
HA + H2O → H3O+ + A (A-5)
which, for convenience, we write in its abbreviated form
HA → H+ + A (A-5)
whose equilibrium constant is
(A-6)
This value of Ka is, of course, numerically identical to that for Eq A-5.
B. pH and proton free energy
This section may be of interest to readers who have studied elementary thermodynamics, and have some familiarity with Gibbs free energy. it is not required for understanding and using the concept of PFE.
From elementary thermodynamics, the driving force of a chemical reaction is given by the standard free energy change:
ΔG° = –RT ln K (B-1)
Solving this equation for K yields
(B-2)
For a weak acid we know that
(B-3)
Substituting Eq B-2 into this expression, we get
(B-4)
Next, we take the negative log of each term, recalling that ln x = 2.3 log x:
(B-5)
or
(B-6)
For the special case in which the concentrations of the conjugate pairs [H+] and [HA] are equal, pH = pKa, and the rightmost term above disappears, leaving
(B-7)
Thus the pH is a direct measure of the average proton free energy in a solution.
Example $1$: PFE level of acetic acid
Construct a PFE diagram showing a) H3O+, b) acetic acid HAc (pKa = 4.76), and c) H2O.
Solution
Solving Eq A13 for ΔG° yields ΔG° = (2.3RT) × pKa, or, at 25°C, ΔG° = (2.3 × 8.314 J K–1 mol–1) × pKa = 5698 J K–1 mol–1 × pKa
a) The proton in H3O+ is already at the level of that in the standard base H2O (Eq A5), so for this transfer is [by definition] ΔG° = 5698 J K–1 mol–1 × 0 = 0
b) The increase in the free energy required to transfer a proton from its level in HAc to that in H2O is ΔG° = 5698 J K–1 mol–1 × 4.76 = 27.1 kJ mol–1
c) To transfer a proton from H2O to another H2O (autoprotolysis), its free energy must increase by 5698 J K–1 mol–1 × 14.0 = 79.8 kJ mol–1 | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.01%3A_Introduction_to_Acid_Base_Equilibria.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts that have been presented above.
• Estimate the pH of a solution of a strong acid or base, given its concentration.
• Explain the distinction between the concentration of a substance and its activity in a solution.
• Describe the origin and effects of ion-pairing in concentrated solutions of strong acids.
• Name and write formulas for the four major strong acids.
To a good approximation, strong acids, in the forms we encounter in the laboratory and in much of the industrial world, have no real existence; they are all really solutions of $\ce{H3O^{+}}$. So if you think about it, the labels on those reagent bottles you see in the lab are not strictly true! However, if the strong acid is highly diluted, the amount of $\ce{H3O^{+}}$ it contributes to the solution becomes comparable to that which derives from the autoprotolysis of water. Under these conditions, we need to develop a more systematic way of working out equilibrium concentrations.
At Moderate Concentrations, Forget About Equilibria
A strong acid, you will recall, is one that is stronger than the hydronium ion $\ce{H3O^{+}}$. This means that in the presence of water, the proton on a strong acid such as HCl will "fall" into the "sink" provided by H2O, converting the latter into its conjugate acid $\ce{H3O^{+}}$. In other words, $\ce{H3O^{+}}$ is the strongest acid that can exist in aqueous solution.
As we explained in the preceding lesson, all strong acids appear to be equally strong in aqueous solution because there are always plenty of H2O molecules to accept their protons. This is called the "leveling effect".
This greatly simplifies our treatment of strong acids because there is no need to deal with equilibria such as for hydrochloric acid
$\ce{HCl + H_2O → H_3O^{+} + Cl^{–}}$
The equilibrium constants for such reactions are so overwhelmingly large that we can usually consider the concentrations of acid species such as "HCl" to be indistinguishable from zero. As we will see further on, this is not strictly true for highly-concentrated solutions of strong acids (Figure $3$).
Over the normal range of concentrations we commonly work with (indicated by the green shading on this plot), the pH of a strong acid solution is given by the negative logarithm of its concentration in mol L–1. Note that in very dilute solutions, the plot levels off, showing that this simple relation breaks down; there is no way you can make the solution alkaline by diluting an acid!
Example $1$
What will be the pH of a 0.025 mol/l solution of hydrochloric acid?
Solution
If we assume all the hydronium concentration originates from the added acid. So
$\ce{H3O^{+}}$
and we just find the negative logarithm of the concentration
$pH = -\log_{10} [\ce{H3o^{+}}] = –\log_{10} 0.025 = 1.6$
Since this pH is so far away from 7, our assumption is reasonable.
Major Strong Acids
Mineral acids are those that are totally inorganic. Not all mineral acids are strong; boric and carbonic acids are common examples of very weak ones. However, in ordinary usage, the term often implies one of those described below. With the exception of perchloric acid, which requires special handling, these are all widely used in industry and are almost always found in chemistry laboratories. Most have been known since ancient times
Table $1$: Mineral Acid properties
acid name pKa base pKb
HClO4 perchloric acid ~ –7 ClO4 ~ 21
HCl hydrochloric acid ~ –7 Cl ~ 17
H2SO4 sulfuric acid ~ –7 HSO4 ~ 17
HNO3 nitric acid –1 NO3 15
H3O+ hydronium ion H2O 14
You should know the names and formulas of all four of these widely-encountered strong mineral acids.
• Hydrochloric acid $HCl$: In contrast to the other strong mineral acids, no pure compound "hydrochloric acid" exists. One of the most commonly used acids, hydrochloric acid is usually sold as a constant-boiling 37% solution of hydrogen chloride gas in water, making its concentration about 10 M. Although higher concentrations are possible, the high partial pressures of HCl in equilibrium with the solution makes them difficult to store, ship and work with. Commercial-grade hydrochloric acid is often called muriatic acid.
• Nitric acid $HNO_3$: Concentrated nitric acid is a 68% constant-boiling solution of HNO3 in water, corresponding to about 11M. Although the acid itself is colorless, its slow decomposition into NO2 (especially in the presence of light) often results in a yellow or orange color. More concentrated solutions are sold as "fuming nitric acid"; one form, available as "white fuming nitric acid" or "anhydrous nitricacid" contains 97.5% HNO3 and only 2% water; the remainder consists of dissolved NO2. Pure HNO3 forms crystals that melt at –42°C. In addition to being a strong acid, nitric acid at high concentrations acts as a powerful oxidizing agent, a property that accounts for its use as a rocket fuel.
• Perchloric acid $HClO_4$: This is the strongest of the mineral acids, thanks to the electron-withdrawing action of the oxygen atoms which make it energetically easier for HClO4 to lose its proton. Its usual commercial form is a constant-boiling 72.5% solution in water. Concentrated perchloric acid is a powerful oxidizing agent that can react explosively with organic materials and certain metals. Its use in the laboratory requires special handling. A 1947 explosion of a 1000-L vat of HClO4 in a Los Angeles electroplating plant killed 17 people and damaged over 250 homes and other buildings.
• Sulfuric acid $H_2SO_4$: Sulfuric acid is by far the most industrially important acid, and is also the most concentrated one available. "Concentrated" sulfuric acid contains 98% by weight of H2SO4; its density of 1.83 kg/L and oil-like viscosity reflect this high concentration. "100% H2SO4" (which can be prepared However, is not stable) actually contains a variety of other species, all in equilibrium with H2SO4. These include H2S2O7, H2S4O13, and H2SO4's autoprotolysis products H3SO4 and HSO4+.
Super acids
There is a class of super acids that are stronger than some of the common mineral acids. According to the classical definition, a superacid is an acid with an acidity greater than that of 100% pure sulfuric acid. Some, like fluorosulfuric acid $FSO_3H$ are commercially available. Strong superacids are prepared by the combination of a strong Lewis acid and a strong Brønsted acid. The strongest known super acid is fluoroantimonic acid ($H_2FSbF_6$). This acid is so corrosive that the fumes alone will dissolve entire fume hoods, glass and plastic beakers, human skin, bone and most synthetic compounds.
Strong bases
The only strong bases that are commonly encountered are solutions of Group 1 hydroxides, mainly NaOH and KOH. Unlike most metal hydroxides, these solids are highly soluble in water, and can thus yield concentrated solutions of hydroxide ion, the strongest base that can exist in water — the ultimate aquatic proton sink. Sodium hydroxide is by far the most important of these strong bases; its common names "lye", "caustic soda" (or, in industry, often just "caustic"), reflect the diverse uses of NaOH.
Solid NaOH is usually sold in the form of pellets. When exposed to air, they become wet (deliquescence) and absorb CO2, becoming contaminated with sodium carbonate. NaOH is the most soluble of the Group 1 hydroxides, dissolving in less than its own weight of water (111 g / 100 ml) to form a 2.8 M/L solution at 20°C. However, as with strong acids, the pH of such a solution cannot be reliably calculated from such a high concentration.
Acids at High Concentrations
At higher concentrations, intermolecular interactions and ion-pairing can cause the effective concentration (known as the activity) of $\ce{H3O^{+}}$ to deviate from the value corresponding to the nominal or "analytical" concentration of the acid. Activities are important because only these work properly in equilibrium calculations. Also, pH is defined as the negative logarithm of the hydrogen ion activity, not its concentration. The relation between the concentration of a species and its activity is expressed by the activity coefficient $\gamma$:
$a = \gamma C$
As a solution becomes more dilute, $\gamma$ approaches unity. At ionic concentrations not exceeding about 2 M, concentrations of typical strong acids can generally be used in place of activities without serious error. Note that activities of single ions other than $\ce{H3O^{+}}$ cannot be determined, so activity coefficients in ionic solutions are always the average, or mean, of those for the ionic species present. This quantity is denoted as $\gamma_±$.
Table $2$: Mean ionic activity coefficient of HCl at 25° C
molarity $\gamma_±$
0.0005 0.975
0.01 0.904
0.10 0.796
1 0.809
2 1.01
5 2.38
10 10.44
12 17.25
Because activities of single ions cannot be measured, these mean values are the closest we can get to $\{H^+\}$ in solutions of strong acids.
Activity is Important for Concentrated Samples
Activity is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater.
In a 12 M solution of hydrochloric acid, for example, the mean ionic activity coefficient is 17.25. This means that under these conditions with [H3O+] = 12 M, the activity {H3O+} = 12 × 17.25 = 207, corresponding to a pH of about –2.3, instead of –1.1 as might be predicted if concentrations were being used.
These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl(g) over a concentrated hydrochloric acid solution even though this acid is supposedly 100% dissociated. It turns out that the activity {HCl} (which represents the "escaping tendency" of HCl(g) from the solution) is almost 49,000 for a 10 M solution! The source of this gas is best described as the result of ion pairing.
Similarly, in a solution prepared by adding 0.5 mole of the very strong acid HClO4 to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 M. This does not mean that the acid is only 80% dissociated; there is no evidence of HClO4 molecules in the solution. What has happened is that about 20% of the $\ce{H3O^{+}}$ and ClO4 ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces (Figure $4$).
Ion-pairing reduces effective dissociation at high concentrations. If you have worked with concentrated hydrochloric acid in the lab, you will have noticed that the choking odor of hydrogen chloride gas is very apparent. How can this happen if this strong acid is really "100 percent dissociated" as all strong acids are said to be?
At very high concentrations, too few H2O molecules are available to completely fill the extended hydration shells that normally help keep the ions apart, reducing the fraction of "free" $\ce{H3O^{+}}$ ions capable of acting independently. Under these conditions, the term "dissociation" begins to lose its meaning. Although the concentration of HCl(aq) is never very high, its own activity coefficient can be as great as 2000 (Table $2$), which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable.
Example $2$
Estimate the pH of a 10.0 M solution of hydrochloric acid in which the mean ionic activity coefficient - is 10.4.
Solution
pH = – log {H+} ≈ – log (10.4 × 10.0) = – log 104 = – 2.0
Compare this result with what you would get by using – log[H+]
Systematic treatment of strong acids
In this section, we will derive expressions that relate the pH of solution of a strong acid to its concentration in a solution of pure water. We will use hydrochloric acid as an example. When HCl gas is dissolved in water, the resulting solution contains the ions H3O+, OH, and Cl−, However, except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions.
To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them. These relations are obtained by observing that certain conditions must always be true in any solution of HCl. These are:
1. The autoprotolysis equilibrium of water must always be satisfied:
$[H_3O^+][OH^–] = K_w \label{4-1}$
2. For any acid-base system, one can write a mass balance equation that relates the concentrations of the various dissociation products of the substance to its “nominal concentration”, which we designate here as Ca. For a solution of HCl, this equation would be
$[HCl] + [Cl^–] = C_a \label{4-2}$
However, since HCl is a strong acid and therefore no "HCl" exists in the solution, we can neglect the first term, so the mass balance equation becomes simply
$[Cl^–] = C_a \label{4-3}$
3. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle.
$[H_3O^+] = [OH^–] + [Cl^–] \label{4-4}$
The next step is to combine these three equations into a single expression that relates the hydronium ion concentration to $C_a$. This is best done by starting with an equation that relates several quantities, such as Equation $\ref{4-4}$, and substituting the terms that we want to eliminate. Thus we can get rid of the [Cl−] term by substituting Equation $\ref{4-3}$ into Equation $\ref{4-4}$ :
$[H_3O^+] = [OH^–] + C_a \label{4-5}$
The [OH] term can be eliminated by use of Equation $\ref{4-3}$:
$[H_3O^+] = C_a + \dfrac{K_w}{[H_3O^+]} \label{4-6}$
This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. Notice that Equation $\ref{4-6}$ is a quadratic equation. Recalling that Kw = 10–14, it is apparent that the final term of the above equation will ordinarily be very small in comparison to the other terms, so it can be ordinarily be dropped, yielding the simple relation
$[H_3O^+] \approx C_a \label{4-7}$
Only in extremely dilute solutions, around 10–6 M or below (where the plot curves), does this approximation become untenable. However, even then, the effect is tiny. After all, the hydronium ion concentration in a solution of a strong acid can never fall below 10–7 M; no amount of dilution can make the solution alkaline! So for almost all practical purposes, Equaton $\ref{4-7}$ is all you will ever need for a solution of a strong acid. | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.02%3A_Strong_Monoprotic_Acids_and_Bases.txt |
The numerical examples and the images in this section have some problems (source text with comments).
Learning Objectives
Make sure you thoroughly understand the following essential concepts that have been presented above.
• Explain the difference between the terms Ca and [HA] as they relate to an aqueous solution of the acid HA.
• Define degree of dissociation and sketch a plot showing how the values of $alpha$ for a conjugate pair HA and A relate to each other and to the pKa.
• Derive the relationship between acid concentration and dissociation constant to the hydrogen ion concentration in a solution of a weak acid.
• Re-write the above relation in polynomial form.
• Derive the approximation that is often used to estimate the pH of a solution of a weak acid in water.
• Calculate the pH of a solution of a weak monoprotic weak acid or base, employing the "five-percent rule" to determine if the approximation 2-4 is justified.
• Predict whether an aqueous solution of a salt will be acidic or alkaline, and explain why by writing an appropriate equation.
Most acids are weak; there are hundreds of thousands of them, whereas there are fewer than a dozen strong acids. We can treat weak acid solutions in much the same general way as we did for strong acids. The only difference is that we must now take into account the incomplete "dissociation"of the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which salts of the acid are present. The latter mixtures are known as buffer solutions and are extremely important in chemistry, physiology, industry and in the environment. To keep our notation as simple as possible, we will refer to “hydrogen ions” and [H+] for brevity, and, wherever it is practical to do so, will assume that the acid HA "ionizes" or "dissociates" into H+ and its conjugate base A−.
1 Aqueous solutions of weak acids or bases
A weak acid (represented here as HA) is one in which the reaction
$HA \rightleftharpoons A^– + H^+ \label{1-1}$
is incomplete. This means that if we add 1 mole of the pure acid HA to water and make the total volume 1 L, the equilibrium concentration of the conjugate base A will be smaller (often much smaller) than 1 M/L, while that of undissociated HA will be only slightly less than 1 M/L. Equation $\ref{1-1}$ tells us that dissociation of a weak acid HA in pure water yields identical concentrations of its conjugate species. Let us represent these concentrations by x. Then, in our "1 M " solution, the concentration of each species is as shown here:
(1-2)
When dealing with problems involving acids or bases, bear in mind that when we speak of "the concentration", we usually mean the nominal or analytical concentration which is commonly denoted by Ca. For example, for a solution made by combining 0.10 mol of pure formic acid HCOOH with sufficient water to make up a volume of 1.0 L, Ca = 0.10 M. However, we know that the concentration of the actual species [HCOOH] will be smaller the 0.10 M because some it ends up as the formate ion HCOO. It will, of course, always be the case that the sum
$[HCOOH] + [HCOO^–] = C_a$
For the general case of an acid HA, we can write a mass balance equation
$C_a = [HA] + [A^–] \label{1-3}$
which reminds us the "A" part of the acid must always be somewhere!
For a strong acid such as hydrochloric, its total dissociation means that [HCl] = 0, so the mass balance relationship in Equation $\ref{1-3}$ reduces to the trivial expression Ca = [Cl-]. Any acid for which [HA] > 0 is by definition a weak acid.
Similarly, for a base B we can write
$C_b = [B] + [HB^+] \label{1-4}$
Note
"Concentration of the acid" and [HA] are typical not the same.
Equilibrium concentrations of the acid and its conjugate base
According to the above equations, the equilibrium concentrations of A and H+ will be identical (as long as the acid is not so weak or dilute that we can neglect the small quantity of H+ contributed by the autoprotolysis of H2O). Let us represent these concentrations by x. Then, in a solution containing 1 M/L of a weak acid, the concentration of each species is as shown here:
(1-5)
Substituting these values into the equilibrium expression for this reaction, we obtain
$\dfrac{[A^-][H^+]}{[HA]} = \dfrac{x^2}{1-x} \label{1-6}$
In order to predict the pH of this solution, we must solve for x. The presence of terms in both x and x 2 here tells us that this is a quadratic equation. In most practical cases in which Ka is 10–4 or smaller, we can assume that x is much smaller than 1 M, allowing us to make the simplifying approximation
$(1 – x) \approx 1 \label{1-7}$
so that
$x^2 \approx Ka\] and thus $x = \sqrt{K_a} \label{1-8}$ This approximation will not generally be valid when the acid is very weak or very dilute. Solutions of arbitrary concentration The above development was for a solution made by taking 1 mole of acid and adding sufficient water to make its volume 1.0 L. In such a solution, the nominal concentration of the acid, denoted by Ca, is 1 M. We can easily generalize this to solutions in which Ca has any value: (1-9) The above relation is known as a "mass balance on A". It expresses the simple fact that the "A" part of the acid must always be somewhere — either attached to the hydrogen, or in the form of the hydrated anion A. The corresponding equilibrium expression is (1-10) and the approximations (when justified) 1-3a and 1-3b become (Cax) ≈ Ca (1-11) x ≈ (Ca Ka)½(1-12) Example \(1$: Aproximate pH of an acetic acid solution
Estimate the pH of a 0.20 M solution of acetic acid, Ka = 1.8 × 10–5.
Solution
For brevity, we will represent acetic acid CH3COOH as HAc, and the acetate ion by Ac. As before, we set x = [H+] = [Ac], neglecting the tiny quantity of H+ that comes from the dissociation of water.
Substitution into the equilibrium expression yields
The rather small value of Ka suggests that we can drop the x term in the denominator, so that
(x2 / 0.20) ≈ 1.8E-5 or x ≈ (0.20 × 1.8E–5)½ = 1.9E-3 M
The pH of the solution is
–log (1.9E–3) = 2.7
Degree of dissociation
Even though we know that the process HA → H+ + A does not correctly describe the transfer of a proton to H2O, chemists still find it convenient to use the term "ionization" or "dissociation". The "degree of dissociation" (denoted by $\alpha$ of a weak acid is just the fraction
$\alpha = \dfrac{[\ce{A^{-}}]}{C_a} \label{1-13}$
which is often expressed as a per cent ($\alpha$ × 100).
Degree of dissociation depends on the concentration
It's important to understand that whereas Ka for a given acid is essentially a constant, $\alpha$ will depend on the concentration of the acid. Note that these equations are also valid for weak bases if Kb and Cb are used in place of Ka and Ca.
This can be shown by substituting Eq 5 into the expression for Ka:
(1-14)
Solving this for $\alpha$ results in a quadratic equation, but if the acid is sufficiently weak that we can say (1 – ) ≈ 1, the above relation becomes
(1-15)
the amount of HA that dissociates varies inversely with the square root of the concentration; as Ca approaches zero, $\alpha$ approaches unity and [HA] approaches Ca.
This principle is an instance of the Ostwald dilution law which relates the dissociation constant of a weak electrolyte (in this case, a weak acid), its degree of dissociation, and the concentration.
A common, but incorrect explanation of this law in terms of the Le Chatelier principle states that dilution increases the concentration of water in the equation HA + H2O → H3O+ + A, thereby causing this equilibrium to shift to the right. The error here is that [H2O] in most aqueous solutions is so large (55.5 M) that it can be considered constant; this is the reason the [H2O] term does not appear in the expression for Ka. Another common explanation is that dilution reduces [H3O+] and [A], thus shifting the dissociation process to the right. However, dilution similarly reduces [HA], which would shift the process to the left. In fact, these two processes compete, but the former has greater effect because two species are involved.
It is probably more satisfactory to avoid Le Chatelier-type arguments altogether, and regard the dilution law as an entropy effect, a consequence of the greater dispersal of thermal energy throughout the system. This energy is carried by the molecular units within the solution; dissociation of each HA unit produces two new particles which then go their own ways, thus spreading (or "diluting") the thermal energy more extensively and massively increasing the number of energetically-equivalent microscopic states, of which entropy is a measure. (More on this here)
Degree of dissociation depends on the pH
Plots of this kind are discussed in more detail in the next lesson in this set under the heading ionization fractions.
The Le Chatelier principle predicts that the extent of the reaction
$\ce{HA → H^{+} + A^{–} }$
will be affected by the hydrogen ion concentration, and thus by the pH. This is illustrated here for the ammonium ion. Notice that when the pH is the same as the pKa, the concentrations of the acid- and base forms of the conjugate pair are identical.
Example $1$: percent dissociation
A 0.75 M solution of an acid HA has a pH of 1.6. What is its percent dissociation?
Solution
The dissociation stoichiometry HA → H+ + AB tells us the concentrations [H+] and [A] will be identical. Thus [H+] = 10–1.6 = 0.025 M = [A]. The dissociation fraction
$α = \dfrac{[\ce{A^{–}}]}{[\ce{HA}]} = \dfrac{0.025}{0.75} = 0.033$
and thus the acid is 3.3% dissociated at 0.75 M concentration.
Sometimes the percent dissociation is given, and Ka must be evaluated.
Example $3$: Ka from degree of dissociation
A weak acid HA is 2 percent dissociated in a 1.00 M solution. Find the value of Ka.
Solution
The equilibrium concentration of HA will be 2% smaller than its nominal concentration, so [HA] = 0.98 M, [A] = [H+] = 0.02 M. Substituting these values into the equilibrium expression gives
"Concentration of the acid" and [HA] are not the same
When dealing with problems involving acids or bases, bear in mind that when we speak of "the concentration", we usually mean the nominal or analytical concentration which is commonly denoted by Ca.
So for a solution made by combining 0.10 mol of pure formic acid HCOOH with sufficient water to make up a volume of 1.0 L, Ca = 0.10 M. However, we know that the concentration of the actual species [HCOOH] will be smaller the 0.10 M because some it ends up as the formate ion HCOO. However, it will always be the case that the sum
[HCOOH] + [HCOO] = Ca.
For the general case of an acid HA, we can write a mass balance equation
Ca = [HA] + [A](1-16)
which reminds us the "A" part of the acid must always be somewhere!
Similarly, for a base B we can write
Cb = [B] + [HB+](1-17)
Degree of dissociation varies inversely with the concentration
If we represent the dissociation of a Ca M solution of a weak acid by
(1-18)
then its dissociation constant is given by
(1-19)
Because the Ca term is in the denominator here, we see that the amount of HA that dissociates varies inversely with the concentration; as Ca approaches zero, [HA] approaches Ca.
If we represent the fraction of the acid that is dissociated as
(1-20)
then Eq 8 becomes
(1-21)
If the acid is sufficiently weak that x does not exceed 5% of Ca,
the -term in the denominator can be dropped, yielding
KaCa 2 (1-22)
Note that the above equations are also valid for weak bases if Kb and Cb are used in place of Ka and Ca.
Example $1$: effects of dilution
Compare the percent dissociation of 0.10 M and .0010 M solutions of boric acid ($K_a = 3.8 \times 10^{-10}$).
Solution
Boric acid is sufficiently weak that we can use the approximation of Eq 1-22 to calculate a:= (5.8E–10 / .1)½ = 7.5E-5; multiply by 100 to get .0075 % diss. For the more dilute acid, a similar calculation yields 7.6E–4, or 0.76%.
2 Carrying out acid-base calculations
In Problem Example 1, we calculated the pH of a monoprotic acid solution, making use of an approximation in order to avoid the need to solve a quadratic equation. This raises the question: how "exact" must calculations of pH be? It turns out that the relation between pH and the nominal concentration of an acid, base, or salt (and especially arbitrary mixtures of these) can become quite complicated, requiring the solution of sets of simultaneous equations.
However, for almost all practical applications, one can make some approximations that simplify the math without detracting significantly from the accuracy of the results.
Equilibrium constants are rarely exactly known
As we pointed out in the preceding lesson, the "effective" value of an equilibrium constant (the activity) will generally be different from the value given in tables in all but the most dilute ionic solutions. Even if the acid or base itself is dilute, the presence of other "spectator" ions such as Na+ at concentrations much in excess of 0.001 M can introduce error.
The usual advice is to consider Ka values to be accurate to ±5 percent at best, and even more uncertain when total ionic concentrations exceed 0.1 M. As a consequence of this uncertainty, there is generally little practical reason to express the results of a pH calculation to more than two significant digits.
Finding the pH of a solution of a weak monoprotic acid
This is by far the most common type of problem you will encounter in a first-year Chemistry class. You are given the concentration of the acid, expressed as Ca moles/L, and are asked to find the pH of the solution. The very important first step is to make sure you understand the problem by writing down the equation expressing the concentrations of each species in terms of a single unknown, which we represent here by x:
(2-1)
Substituting these values into the expression for $K_a$, we obtain
(2-2)
Don't bother to memorize these equations! If you understand the concept of mass balance on "A" expressed in (2-1), and can write the expression for Ka, you just substitute the x's into the latter, and you're off! If you feel the need to memorize stuff you don't need, it is likely that you don't really understand the material — and that should be a real worry!
In order to predict the pH of this solution, we must first find [H+], that is, x. The presence of terms in both x and x 2 here tells us that this is a quadratic equation. This can be rearranged into x 2 = Ka (1 – x) which, when written in standard polynomial form, becomes the quadratic
$[\ce{H^{+}}]^2 – C_a [H^{+}] – K_w = 0 \label{2-3}$
However, don't panic! As we will explain farther on, in most practical cases we can make some simplifying approximations which eliminate the need to solve a quadratic. And when, as occasionally happens, a quadratic is unavoidable, we will show you some relatively painless ways of dealing with it.
How to deal with Quadratic Equations
What you do will depend on what tools you have available. If you are only armed with a simple calculator, then there is always the venerable quadratic formula that you may have learned about in high school, but if at all possible, you should avoid it: its direct use in the present context is somewhat laborious and susceptible to error.
Use of the standard quadratic formula on a computer or programmable calculator can lead to weird results!
The reason for this is that if b2 >> |4ac|, one of the roots will require the subtraction of two terms whose values are very close; this can lead to considerable error when carried out by software that has finite precision.
One can get around this by computing the quantity
$Q = –\dfrac{b + \pm (b) \sqrt{ b^2 – 4ac}}{2}$
from which the roots are x1= Q /a and x2 = c /Q. (See any textbook on numerical computing for more on this and other metnods.)
However, who want's to bother with this stuff in order to solve typical chemistry problems? Better to avoid quadratics altogether if at all possible!
Remember: there are always two values of x (two roots) that satisfy a quadratic equation. For all acid-base equilibrium calculations that are properly set up, these roots will be real, and only one will be positive; this is the one you take as the answer.
Approximations, judiciously applied, simplify the math
We have already encountered two of these approximations in the examples of the preceding section:
1. In all but the most dilute solutions, we can assume that the dissociation of the acid HA is the sole source of H+, with the contribution due to water autoprotolysis being negligible.
2. We were able to simplify the equilibrium expressions by assuming that the x-term, representing the quantity of acid dissociated, is so small compared to the nominal concentration of the acid Ca that it can be neglected. Thus in Problem Example 1, the term in the denominator that has the form (0.1 - x) , representing the equilibrium concentration of the undissociated acid, is replaced by 0.1.
Most people working in the field of practical chemistry will never encounter situations in which the first of these approximations is invalid. This is not the case, however, for the second one.
Should I drop the x, or forge ahead with the quadratic form?
If the acid is fairly concentrated (usually with Ca > 10–3 M) and sufficiently weak that most of it remains in its protonated form HA, then the concentration of H+ it produces may be sufficiently small that the expression for Ka reduces to
Ka [H+]2 / Ca
so that
[H+] (KaCa)½(2-4)
This can be a great convenience because it avoids the need to solve a quadratic equation. However, it also exposes you to the danger that this approximation may not be justified.
The usual advice is that if this first approximation of x exceeds 5 percent of the value it is being subtracted from (0.10 in the present case), then the approximation is not justified. We will call this the "five percent rule".
Example $5$:
Problem Example 5 - pH and degree of dissociation
1. Estimate the pH of a 0.20 M solution of acetic acid, Ka = 1.8 × 10–5.
2. What percentage of the acid is dissociated?
Solution
For brevity, we will represent acetic acid CH3COOH as HAc, and the acetate ion by Ac. As before, we set x = [H+] = [Ac], neglecting the tiny quantity of H+ that comes from the dissociation of water.
Substitution into the equilibrium expression yields
$K_a = x^2/(0.20 - x) Can we simplify this by applying the approximation 0.20 – x ≈ 0.20 ? Looking at the number on the right side of this equation, we note that it is quite small. This means the left side must be equally small, which requires that the denominator be fairly large, so we can probably get away with dropping x. Doing so yields (x2 / 0.20) = 1.8E-5 or x = (0.20 × 1.8E–5)½ = 1.9E-3 M The "5 per cent rule" requires that the above result be no greater than 5% of 0.20, or 0.010. Because 0.0019 meets this condition, we can set x = [H+] ≈ 1.9 × 10–3 M, and the pH will be –log (1.9 × 10–3) = 2.7 b) Percent dissociation: 100% × (1.9 × 10–3 M) / (0.20 M) = 0.95% This plot shows the combinations of Ka and Ca that generally yield satisfactory results with the approximation of Eq 4. Weak bases are treated in an exactly analogous way: Example $1$: Weak base Methylamine CH3NH2 is a gas whose odor is noticed around decaying fish. A solution of CH3NH2 in water acts as a weak base. A 0.10 M solution of this amine in water is found to be 6.4% ionized. Use this information to find \Kb and pKb for methylamine. Solution: When methylamine "ionizes", it takes up a proton from water, forming the methylaminium ion: CH3NH2 + H2O → CH3NH3+ + OH Let x = [CH3NH3+] = [OH] = .064 × 0.10 = 0.0064 [CH3NH2] = (0.10 – .064) = 0.094 Substitute these values into equilibrium expression for \Kb: \[K_b = \frac{0.0064^2}{0.094}$
To make sure we can stop here, we note that (3.6E4 / .01) = .036; this is smaller than .05, so we pass the 5% rule and can use the approximation and drop the
x-term in the denominator.
pKb = – log \Kb = – log (4.4 × 10–10) = 3.36
However, one does not always get off so easily!
Example $6$: pH of a chloric acid solution
With a Ka of 0.010, HClO2 is one of the "stronger" weak acids, thanks to the two oxygen atoms whose electronegativity withdraws some negative charge from the chlorine atom, making it easier for the hydrogen to depart as a proton.
Find the pH of a 0.10 M solution of chloric acid in pure water.
$0.010 = \frac{x^2}{0.10 - x}$ (i)
The approximation 0.10 – x ≈ 0.10 gives us
x ≈ (Ka Ca)½ = (0.010 ×0.10)½ = (.001)½ = .032 (ii)
The difficulty, in this case, arises from the numerical value of Ka differing from the nominal concentration 0.10 M by only a factor of 10. As a result,
x / Ca = .032 / 0.10 = 0.32 which clearly exceeds the 5% limit; we have no choice but to face the full monte of the quadratic solution. (see Problem Example 8 below).
Successive approximations will get you there with minimal math
In the method of successive approximations, you start with the value of [H+] (that is, x) you calculated according to (2-4), which becomes the first approximation. You then substitute this into (2-2), which you solve to get a second approximation. This cycle is repeated until differences between successive answers become small enough to ignore.
Example $1$: Method of successive approximations
Estimate the pH of a 0.10 M aqueous solution of HClO2, Ka = 0.010, using the method of successive approximations.
Solution: The equilibrium condition is
$K_{\mathrm{a}} = \dfrac{x \cdot x}{c_{\mathrm{a}} - x}$
We solve this for x (neglecting the x in the denominator), resulting in the first approximation x1, and then successively plug each result into the previous equation, yielding approximations x2 and x3:
$x_1 = \sqrt{K_{\mathrm{a}} \cdot c_\mathrm{a}} =0.032$
$x_2 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_1)} =0.026$
$x_3 = \sqrt{K_{\mathrm{a}} \cdot (c_\mathrm{a} - x_2)} =0.027$
The last two approximations x2 and x3 are within 5% of each other.
Note that if we had used x1 as the answer, the error would have been 18%.
(An exact numeric solution yields the roots 0.027016 and –0.037016)
Use a graphic calculator or computer to find the positive root
The real roots of a polynomial equation can be found simply by plotting its value as a function of one of the variables it contains.
In this example, the pH of a 10–6 M solution of hypochlorous acid (HOCl, Ka = 2.9E–8) was found by plotting the value of
y = ax2 + bx + c, whose roots are the two values of x that correspond to y = 0.
This method generally requires a bit of informed trial-and-error to make the locations of the roots visible within the scale of the axes.
Be lazy, and use an on-line quadratic equation solver
If you google "quadratic equation solver", you will find numerous on-line sites that offer quick-and-easy "fill-in-the-blanks" solutions.
Unfortunately, few of these will be useful for acid-base problems involving numbers that must be expressed in "E-notation" (e.g., 2.7E-11.) Of those that do, the one at the MathIsFun site is highly recommended; others can be found here and at the Quad2Deg site.
If you can access a quad equation solver on your personal electronic device or through the Internet, this is quick and painless. All you need to do is write the equation in polynomial form ax2 + bx + c = 0, insert values for
a, b, and c, and away you go!
This is so easy, that many people prefer to avoid the "5% test" altogether, and go straight to an exact solution. However, make sure can do the 5%-thing for exams where Internet-accessible devices are not permitted!
Example $1$: chloric acid, again
Estimate the pH of a 0.10 M aqueous solution of HClO2, Ka = 0.010.
Solution
The reaction equation HClO2 → H+ + ClO2– defines the equilibrium expression
$0.010 = \dfrac{x \cdot x}{0.10 - x}$
Multiplying by the denominator yields
x2 = 0.010 × (0.10 – x) = .0010 – .01 x which we arrange into standard polynomial form:
x2 + 0.01 x – 0.0010 = 0
Entering the coefficients {1 .01 –.001} into an online quad solver yields the roots
.027 and –.037. Taking the positive one, we have [H+] = .027 M;
the solution pH is – log .027 = 1.6.
Note: a common error is to forget to enter the minus sign for the last term; try doing this and watch the program blow up!
Avoid math altogether and make a log-C vs pH plot
This is not only simple to do (all you need is a scrap of paper and a straightedge), but it will give you far more insight into what's going on, especially in polyprotic systems. All explained in Section 3 of the next lesson.
Solutions of salts
Most salts do not form pH-neutral solutions
Salts such as sodium chloride that can be made by combining a strong acid (HCl) with a strong base (NaOH, KOH) have a neutral pH, but these are exceptions to the general rule that solutions of most salts are mildly acidic or alkaline.
Salts of a strong base and a weak acid yield alkaline solutions.
"Hydro-lysis" literally means "water splitting", as exemplified by the reaction A + H2O → HA + OH. The term describes what was believed to happen prior to the development of the Brønsted-Lowry proton transfer model.
This important property has historically been known as hydrolysis — a term still used by chemists.
Some examples:
Potassium Cyanide
KCN can be thought of the salt made by combining the strong base KOH with the weak acid HCN: K+ + OH → KOH. When solid KCN dissolves in water, this process is reversed, yielding a solution of the two ions. K+, being a "strong ion", does not react with water. However, the "weak ion" CN, being the conjugate base of a weak acid, will have a tendency to abstract protons from water: CN + H2O → HCN + OH, causing the solution to remain slightly alkaline.
Sodium bicarbonate
NaHCO3 (more properly known as sodium hydrogen carbonate) dissolves in water to yield a solution of the hydrogen carbonate ion HCO3. This ion is an ampholyte — that is, it is both the conjugate base of the weak carbonic acid H2CO3, as well as the conjugate acid of the carbonate ion CO32:
The HCO3 ion is therefore amphiprotic: it can both accept and donate protons, so both processes take place:
However, if we compare the Ka and Kb of HCO3, it is apparent that its basic nature wins out, so a solution of NaHCO3 will be slightly alkaline. (The value of pKb is found by recalling that Ka + Kb = 14.)
Sodium acetate
A solution of CH3COONa serves as a model for the strong-ion salt of any organic acid, since all such acids are weak: CH3COO + H2O → CH3COOH + OH
Salts of most cations (positive ions) give acidic solutions
The protons can either come from the cation itself (as with the ammonium ion NH4+), or from waters of hydration that are attached to a metallic ion.
This latter effect happens with virtually all salts of metals beyond Group I; it is especially noticeable for salts of transition ions such as hexaaquoiron(III) ("ferric ion"):
Fe(H2O)63+ → Fe(H2O)5OH2+ + H+ (2-5)
This comes about because the positive field of the metal enhances the ability of H2O molecules in the hydration shell to lose protons. In the case of the hexahyrated ion shown above, a whose succession of similar steps can occur, but for most practical purposes only the first step is significant.
Example $10$: Aluminum chloride solution
Find the pH of a 0.15 M solution of aluminum chloride.
Solution
The aluminum ion exists in water as hexaaquoaluminum Al(H2O)63+, whose pKa = 4.9, Ka = 10–4.9 = 1.3E–5. Setting x = [H+] = [Al(H2O)5OH 2+], the equilibrium expression is
Using the above approximation, we get
x ≈ (1.96E–6)½ = 1.4E–3, corresponding to pH = 2.8.
Finally, we compute x/Ca = 1.4E–3 ÷ 0.15 = .012 confirming that we are within the "5% rule".
The only commonly-encountered salts in which the proton is donated by the cation itself are those of the ammonium ion:
$\ce{NH_4^{+}→ NH)3(aq) + H^{+}\lable{2-6}$
Example $11$: Ammonium chloride solution
Calculate the pH of a 0.15 M solution of NH4Cl. The ammonium ion Ka is 5.5E–10.
Solution
According to Eq 6 above, we can set [NH3] = [H+] = x, obtaining the same equilibrium expression as in the preceding problem.
Because Ka is quite small, we can safely use the approximation 0.15 - 1 ≈ .015, which yields pH = –log 0.90E–5 = 5.0
Most salts of weak acids form alkaline solutions
As indicated in the example, such equilibria strongly favor the left side; the stronger the acid HA, the less alkaline the salt solution will be.
Because an ion derived from a weak acid such as HF is the conjugate base of that acid, it should not surprise you that a salt such as NaF forms an alkaline solution, even if the equilibrium greatly favors the left side::
F + H2O HF + OH(2-7)
Example $1$:
Solution of sodium fluoride
Find the pH of a 0.15 M solution of NaF. (HF Ka = 6.7E–4)
Solution: The reaction is F- + H2O = HF + OH; because HF is a weak acid, the equilibrium strongly favors the right side. The usual approximation yields
However, on calculating x/Ca = .01 ÷ 0.15 = .07, we find that this does not meet the "5% rule" for the validity of the approximation. We therefore expand the equilibrium expression
into standard polynomial form x2 + 6.7E–4 x – 1.0E–4 = 0 and enter the coefficients {1 6.7E–4 –.0001} into a quadratic solver. This yields the positive root x = 0.0099 which turns out to be sufficiently close to the approximation that we could have retained it after all.. perhaps 5% is a bit too restrictive for 2-significant digit calculations!
What about a salt of a weak acid and a weak base?
A salt of a weak acid gives an alkaline solution, while that of a weak base yields an acidic solution. What happens if we dissolve a salt of a weak acid and a weak base in water? Ah, this can get a bit tricky! Nevertheless, this situation arises very frequently in applications as diverse as physiological chemistry and geochemistry.
As an example of how one might approach such a problem, consider a solution of ammonium formate, which contains the ions NH4+ and HCOO-. Formic acid, the simplest organic acid, has a pKa of 3.7; for NH4+, pKa = 9.3.
Three equilibria involving these ions are possible here; in addition to the reactions of the ammonium and formate ions with water, we must also take into account their tendency to react with each other to form the parent neutral species:
NH4+ → NH3 + H+ K1 = 10–9.3
HCOO + H2O → HCOOH + OH K2 = (1O–14/10–3.7) = 10–10.3
NH4+ + HCOO → NH3 + HCOOH K3
Inspection reveals that the last equation above is the sum of the first two,plus the reverse of the dissociation of water
H+ + OH → H2O K4 = 1/Kw
The value of K3 is therefore
(2-8)
A rigorous treatment of this system would require that we solve these equations simultaneously with the charge balance and the two mass balance equations. However, because K3 is several orders of magnitude greater than K1 or K2, we can greatly simplify things by neglecting the other equilibria and considering only the reaction between the ammonium and formate ions.
Notice that the products of this reaction will tend to suppress the extent of the first two equilibria, reducing their importance even more than the relative values of the equilibrium constants would indicate.
Example $1$:
ammonium formate
Estimate the pH of a 0.0100 M solution of ammonium formate in water.
Solution: From the stoichiometry of HCOONH4,
[NH4+] = [HCOO] and [NH3] = [HCOOH] (i)
then, from Eq 8 above,
(ii)
in which Kb is the base constant of ammonia, Kw /10–9.3.
From the formic acid dissociation equilibrium we have
(iii)
We now rewrite the expression for K3
(iv)
which yields
(v)
and thus the pH is 6.5
What is interesting about this last example is that the pH of the solution is apparently independent of the concentration of the salt. If Ka = Kb, then this is always true and the solution will be neutral (neglecting activity effects in solutions of high ionic strength). Otherwise, it is only an approximation that remains valid as long as the salt concentration is substantially larger than the magnitude of either equilibrium constant. Clearly, the pH of any solution must approach that of pure water as the solution becomes more dilute.
Salts of analyte ions
Polyprotic acids form multiple anions; those that can themselves donate protons, and are thus amphiprotic, are called analytes. The most widely known of these is the bicarbonate (hydrogen carbonate) ion, HCO3, which we commonly know in the form of its sodium salt NaHCO3 as baking soda.
The other analyte series that is widely encountered, especially in biochemistry, is those derived from phosphoric acid:
The solutions of analyte ions we most often need to deal with are the of "strong ions", usually Na+, but sometimes those of Group 2 cations such as Ca2+.
The exact treatment of these systems is generally rather complicated, but for the special cases in which the successive Ka's of the parent acid are separated by several orders of magnitude (as in the two systems illustrated above), a series of approximations reduces the problem to the simple expression
(2-9)
which, you will notice, as with the salt of a weak acid and a weak base discussed in the preceding subsection predicts that the pH is independent of the salt's concentration. This, of course, is a sure indication that this treatment is incomplete. Fortunately, however, it works reasonably well for most practical purposes, which commonly involve buffer solutions.
Exact treatment of solutions of weak acids and bases
When dealing with acid-base systems having very small Ka's, and/or solutions that are extremely dilute, it may be necessary to consider all the species present in the solution, including minor ones such as OH.
This is almost never required in first-year courses. However, for students in more advanced courses, this "comprehensive approach" (as it is often called) illustrates the important general methodology of dealing with more complex equilibrium problems. It also shows explicitly how making various approximations gradually simplifies the treatment of more complex systems.
In order to keep the size of the present lesson within reasonable bounds (and to shield the sensitive eyes of beginners from the shock of confronting simultaneous equations), this material has been placed in a separate lesson.
3 Solutions of polyprotic acids
A diprotic acid H2A can donate its protons in two steps:
H2A → HA →HA
and similarly, for a tripotic acid H3A:
H3A → H2A → HA2 → A3–
In general, we can expect Ka2 for the "second ionization" to be smaller than Ka1 for the first step because it is more difficult to remove a proton from a negatively charged species. The magnitude of this difference depends very much on whether the two removable protons are linked to the same atom, or to separate atoms spaced farther apart.
Some polyprotic acids you should know
These acids are listed in the order of decreasing Ka1. The numbers above the arrows show the successive Ka's of each acid.
• Notice how successive Ka's of each acid become smaller, and how their ratios relate to the structures of each acid.
• Sulfuric acid is the only strong polyprotic acid you are likely to encounter.
• Sulfurous acid has never been detected; what we refer to as H2SO3 is more properly represented as a hydrated form of sulfur dioxide: which dissociates into the hydrosulfite ion:
SO2·H2O HSO3 + H+ . The ion HSO3 exists only in solution; solid bisulfite salts are not known.
• Similarly, pure carbonic acid has never been isolated, but it does exist as a minority species in an aqueous solution of CO2: [CO2(aq)] = 650[H2CO3]. The formula H2CO3 ordinarily represents the combination CO2(aq) and "true" H2CO3. The latter is actually about a thousand times stronger than is indicated by the pKa of 6.3, which is the weighted average of the equilibrium mixture.
pH of a polyprotic acid (LindaHanson, 17 min)
Example $1$: Comparison of two diprotic acids
Compare the successive pKa values of sulfuric and oxalic acids (see their structures in the box, above right), and explain why they should be so different.
Solution: The two pKa values of sulfuric acid differ by 3.0 – (–1.9) = 4.9, whereas for oxalic acid the difference is 1.3 – (–4.3) = 3.0. That's a difference of almost 100 between the two Ka's. Removal of a second proton from a molecule that already carries some negative charge is always expected to be less favorable energetically. In sulfuric acid, the two protons come from –OH groups connected to the same sulfur atom, so the negative charge that impedes loss of the second proton is more localized near the site of its removal. In oxalic acid, the two protons are removed from –OH groups attached to separate carbon atoms, so the negative charge of the mono-negative ions will exert less restraint on loss of the second proton.
Solutions of polyprotic acids in pure water
With the exception of sulfuric acid (and some other seldom-encountered strong diprotic acids), most polyprotic acids have sufficiently small Ka1 values that their aqueous solutions contain significant concentrations of the free acid as well as of the various dissociated anions.
Thus for a typical diprotic acid H2A, we must consider the three coupled equilibria
$\ce{H_2A → H^+ + HA^{–} }\,\,\, K_1$
$\ce{HA^{–} → H^{+} + HA^{2–}} \,\,\,K_2$
$\ce{H_2O → H+ + OH– } \,\,\,K_w$
An exact treatment of such a system of four unknowns [H2A], [HA], [A2–] and [H+] requires the solution of a quartic equation. If we include [OH], it's even worse! In most practical cases, we can make some simplifying approximations:
• Unless the solution is extremely dilute or K1 (and all the subsequent K's) are extremely small, we can forget that any hydroxide ions are present at all.
• If $K_1$ is quite small, and the ratios of succeeding K's are reasonably large, we may be able, without introducing too much error, to neglect the other K's and treat the acid as monoprotic.
In addition to the three equilibria listed above, a solution of a polyprotic acid in pure water is subject to the following two conditions:
Material balance: although the distribution of species between the acid form H2A and its base forms HAB and A2– may vary, their sum (defined as the "total acid concentration" Ca is a constant for a particular solution:
Ca = [H2A] + [HA] + [A2–]
Charge balance: The solution may not possess a net electrical charge:
[H3O+] = [OH] + [HA] + 2 [A2–]
Why do we multiply [A2–] by 2? Two moles of H3O+ are needed in order to balance out the charge of 1 mole of A2–.
Simplified treatment of polyprotic acid solutions
The calculations shown in this section are all you need for the polyprotic acid problems encountered in most first-year college chemistry courses. More advanced courses may require the more exact methods in Lesson 7.
Because the successive equilibrium constants for most of the weak polyprotic acids we ordinarily deal with diminish by several orders of magnitude, we can usually get away with considering only the first ionization step. To the extent that this is true, there is nothing really new to learn here. However, without getting into a lot of complicated arithmetic, we can often go farther and estimate the additional quantity of H+ produced by the second ionization step.
Example $14$: Solution of CO2 in water
a) Calculate the pH of a 0.050 M solution of CO2 in water.
For H2CO3, K1 = 10–6.4 = 4.5E–7, K2 = 10–10.3 = 1.0E–14.
b) Estimate the concentration of carbonate ion CO32 in the solution.
Solution
a) Because K1 and K2 differ by almost four orders of magnitude, we will initially neglect the second dissociation step. Because this latter step produces only a tiny additional concentration of H+, we can assume that [H+] = [HCO3] = x:
(i)
Can we further simplify this expression by dropping the x in the denominator? Let's try:
x = [0.05 × (4.5E–7)]½ = 1.5E–4.
Applying the "5-percent test", the quotient x/Ca must not exceed 0.05. Working this out yields (1.5E–4)/(.05) = .003, so we can avoid a quadratic.
x = [H+] ≈ (KaCa)½ = [(4.5E–7) × .01]½ = (.001)½ = 0.032 M,
and the
pH = – log .032 = 1.5.
b) We now wish to estimate [CO32] ≡ x.
Examining the second dissociation step, it is evident that this will consume x mol/L of HCO3, and produce an equivalent amount of H+ which adds to the quantity we calculated in (a).
(ii)
Owing to the very small value of K2 compared to K1, we can assume that the concentrations of HCO3 and H+ produced in the first dissociation step will not be significantly altered in this step. This allows us to simplify the equilibrium constant expression and solve directly for [CO32]:
(iii)
It is of course no coincidence that this estimate of [CO32] yields a value identical with K2; this is entirely a consequence of the simplifying assumptions we have made. Nevertheless, as long as K2 << K1 and the solution is not highly dilute, the result will be sufficiently accurate for most purposes.
Sulfuric acid: A Special Case
Although this is a strong acid, it is also diprotic, and in its second dissociation step it acts as a weak acid. Because sulfuric acid is so widely employed both in the laboratory and industry, it is useful to examine the result of taking its second dissociation into consideration.
Example $1$: solution of H2SO4
Estimate the pH of a 0.010 M solution of H2SO4. K1 = 103, K2 = 0.012
Solution: Because K1 > 1, we can assume that a solution of this acid will be completely dissociated into H3O+ and bisulfite ions HSO4. Thus the only equilibrium we need to consider is the dissociation of a 0.010 M solution of bisulfite ions.
HSO4 + H2O → SO42 + H3O+
Setting [H+] = [SO42] = x, and dropping x from the denominator, yields
x ≈ (0.010 x .012)½ = (1.2E–4)½ = 0.0011
Applying the "five percent rule", we find that x / Ca = .0011/.01 = .11 which is far over the allowable error, so we must proceed with the quadratic form. Rewriting the equilibrium expression in polynomial form gives
x2 + 0.022x – 1.2E–4 = 0
Inserting the coefficients {1 .022 .000012} into a quad-solver utility yields the roots 4.5E–3 and –0.0027. Taking the positive root, we obtain
pH = – log (.0045) = 2.3
Thus the second "ionization" of H2SO4 has only reduced the pH of the solution by 0.1 unit from the value (2.0) it would theoretically have if only the first step were considered.
4 Amino acids and Zwitterions
Zwitterions: molecular hermaphrodites
Amino acids, the building blocks of proteins, contain amino groups –NH2 that can accept protons, and carboxyl groups –COOH that can lose protons. Under certain conditions, these events can occur simultaneously, so that the resulting molecule becomes a “double ion” which goes by its German name Zwitterion.
Glycine: the simplest amino acid
The simplest of the twenty natural amino acids that occur in proteins is glycine H2N–CH2–COOH, which we use as an example here. Solutions of glycine are distributed between the acidic-, zwitterion-, and basic species:
(3-1)
Although the zwitterionic species is amphiprotic, it differs from a typical ampholyte such as HCO3 in that it is electrically neutral owing to the cancellation of the opposite electrical charges on the amino and carboxyl groups.
Amino acids are the most commonly-encountered kind of zwitterions, but other substances, such as quaternary ammonium compounds, also fall into this category. For more on Zwitterions, see this Wikipedia article or this UK ChemGuide page.
In the following development, we use the abbreviations H2Gly+ (glycinium), HGly (zwitterion), and Gly (glycinate) to denote the dissolved forms.
The two acidity constants are
(3-2)
If glycine is dissolved in water, charge balance requires that
$H_2Gly^+ + [H^+] \rightleftharpoons [Gly^–] + [OH^–] \label{3-3}$
Substituting the equilibrium constant expressions (including that for the autoprotolysis of water) into the above relation yields
(3-4)
If Eqs ii and iii in this Problem Example are recalculated for a range of pH values, one can plot the concentrations of each species against pH for 0.10 M glycine in water:
This distribution diagram shows that the zwitterion is the predominant species between pH values corresponding to the two pKas given in Equation $\ref{3-1}$.
Example $3$: Glycine solution speciation
Calculate the pH and the concentrations of the various species in a 0.100 M solution of glycine.
Solution
Substitution into Eq 4 above yields
(i)
The concentrations of the acid and base forms are found from their respective equilibrium constant expressions (Eqs 2):
(ii)
(iii)
The small concentrations of these singly-charged species in relation to Ca = 0.10 shows that the zwitterion is the only significant glycine species in the solution. | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.03%3A_Finding_the_pH_of_weak_Acids_Bases_and_Salts.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• State the values of Ca and Cb after a weak monoprotic acid or base has been partially neutralized with strong base or acid.
• Define the ionization fraction, and calculate the percent ionization, of a Ca M solution of a weak monoprotic acid.
• Sketch out a rough plot showing the distribution of conjugate species in a solution of a weak acid having a given value of Ka.
• Define a buffer solution, and explain how it works.
• Calculate the approximate pH of a solution of a weak monoprotic acid having a given pKa that has been partially neutralized by addition of a strong base.
• Specify how to make a given volume of a buffer solution having a certain pH, starting with a weak acid and sodium hydroxide.
• Define the buffer index, and explain its significance and the conditions under which its value is maximized.
• Sketch out a –log C vs pH (Sillén) plot for a weak monoprotic acid/base system having a given nominal concentration and pKa, and use it to estimate the pH of the solution.
We often tend to regard the pH as a quantity that is dependent on other variables such as the concentration and strength of an acid, base or salt. But in much of chemistry (and especially in biochemistry), we find it more useful to treat pH as the "master" variable that controls the relative concentrations of the acid- and base-forms of one or more sets of conjugate acid-base systems. In this lesson, we will explore this approach in some detail, showing its application to the very practical topics of buffer solutions, as well as the use of a simple graphical approach that will enable you to estimate the pH of a weak monoprotic or polyprotic acid or base without doing any arithmetic at all!
When the pH takes control
If we add 0.2 mol of sodium hydroxide to a solution containing 1.0 mol of a weak acid HA, then an equivalent number of moles of that acid will be converted into its base form A. The resulting solution will contain 0.2 mol of A and 0.8 mol of HA. Note that because we are discussing stoichiometry here, we are interested in quantities (moles) of reactants, not concentrations of reactants.
The important point to understand here is that we will end up with a “partly neutralized” solution in which both the acid and its conjugate base are present in significant amounts. Solutions of this kind are far more common than those of a pure acid or a pure base, and it is very important that you have a thorough understanding of them.
Example $1$: Partly neutralized acid
To a solution containing 0.010 mole of acetic acid (HAc), we add 0.002 mole of sodium hydroxide. If the volume of the final solution is 100 ml, find the values of Ca, Cb, and the total system concentration Ct.
Solution
The added hydroxide ion, being a strong base, reacts completely with the acetic acid, leaving 0.010 – 0.002 = 0.008 mole of HAc and 0.002 mole of acetate ion Ac. The final concentrations are
$C_a = \dfrac{0.008 \;mol}{0.10\; L} = 0.08 \; M \nonumber$
$C_b = \dfrac{0.002 \; mol}{ 0.10 \; L} = 0.02 M \nonumber$
$C_t = \dfrac{0.010 \;mol}{0.10\; L} = 0.10\; M. \nonumber$
Note that this solution would be indistinguishable from one prepared by combining Ca = 0.080 mole of acetic acid with Cb = 0.020 mole of sodium acetate and adjusting the volume to 100 ml.
Thus starting with a solution of a pure weak acid or weak base in water, we can add sufficient strong base or strong acid, respectively, to adjust the ratio of the conjugate species — that is, the ratios [HA]/[A] in the case of an acid, or [B]/[BH+] for a base, to any value we want.
Ionization Fractions
To express the relative concentrations of the protonated and deprotonated forms of an acid-base system present in a solution, we could use the simple ratio [HA]/[A] (or its inverse), but this suffers from the drawback of yielding an indeterminate result when the concentration in the denominator is zero. For many purposes it is more convenient to use the ionization fractions
$\alpha_0 = \dfrac{[HA]}{[HA]+A^-]} = \dfrac{[HA]}{C_a} \label{1-7a}$
$\alpha_1 = \dfrac{[A^-]}{[HA]+A^-]} = \dfrac{[A^-]}{C_a} \label{1-7b}$
The fraction $\alpha_1$ is also known as the degree of dissociation of the acid. By making appropriate substitutions using the relation
$[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{1-8}$
we can express the ionization fractions as functions of the pH:
$\alpha_0 = \dfrac{[H^+]}{K_a+[H^+]} \label{1-9a}$
$\alpha_1 = \dfrac{K_a}{K_a+[H^+]} \label{1-9b}$
Notice that the the values for both of these functions are close to zero or unity except within the pH range pKa ± 1 (Figure $1$).
Plots of the $\alpha$ functions vs. pH for several systems are shown below. Notice the crossing points where [HA] = [A] when [H+] = Ka that corresponds to unit value of the quotient in the Henderson-Hasselbalch approximation.
The ionization fractions of a series of acids over a wide pH range can conveniently be summarized as shown below.
If you know the pKa of an acid, you can easily sketch out its ionization fraction plot. The extreme tops and bottoms can only be estimated, but the rest of the plots are essentially straight lines.
As valuable as these plots are for showing how the distribution of conjugate species varies with the pH, they suffer from two drawbacks:
• Plots of$\alpha$ cover only a single order of magnitude, but the actual concentrations, which we are often more interested in, vary over a far greater range from pH 0 to 14.
• They are of little help if one wishes to estimate the pH of a solution of the acid or base, or of one of its salts.
Both of these limitations are readily overcome by the use of easily-constructed logarithmic plots which we describe in the following section.
Buffer Solutions
The word buffer has several meanings in English, most of them referring (in its verb form) to cushion, shield, protect, or counteract an adverse effect. In chemistry, it refers specifically to a solution that resists a change in pH when acid or base is added. A buffer (or buffered) solution is one that resists a change in its pH when H+ or OH ions are added or removed owing to some other reaction taking place in the same solution. Buffer solutions are essential components of all living organisms.
• Our blood is buffered to maintain a pH of 7.4 that must remain unchanged as metabolically-generated CO2 (carbonic acid) is added and then removed by our lungs.
• Buffers in the oceans, in natural waters such as lakes and streams, and within soils help maintain their environmental stability against acid rain and increases in atmospheric CO2.
• Many industrial processes, such as brewing, require buffer control, as do research studies in biochemistry and physiology that involve enzymes, are active only within certain pH ranges.
The essential component of a buffer system is a conjugate acid-base pair whose concentration is fairly high in relation to the concentrations of added H+ or OH it is expected to buffer against. A simple buffer system might be a 0.2 M solution of sodium acetate; the conjugate pair here is acetic acid HAc and its conjugate base, the acetate ion Ac. The idea is that this conjugate pair "pool" will be available to gobble up any small (≤ 10–3 M) addition of H+ or OH that may result from other processes going on in the solution.
Exercise $1$
Sketch out a similar diagram showing what happens when you remove H+ or OH.
Answer
When this happens, the ratio [HAc]/[Ac], will remain substantially unchanged, as will the pH, as you will see below.
You can also think of the process depicted above in terms of the Le Chatelier principle: addition of H+ to the solution suppresses the dissociation of HAc, partially counteracting the effect of the added acid, as illustrated by the equation at the bottom left of the above diagram.
The Henderson-Hasselbalch Approximation
To develop this more quantitatively, we will consider the general case of a weak acid HA to which a quantity Cb of strong base has been added; think, for example, of acetic acid which has been partially neutralized by sodium hydroxide, yielding the same conjugate pair described above, although not necessarily identical concentrations. The mass balance for such a system would be
$[HA] + [A^–] = C_a + C_b = C_T \label {2-1}$
in which Ct denotes the total concentration of all species in the solution. Because we added a strong (completely dissociated) base NaOH to the acid, we also note that
$C_b = [Na^+] \label{2-2}$
Recalling the equilibrium expression for a weak acid
$K_a = \dfrac{[H^+][A^-]}{[HA]} \label{2-3}$
We can solve this for [H+]:
$[H^+] = K_a \dfrac{[HA]}{[A^-]} \label{2-4}$
Re-writing this in terms of negative logarithms, this becomes
$-\log [H_3O^+] = -\log K_a - log [HA] + \log[A^-] \label{2-5}$
or, since $pK_a = –\log K_a$, we invert the ratio to preserve the positive sign:
$pH = pK_a + \log \dfrac{[A^-]}{[HA]} \label{2-6}$
This equation is known as the Henderson-Hasselbalch Approximation. It tells us that the pH of a solution containing a weak acid-base system controls the relative concentrations of the acid and base forms of that system.
It is interesting to note that the H-H equation was not developed by chemists!
• Lawrence Henderson (1878-1942) was an American medical doctor who taught at Harvard and who studied the acidity of blood and its relation to respiration. In 1908 he worked out the relation shown in Equation $\ref{2-3}$.
• In 1916, K. Hasselbalch, a physiologist at U. of Copenhagen, derived the logarithmic form in Equation $\ref{2-6}$.
You may wonder why these two equations, whose derivation we now consider almost trivial, should have immortalized the names of these two scientists. The answer is that the theory of chemical equilibrium was still developing in the early 1900's, and had not yet made its way into chemistry textbooks. Even the concept of pH was unknown until Sørenson's work appeared in 1909. It was the mystery (and medical necessity) of understanding why shortness of breath made the blood more alkaline, and too-rapid breathing made it more acidic, that forced the work of H&H into modern Chemistry.
Of special interest is the case in which the pH of a solution of an acid-base system is set to the value of its pKa. According to the above equation, when pH = pKa, the log term becomes zero, so that the ratio [AB] / [HA] = 100 = 1, meaning that [HA] = [AB]. In other words, when the pH of a solution is set to the value of the pKa of an acid-base pair, the concentrations of the acid- and base forms will be identical. This condition can be represented schematically on a proton-free energy diagram:
This says, in effect, that at when the pH of a solution containing both the acid and it conjugate base is made identical to the acid pKa, the forms HA and A possess identical free energies, and will therefore be present in equal concentrations.
Note
When the pH of a solution is set to the value of the pKa of an acid-base pair, the concentrations of the acid- and base forms will be identical.
Making and using Buffer Solutions
Buffers are generally most effective when the buffer system pKa is not too far from the target pH. Under these conditions, both [HA] and [A] are large enough to compensate for the withdrawal or addition, respectively, of hydrogen ions.
Table $1$: Common buffer Systems
acid formulas pKa pH range
phosphoric H2PO4 / HPO4 2.16 1-3
carbonic H2CO3* / HCO3 3.75 3-5
acetic HCOOH / HCOO 4.75 4-6
dihydrogen phosphate H2PO4 / HPO42 7.21 6-8
boric H3BO3 / H2BO3 9.24 8-10
ammonium NH4+ / NH3 9.25 8-10
bicarbonate HCO3 / CO32 10.3 9-11
monohydrogen phosphate HPO42 / PO43 12.3 11-13
Equation $\ref{2-4}$ above and its logarithmic equivalent in Equation $\ref{2-6}$ are of limited use in calculations because the exact values [A] and [HA] are known only for the special case when the pH of the solution is identical to the pKa. Most buffer solutions will be adjusted to other pH's, and of course, once the buffered solution begins doing it work by counteracting the effect of additions of H+ or OH, both [A] and [HA] will have changed. These two equations are so widely used in practical chemistry (and especially in biochemistry) that they are worth committing to memory.
increases [H+] and decreases the pH. So you can always arrange the equation to make it come out right. For these reasons, it is useful to rewrite the above two equations by replacing the equilibrium concentrations [A] and [HA] with the known initial values Ca and Cb:
$[H^+] \approx K_a \dfrac{C_a}{C_b} \label{2-7}$
$pH \approx pK_a + \log \dfrac{C_b}{C_a} \label{2-8}$
Although these are approximations, they are usually justified because useful buffer systems are always significantly more concentrated than those responsible for adding or removing hydrogen ions. Also, at these higher ionic concentrations, the Ka of the buffer system will seldom be precisely known anyway. Don't expect actual buffer pH's to match calculations to better than 5%.
Example $2$: Buffer Activity
Compare the effects of adding 1.0 mL of 2.5 M HCl to
1. 100 mL of pure water
2. 100 mL of 0.20 M sodium dihydrogen phosphate solution.
Solution
a)
1. quantity of HCl added: (1.0 mL) × (2.5\ mM/mL) = 2.5 mMol = 0.0025 mol
2. concentration of H+ in solution: (.0025 mol) ÷ (.101 L) = .025 M
3. pH = – log (.025) = 1.60, change in pH = 1.60 – 7.00 = –5.4
b) (From Table $1$, the pKa of H2PO4 is 7.21)
1. Initial Ca (H2PO4) = Cb (HPO42) = 0.20 M
2. Initial pH of buffer solution: pH = pKa + log (Cb/Ca) = 7.21 + 0 = 7.21
3. Added HCl will convert .0025 mol of HPO42 to H2PO4,
changing Ca to (.20 + .0025) mol ÷ (.101 L) = 2.00 M, and
reducing Cb to (.20 – .0025) mol ÷ (.101 L) = 1.95 M
4. Resulting pH = 7.21 + log (1.95 / 2.00) = 7.21 –.011 = 7.00
5. Change in pH = 7.00 – 7.21 = –.21
Note that in both cases the pH is reduced (as it must be if we are adding acid!), but the change is far less in the buffered solution.
Example $3$: Buffer Preparation
How would you prepare 200 mL of a buffer solution whose pH is 9.0?
Solution
The first step would be to select a conjugate acid-base pair whose pKa is close to the desired pH. Looking at the above table of pKa values, either boric acid or ammonium ion would be suitable. For this example, we will select the NH4+/NH3 system, Ka = 5.5E–10, pKa = 9.25.
2) We will use Equation $\ref{2-7}$ to determine the ratio Ca/Cb required to set [H+] to the desired value of 1.00 × 10–9. To do this, write out the complete equilibrium constant expression for the acid NH4+: NH4+ → NH3 + H+, and then solve it to find the required concentration ratio:
$K_a = \dfrac{[NH_3][H^+]}{[NH_4^+]} = 5.5 \times 10^{-10} \nonumber$
$\dfrac{C_a}{C_b} = \dfrac{[NH_3]}{[NH_4^+]} = \dfrac{5.5 \times 10^{-10}}{[H^+]} = \dfrac{5.5 \times 10^{-10}}{1.00 \times 10^{-9}} = \dfrac{5.5 \times 10^{-10}}{10.00 \times 10^{-10}} = 0.55 \nonumber$
3) The easiest (if not particularly elegant) way to work this out is to initially assume that we will dissolve 1.00 mole of solid NH4Cl in water, add sufficient OH to partially neutralize some of this acid according to the reaction equation NH4+ + OH → NH3 + H2O, and then add sufficient water to make the volume 1.00 L.
mass of ammonium chloride:
(1 mol) × (53.5 g Mol) = 53.8 g
let x = moles of NH4+ that must be converted to NH3.
Then Ca/Cb = (1-x)/x = .55; solving this gives x = .64, 1–x = .36.
4) So to make 1 L of the buffer, dissolve .36 × 53.8 g = 19.3 g of solid NH4Cl in a small quantity of water. Add .64 mole of NaOH (most easily done from a stock solution), and then sufficient water to make 1.00 L.
5) To make 200 mL of the buffer, just multiply each of the above figures by 0.200.
Pitfalls of the Henderson-Hasselbalch Approximation
The Henderson-Hasselbalch Approximation is widely used in practical calculations. What most books do not tell you is that Eqution $\ref{2-4}$ is no more than an “approximation of an approximation” which can give incorrect and misleading results when applied to situations in which the simplifying assumptions are not valid. An exact treatment of conjugate acid-base pairs, including a correct derivation of the Henderson-Hasselbalch equation, is given in the chapter on Exact Calculations.
The Henderson-Hasselbalch Approximation is only valid for fairly high concentrations
The approximations that lead to the H-H equation limit its reliable use to values of Ca and Cb that are within an order of magnitude of each other, and are fairly high. Also, the pKa of the acid should be moderate.
The shaded portion of this set of plots indicates the values of Ca and Ka that yield useful results. Clearly, the smaller the buffer concentration, the narrower the range of useable acid pKas. Most buffer solutions tend to be fairly concentrated, with Ca and Cb typically around 0.01 - 0.1 M. Thus a buffer based on a .01 M solution of an acid such as chloric (HClO3) with pKa of 1.9 will fall just outside the "safe" boundary near the upper left part of the diagram.
$\underbrace{[H^+] = K_a \dfrac{[HA]}{[A^-]}}_{\text{exact}} \label{2-9a}$
Equation $\ref{2-9a}$ is simply a re-writing of the equilibrium constant expression, and is therefore always true. Of course, without knowing the actual equilibrium values of [HA] and [A], this relation is of little direct use in pH calculations.
$\underbrace{[H^+] \approx K_a \dfrac{C_a}{C_b}}_{\text{approximate}} \label{2-9b}$
Equation $\ref{2-9b}$ is never true, but will yield good results if the acid is sufficiently weak in relation to its concentration to keep the [H+] from being too high. Otherwise, the high [H+] will convert a significant fraction of the A into the acid form HA, so that the ratio [HA]/[A] will differ from Ca /Cb in the above two equations. Consumption of H+ by the base will also raise the pH above the predicted value as we saw in the preceding problem example.
Buffer Index
The terms buffer intensity and buffer capacity are commonly employed as synonyms for buffer index, but in some contexts, buffer capacity denotes the quantity of strong acid or strong base which alters the buffer's pH by 1 unit. (see below). How effective is a given buffer system in resisting changes in the pH? The most direct expression of this is the rate of change of the pH as small quantities of strong acid or base are added to the system: ΔC/Δ(pH). Expressed in calculus notation, this is the buffer index, defined as
$\beta = \left| \dfrac{dC}{d(pH)} \right| \label{2-19}$
C here refers to the concentration of strong acid or base added to the solution. Because added acid or base affects the pH in opposite ways, we take the absolute value of this function in order to ensure that β is always positive. The value of β can be calculated analytically from Ca, Cb, Ka, Kb and [H+]. By taking the second derivative of β, it can be shown that the buffer index has a maximum value when the pH = pKa.
This buffer index plot for a 0.10 M solution of sodium acetate is typical, and confirms that buffering is most efficient within about ±1 pH unit of pKa. But what about the even greater buffering that apparently occurs at the two extremes of pH (orange shading)? This is due to the buffering associated with the water itself, and will be seen in all aqueous buffer solutions.
Pure water is buffered by the H3O+/H2O conjugate pair at very low pH, and by H2O/OH- at high pH. This is easily understood if you think about adding some strong acid acid to pure water; even one drop of HCl will send the pH shooting down toward 0. If you continue adding acid, the pH will not drop significantly below 0, because there won't be enough free water molecules remaining to hydrate the HCl to produce H3O+ ions — thus the solution is strongly buffered.
Note that if we were to subtract the effect of the NaAc buffering from the above plot, the remaining plot for water itself would exhibit a minimum at pH 7, where both [H3O+] and [OH] share a common minimum value.
This alternative view shows how the distribution fractions of HAc and Ac relate to the effective buffering range of this conjugate pair, which is conventionally defined as ±1 pH unit of the pKa. The term buffer capacity is an alternative means of expressing the ability of a buffer system to absorb the addition of strong acid or base without causing the pH to deviate by more than one unit from that of the pure buffer.
In this example, the buffer capacities for addition of acid and base will differ because the buffer pH has been adjusted to a value that differs from its pKa.
The Importance of buffer concentration
Buffer systems must be appreciably more concentrated than the concentrations of strong acid or base they are required to absorb while still remaining within the desired pH range. Once the added acid or base has consumed most of one or other of the conjugate species comprising the buffer, the pH will no longer be stabilized. And of course, very small buffer concentrations will approach the pH of pure water.
A more complete view: Log-C vs pH plots
These plots are often referred to as Sillén plots. Lars Gunnar Sillén (1916-1970) was a Swedish chemist who studied the distribution of ionic species in aqueous solutions and especially in the oceans. Because the concentrations of conjugate species can vary over many orders of magnitude, it is far more useful to express them on a logarithmic scale. Since pH is already logarithmic, one can obtain a "bird's-eye view" of an acid-base system in a compact log-log plot.
Even better, these plots are easily constructed without any calculations or arithmetic — or even any graph paper — any scrap paper, even the back of an envelope, will be sufficient. A ruler or other straightedge will, however, give more accurate results. In addition, you can use these plots to estimate the pH of a solution of a monoprotic or polyprotic acid or base without contending with quadratic or higher-order equations — or doing any arithmetic at all! The results will not be as precise as you would get from a proper numerical solution, but given the uncertainties of how equilibrium constants are affected by the presence of other ions in the solution, this is rarely a problem.
Aside from these advantages, the use of log-C vs pH plots will afford you an insight into the chemistry of acid-base systems that cannot be obtained simply by doing numerical calculations. The basic form of the plot, and the starting point for any use of such plots, looks like this:
If you examine this plot carefully, you will see that it is nothing more than a definition of pH and pOH, as well as a definition of a neutral solution at pH 7. Notice, for example, that when the pH is 4.0, [H+] = 10–4 M and [OH] = 10–10 M.
A simple example: acetic acid solution
Here is a log C plot for a 10–3 M solution of acetic acid ("HAc") in water. Although it may look a bit complicated at first, it is really very simple. The heavy maroon line on the left plots the concentration of the acid HAc as a function of pH. The blue line on the right shows how the concentration of the base Ac depends on pH. The horizontal parts of these lines are aligned with "3" on the –log-C axis, corresponding to the 10–3 M nominal concentration (Ca).
How do we know the shapes and placement of the plots for the concentrations of acetic acid [HAc] and the acetate ion [Ac] ? Although both of these concentrations will of course vary with the pH, their deviations from 10–3 M are too small to reveal themselves on a logarithmic plot until the pH approaches the pKa.
• At very low pH, virtually all of the acetate system will be in its acid form (that is, [HAc] = 10–3), and similarly, at high pH, the base form [Ac] = 10–3 M.
• When pH = pKa, the concentrations of the conjugate species are identical, as indicated by the crossing of the lines representing the two concentrations. These concentrations will be half of the total concentration given by Ca, so [HAc] = [Ac] = ½Ca = 0.5E–3 M. Where does this go on the log-C axis? Well, the logarithm of 0.5 is –0.3 (a useful fact to remember!), so the crossing point at is displaced by 0.3 of a log-C unit below the –3 level on the y-axis.
• Because point defines both the pKa and concentration of a particular acid-base system, it is known as the system point.
• What about the slopes of the plots when they bend down? It turns out that these slopes are +1 for
[Ac] and –1 for [HAc]. Since the slopes of the lines for [H+] and [OH] are ±1, we can use these as guidelines; just make the plots of [Ac] and [HA] parallel to them.
• The curved portions of the plot that joins the the horizontal and diagonal parts on either side of the system point can be drawn freehand without undue error; logarithmic plots are very forgiving!
Estimating the pH
Of special interest in acid-base chemistry are the pH values of a solution of an acid and of its conjugate base in pure water; as you know, these correspond to the beginning and equivalence points in the titration of an acid with a strong base. Continuing with the example of the acetic acid system, we show below another plot that is just like the one above, but with a few more lines and numbers added.
pH of acetic acid in pure water. Suppose you would like to find the pH of a 0.001 M solution of acetic acid in pure water. From the equation
$HAc + H_2O \rightleftharpoons H_3O^+ + Ac^– \label{Acetate 2-1}$
we know that equal numbers of moles of hydronium and acetate ions will be formed, so the concentrations of these species should be about the same:
$[H^+] \approx [Ac^–] \label{Acetate 2-2}$
will hold. The equivalence of these two concentrations corresponds to the point labeled on the logC-pH plot; this occurs at a pH of about 4, and this is the pH of a 0.001 M solution of acetic acid in pure water.
pH of sodium acetate in pure water. A 0.001 M solution of NaAc in water corresponds to the composition of a solution of acetic acid that has been titrated to its equivalence point with sodium hydroxide. The acetate ion, being the conjugate base of a weak acid, will undergo hydrolysis according to
$Ac^– + H_2O \rightleftharpoons HAc + OH^– \label{Acetate 2-3}$
which establishes the approximate relation
$[HAc] \approx [OH^–] \label{Acetate 2-4}$
This condition occurs where the right sloping part of the line representing [HAc] intersects the [OH] line at in the plot above. As you would expect for a solution of the conjugate base of a weak acid, this corresponds to an alkaline solution, in this case, at about pH = 7.8.
You will have to admit that estimating a pH in this way is far more painless than an ordinary numerical calculation would be!
It is interesting to show the Sillén plot for a weak acid system with its titration curve, both on the same pH scale. In this example, the titration plot has been turned on its side and reversed in order to illustrate the correspondence of its ƒ=0 (initial), ƒ=0.5 and ƒ=1 (equivalence) points with the corresponding points on the upper plot.
Ammonia - a weak base
Ammonia, unlike most gases, is extremely soluble in water. Solutions of NH3 in water are properly known as aqueous ammonia, NH3(aq); the name "ammonium hydroxide" is still often used, even though there is no evidence to suggest that a species NH4OH exists. Ammonia is the conjugate base of the ammonium ion NH4+. The fact that ammonia is a base has no special significance insofar as the construction and interpretation of the log-C vs pH diagram is concerned; such diagrams always refer to conjugate acid-base systems, rather than to individual acids and bases.
Thus the system point defines the NH4+/NH3 pair (pKa 9.3) at 25°C and a total concentration
$C_T = C_a + C_b = 0.0010\; M. \label{NH3 2-5}$
To estimate the pH of a solution of ammonia in pure water, we make use of the charge balance requirement (known as the proton condition)
$[NH_4^+] + [H^+] = [OH^-] \label{NH3 2-5.5}$
which we simplify by assuming that, because NH4+ is a weak acid, [NH4+] >> [H+] and we can drop the [H+] term; thus the approximation
$[NH_4^+] \approx [OH^-] \label{NH3 2-6}$
which corresponds to the crossing point and a pH of 10.0. Similarly, at the crossing point , [NH3] ≈ [H+]. This comes from the proton condition
$[NH_3] + [H^+] – [OH^–] = 0 \label{NH3 2-7}$
by dropping the [OH] term on the assumption that its value is negligible compared to [H+].
Equation $\ref{NH3 2-7}$ is referred to as the "proton condition". Sorry to sneak it in without warning, but we hope that having piqued your curiosity, you might take the time to look at the following discussion of this important tool for estimating the pH of a solution of a pure acid, base, or a salt.
Mass- and charge balance, and the proton condition
In any aqueous solution of an acid or base, certain conservation conditions are strictly observed. Together with concentration and the Ka or Kb , these put certain constraints on the system that determine the state of the system. The most fundamental of these are conservation of mass and of electric charge, which of course apply to chemical changes of all kinds. We commonly express these as mass balance and charge balance, respectively. (The latter of these is sometimes referred to as the electroneutrality condition.)
Using the above example of the ammonium system as an illustration of this, we are interested in two particular instances of practical importance: what conditions apply to solutions made by dissolving a) pure ammonia, or b) ammonium chloride, in water?
Sillén diagram
Charge balances
CT, NH3 = [NH3] + [NH4+] = 10–2 M
CT, Ac = [HAc] + [Ac] = 10–2 M
NH3: [H+] + [NH4+] = [OH]
NaAc: [Na+] + [H+] = [Ac] + [OH]
OK so far? That's really all we need, but it's usually more convenient to combine these with mass balances on the protons alone, so that we have a single equation for each of the two solutions. The resulting equations are known as the proton conditions for the two solutions.
To avoid a lot of algebra, there is a simple short-cut for writing a proton condition equation:
1. Identify the substance you are starting with — the pure acid, base, or salt, and also H2O itself, that make up the solution whose pH you wish to know at the concentration for which the log C-pH chart is drawn. we call this the proton reference level (if that appellation scares you, it is sometimes referred to as the "basis substance".)
2. On the left side of the proton condition, write concentration expressions for all species that possess protons in excess of the reference level.
3. On the right side, write concentrations of all species that possess fewer protons the reference level.
Let's try it for our ammonia system.
Solution of NH3 in water:
The proton reference level (PRL) is defined by NH3(aq) and H2O.
Proton condition: [H3O+] + [NH4+] = [OH]
(Notice that the substances that define the PRL (in case, H2O and NH3) never appear in the proton condition equation.)
Solution of NH4Cl (or of any other strong-anion ammonium salt) in water:
The PRL is defined by NH4+ and H2O.
Proton condition: [H3O+] = [NH3] + [OH]
(The chloride ion has nothing to do with protons, so it does not appear here.)
This proton condition stuff seems awfully complicated: why do we bother?
Good question: the answer is that what seems complicated at first sometimes turns out to be simpler and easier to understand than any alternative. Think of it this way: acid-base reactions involve the transfer of protons: some protons jump up to higher proton-free energy levels (e.g., NH3 + H2O → NH4+ + OH), others drop down to proton-vacant levels (H3O+ → NH3 → H2O + NH4+). But no matter what happens, the total number of "available" protons (that is, all "dissociable" protons, including those in H3O+) must be conserved — thus the "mass balance on protons". So let's look again at the log C - pH plot for the ammonium system, which we reproduce here for your convenience:
Consider first the solution of ammonium chloride, whose proton condition is given by [H3O+] = [NH3] + [OH]. We know that NH4Cl, being the salt of a weak base and a strong acid, will give a solution that is slightly acidic. In such a solution, [OH] will be quite small so that we can neglect it without too much error. We can then write the proton condition as [H+] [NH3]. On the plot above, this corresponds to point where the lines representing these quantities cross. At this pH of around 6.2, [OH] is about two orders of magnitude smaller than [NH3], so we are justified in dropping it.
Similarly, for a solution of ammonia in water, the proton condition [H3O+] + [NH4+] = [OH] that we worked out above can be simplified to [NH4+] [OH] (point ) with hardly any error at all, since [H+] is here about six orders of magnitude smaller than [NH4+].
In our discussion of the plot for the acetic acid system, we arrived at the proton conditions [H+] [Ac] (for HAc in water) and [H+] [OH] (for NaAc) by simply using the stoichiometries of the reactions. This can work in the simplest systems, but not in the more complicated ones involving polyprotic systems. In general, it is far safer to write out the complete proton condition in order to judge what concentrations, if any, can be dropped.
Ammonium acetate solution: solution of a salt
Having looked at the Sillén diagrams for acetic acid and ammonia, let's examine the log C-pH plot for this salt of acetic acid and ammonia.
You will recall that we dealt with solutions of the salts sodium acetate and ammonium chloride in the two examples described above. What is different here is that both components of the salt CH3COONH4 ("NH4Ac") are "weak" in the sense that their conjugate species NH3 and HAc are also present in significant quantities. This means that we are really dealing with two acid-base systems — the ones shown previously — on the same plot.
Each system retains its own system point, for the acetate system and for the ammonium system. Note that the two plots previously shown were for 10–3 M solutions, so the non-system points for this more concentrated solution occur at different pH values. The pH's of 10–2 M solutions of HAc , NaAc , and NH3 are located by using the same proton conditions as before.
The only new thing here is the point , which corresponds to the ammonium acetate solution in which we are interested. The proton condition for this solution is [H+] + [HAc] = [NH3(aq)] + [OH]. Inspection of the plot reveals that [H+] << [HAc] and [OH] << [NH3(aq)], so the proton condition can be simplified to [HAc] [NH3(aq)], corresponding to the intersection of these two lines at .
Polyprotic systems
Now that you are able to find your way around log C-pH plots that encompass two acid-base systems, the polyprotic systems that we describe below should be no trouble at all.
The Carbonate System
A thorough understanding of the carbonic acid-carbonate system is essential for understanding the chemistry of natural waters and the biochemistry of respiration in humans and other animals. There are two special things about this system that you need to know:
• A solution of carbon dioxide in water is mostly in the form of hydrated CO2, CO2(aq). But a small fraction of dissolved CO2 reacts with water to form carbonic acid, H2CO3. Because this acid cannot be isolated, common practice is to add an asterisk to the formula to designate the "total" CO2 concentration
[CO2 + H2CO3]: thus H2CO3*.
• Because CO2 is a gas, solutions containing carbonate species of all kinds can equilibrate with the atmosphere, which normally contains small quantities of CO2. This means that a solution of sodium bicarbonate that is open to the atmosphere can lose CO2 to the atmosphere at low pH, and absorb it at high pH. A system of this kind is said to be open. Alternatively, a system can be closed to exchange with the atmosphere. Sillén plots for open and closed carbonate systems have very different appearances.
This Sillén plot for the H2CO3* closed system is typical of other diprotic acid systems in that there are three conjugate species and two system points. The 10–3 M concentration plotted here is typical of what is found in groundwaters that are in contact with carbonate-containing sediments.
The pH of a 10–3 M solution of CO2, NaHCO3 and Na2CO3 are indicted by the blue annotations.
solute, (reference level) proton condition approximation
carbon dioxide (H2CO3*) [H+] = [HCO3] + 2[CO32] + [OH] [H+] ≈ [HCO3]
sodium bicarbonate (NaHCO3) [H+] + [H2CO3] = [CO32] + [OH] [H2CO3] ≈ [OH]
sodium carbonate (Na2CO3) [H+] + 2[H2CO3*] + [HCO3] = [OH] [HCO3] ≈ [OH]
You should take a few moments to verify the approximations in the rightmost column. Notice the factors of 2 that multiply some of the concentrations. For example, the "2[CO32]" term in the first line of the table means that the CO32 species is two steps away from the proton reference level H2CO3*, so it would take two hydrogen ions to balance a single carbonate ion.
Another new feature introduced here is the doubling of the slopes of the lines representing [H2CO3*] and [CO32] where they cross the pH values corresponding to the Kas that are one step removed from their own system points and near –log C = 7.
The Phosphate System
This triprotic system is widely used to prepare buffer solutions in biochemical applications. In its fundamental form, there is nothing new here, other than an additional system point corresponding to the pKa of H2PO4.
The pH of a 10–3 M solutions of H3PO4, NaH2PO4, Na2HPO4 and Na3PO4can be estimated by setting up the proton conditions as follows. Unfortunately, these estimates become a bit more complicated owing to the proximity of the three system points. But it's far easier than doing a numerical solution which requires solving a firth-degree polynomial!
solute, (reference level) proton condition approximation
phosphoric acid (H3PO4) [H+] = [H2PO4] + 2[HPO42] + 3[PO43] + [OH] [H+] ≈ [H2PO4]
sodium dihydrogen phosphate (NaH2PO4) [H+] + [H3PO4] = [HPO42] + 2[PO43] + [OH]
[H+] + [H3PO4] ≈
[HPO42]
disodium hydrogen phosphate (Na2HPO4)
[H+] + 2[H2PO4] + [H2PO4] = 3[PO43] + [OH] [H3PO4-] = [PO43] + [OH]
trisodium phosphate (Na3PO4) [H+] + 3[H3PO4] +2 [H2PO4] + [HPO42] = [OH] [HPO42] ≈ [OH]
Because the phosphate plot is rather crowded, we show here a modified one in which the details of the pH estimates for the solutions of H3PO4 and its salts are emphasized.
Estimation of the pH of 10–3 M solutions of phosphoric acid and trisodium phosphate , based on the approximations in the above table are straightforward.
Those for NaH2PO4 and Na2HPO4 , however, are complicated by the presence of two concentration terms on the right side of the respective proton condition approximations.
Unless you are in an advanced course, you may wish to skip the following details. The important thing to know is that graphical estimates of these more problematic systems are fundamentally possible.
Thus for sodium dihydrogen phosphate, if we were to follow the pattern of the other systems, the approximation at the crossing point would be
[H+] = [HPO42], corresponding to point . But inspection of the plot shows that at that pH (4.0), the concentration of H3PO4 is ten times greater than that of H+ (point ). A rough estimate of the pH can be made by constructing a line (shown in red) that is parallel to those for H3PO4 and H+, but is raised up by a factor of 10. This results in the new crossing point .
Similarly, the pH of a disodium hydrogen phosphate solution does not correspond to the crossing point because of the significant value of [HPO43] at this pH. Again, we construct a line above this crossing point that predicts a pH corresponding to point .
Although these stratagems may seem rather crude, it should be noted that the uncertainties associated with them tend to be minimized on a logarithmic plot. In addition, it should be borne in mind that in a solution as concentrated as 0.1M, the pKa values found in tables are not applicable. The alternative of solving for the pH algebraically, taking into consideration activity coefficients for the high ionic concentrations, is rarely worth the effort.
How to draw Sillén diagrams
The relation between the pH of a solution and the concentrations of weak acids and their conjugate species is algebraically rather complicated. But over most of the pH range, simplifying assumptions can be made that, when expressed in logarithmic form, plot as straight lines having integral slopes of 0, ±1, ±2, etc. It is only within the narrow pH range near the pKa that these simplifying assumptions break down, but on a logarithmic plot, one can draw a smooth curve that covers this range without introducing significant error.
Here is your basic stationery — a plot showing the two ions always present in any aqueous solution. In practice, you can usually cut off the bottom part where concentrations smaller than $10^{–10}\, M$.
For any given acid-base system, you need to know the pKa and nominal concentration CT.
Find the point at which these two lines intersect.
Locate and mark the system point on the pKa line. It will be 0.3 of a log-C unit below the concentration line.
This is where [HA]
= [A] = ½ Ct . (Recall that log.5 = .3)
For pH < pKa, [HA] plots as a line through the system point with slope = 0.
For pH > pKa, the line assumes a slope of –1. Use the [H+] and
[OH] lines as guides.
In the pH interval of ±pKa, the slope changes from 0 to +1. You can approximate this with a sloped line as shown, or "prettify" it with a smooth curve.
Finally, draw the lines for the conjugate form [A], using slopes +1 and 0.
What's all this good for?
Suppose you have a solution of NaA whose pH is 8. Estimate the concentration of the acid HA in the solution. All you need do is look at the plot. The [HA] line crosses the line marking this pH at – log C = 6, or [HA] = 10–6. This quantity too small to significantly reduce [A], which remains at approximately CT = 10–6M.
References
1. Boundless outlines on buffer solutions. These summarize the main points and problem examples as presented in major textbooks. They are an excellent aid for pre-exam review. Chemistry: The Central Science (8 subtopics)
2. A collection of practice exercises (with solutions) on buffers and titrations from Bryn Mawr College.
3. The Henderson-Hasselbalch equation: its history and limitations - Henry N. Po and N.M. Senoza, J.Chem. Ed. 78 (11) 2001 (1499-1503)
4. A detailed treatment of log-C vs pH diagrams can be found in the book Acid-base diagrams by Kahlert and Schotz. It is available online in some college libraries. | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.04%3A_Conjugate_Pairs_and_Buffers.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Sketch out a plot representing the titration of a strong monoprotic acid by a strong base, or of a strong base titrated by a strong acid. Identify the equivalence point and explain its significance.
• On the plot referred to above, draw a similar plot that would correspond to the same system at a higher or lower concentration.
• Sketch out a plot representing the titration of a weak monoprotic acid by a strong base, or of a weak base titrated by a strong acid. Identify the equivalence point and half-equivalence points.
• Explain what an acid-base indicator is, and how it works.
• When solutions of some polyprotic acids are titrated with strong base, not all of the equivalence points can be observed. Explain the most common reasons for this.
• Calculate the molarity of a monoprotic acid HA whose titration endpoint occurs after V ml of strong base of a given concentration has been added.
The objective of an acid-base titration is to determine $C_a$, the nominal concentration of acid in the solution. In its simplest form, titration is carried out by measuring the volume of the solution of strong base required to complete the reaction
$\ce{H_nA + n OH- → n A- + n H_2O} \label{0-1}$
in which $n$ is the number of replaceable hydrogens in the acid. The point at which this reaction is just complete is known as the equivalence point. This is to be distinguished from the end point, which is the value we observe experimentally. A replaceable hydrogen atom (sometimes called an "acidic" hydrogen) is one that can be donated to a strong base — that is, to an OH ion. Thus in acetic acid HCOOH, only the hydrogen in the carboxyl group is considered "replaceable".
What we actually measure, of course, is the volume of titrant delivered by the burette. Learning to properly control the stopcock at the bottom of the burette usually requires some instruction and practice, as does the reading of the volume. For highly precise work, the concentration of the titrant itself must be determined in a separate experiment known as "standardization".
Understanding Titration Curves
A plot showing the pH of the solution as a function of the quantity of base added is known as a titration curve. These plots can be constructed by plotting the pH as a function of either the volume of base added, or the equivalent fraction $ƒ$ which is simply the number of moles of base added per mole of acid present in the solution. In most of the titration curves illustrated in this section, we plot pH as a function of $ƒ$. It's worth taking some time to thoroughly familiarize yourself with the general form of a titration curve such as the one shown below, in which a weak acid HA is titrated with a strong base, typically sodium hydroxide.
The equivalence point occurs at the pH at which the equivalent fraction ƒ of base added is unity. At this point, the reaction
$HA + OH^– → AB^– + H_2O \label{1-1}$
is stoichiometrically complete; a solution initially containing n moles of a monoprotic acid HA will now be identical to one containing the same number of moles of the conjugate base A. At the half-equivalence point ƒ = 0.5, the concentrations of the conjugate species are identical: [HA] = [A]. This, of course corresponds to a buffer solution (hence the relatively flat part of the curve) whose pH is the same as pKa.
As base is added beyond ƒ = 1, the pH begins to level off, suggesting that another buffered system has come into play. In this case it involves the solvent (water) and hydroxide ion: {H2O} ≈ {OH-}.
A similar effect is seen at the low-pH side of the curve when a strong acid is titrated, as in the plot for the titration of HCl below. In this case, the buffering is due to {H3O+) ≈ {H2O}.
How can this be? Surely, the concentration of OH, even when the pH approaches 14, cannot be anything like that of [H2O] which will be about 55.5 M in most solutions! This fine point (along with the mention of H2O/OH buffering) is rarely mentioned in elementary courses because the theory behind it involves some rather esoteric elements of solution thermodynamics. However, in case you are curious, note that the curly brackets in {H2O} ≈ {OH} denote activities, not concentrations. And by convention the activity of a pure liquid (H2O in this case) is unity. At a pH of around 12, pOH = 2, [OH] = .01. At this rather high ion concentration, {OH} will be somewhat smaller than this, but the two activities will be similar enough to produce the buffering effect we observe.
The pH of the solution at its equivalence point will be 7 if we are titrating a strong acid with strong base, as in HCl + NaOH → H2O + NaCl. However, if the acid is weak, as in the above plot, the solution will be alkaline. This pH can be calculated from Cb and Kb in a manner exactly analogous to that used for calculating the pH of a solution of a weak acid in water.
It is important to understand that the equivalent fraction ƒ of base that must be added to reach the equivalence point is independent of the strength of the acid and of its concentration in the solution. The whole utility of titration as a means of quantitative analysis rests on this independence; we are in all cases measuring only the total number of moles of “acidic” hydrogens in the sample undergoing titration.
Acid and base strengths determine the shape of the curve
Although the strength of an acid has no effect on the location of the equivalence point, it does affect the shape of the titration curve and can be estimated on a plot of the curve.
The weaker the acid being titrated, the higher the initial pH (at ƒ=0), and the smaller will be the vertical height of the plot near the equivalence point. As we shall see later, this can make it difficult to locate the equivalence point if the acid is extremely weak.
Estimating the acid strength
As shown in plot above, the pKa of a weak acid can be estimated by noting the pH that corresponds to the half-titration point ƒ = 0.5. Recalling that the pH is controlled by the ratio of conjugate species concentrations
$pH = pK_a + \log \dfrac{[A^]}{[HA]} \label{1-2}$
it will be apparent that this equation reduces to pH = pKa when the titration is half complete (that is, when [HA] = [A]), the pH of the solution will be identical to the pKa of the acid. This equation does not work for strong acids owing to the strong buffering that occurs at the very low pH at which ƒ = 0.5.
As indicated here, the buffering has nothing to do with the acid HCl itself (which does not exist as such in water), but rather with its dissociation products H3O+ and OH, "the strongest acid and base that can exist in water."
Monoprotic titration curves
The following two principles govern the detailed shape of a titration curve:
• The stronger the acid or base, the greater will be the slope of the curve near the equivalence point;
• The weaker the acid or base, the greater will the deviation of the pH from neutrality at the equivalence point.
It is important to understand the reasons for these two relations. The second is the simplest to explain. Titration of an acid HA with a base such as NaOH results in a solution of NaA; that is, a solution of the conjugate base A. Being a base, it will react with water to yield an excess of hydroxide ions, leaving a slightly alkaline solution. Titration of a weak base with an acid will have the opposite effect.
The extent of the jump in the pH at the equivalence point is determined by a combination of factors. In the case of a weak acid, for example, the initial pH is likely to be higher, so the titration curve starts higher. Further, the weaker the acid, the stronger will be its conjugate base, so the higher will be the pH at the equivalence point. These two factors raise the bottom part of the titration curve. The upper extent of the curve is of course limited by the concentration and strength of the titrant.
These principles are clearly evident in the above plots for the titrations of acids and bases having various strengths. Notice the blue curves that represent the titration of pure water (a very weak acid) with strong acid or base.
Monoprotic titration curve gallery
When both the titrant and sample are "strong", we get long vertical plots at ƒ = 1. Adding even half a drop of titrant can take us across the equivalence point!
When one of the reactants is weak, the pH changes rapidly at first until buffering sets in.
In (C), the onset of H2O/OH- buffering near ƒ=1 makes the equivalence point more difficult to locate.
"Weak/weak" titrations tend to be problematic as the buffered regions move closer to ƒ=1. The equivalence point pH of 7 in these examples reflects the near-equality of pKa and pKb of the reactants.
Dealing with very weak acids
It can be difficult to reliably detect the equivalence point in the titration of boric acid (pKa = 9.3) or of other similarly weak acids from the shape of the titration curve*. *An interesting student laboratory experiment that employs an auxiliary reagent (mannitol) to make boric acid stronger and thus more readily titratable was described in J. Chem Ed. 2012, 89, 767-770.
The problem here is that aqueous solutions are buffered against pH change at very low and very high pH ranges. An extreme example occurs in the titration of pure water with a strong acid or base. At these extremes of pH the concentrations of H3O+ and of OH are sufficiently great that a competing buffer system (either H3O+/H2O or H2O/OH, depending on whether the solution is highly acidic or highly alkaline) comes into play.
Why we usually use a "strong" titrant
The above plots clearly show that the most easily-detectable equivalence points occur when an acid with is titrated with a strong base such as sodium hydroxide (or a base is titrated with a strong acid.)
In practice, many of the titrations carried out in research, industry, and clinical practice involve mixtures of more than one acid. Examples include natural waters, physiological fluids, fruit juices, wine making, brewing, and industrial effluents. For titrating these kinds of samples, the use of anything other than a strong titrant presents the possibility that the titrant may be weaker than one or more of the "stronger" components in the sample, in which case it would be incapable of titrating these components to completion.
In terms of proton-free energies, the proton source (the acidic titrant) would be unable to deliver an equivalent quantity of protons to the (stronger) component of the mixture.
Polyprotic acids
There will be as many equivalence points as there are replaceable hydrogens in an acid. Thus in the extremely important carbonate system, equivalence points are seen at both ƒ=1 and ƒ=2:
In general, there are two requirements for a clearly discernible jump in the pH to occur in a polyprotic titration:
• The successive Ka's must differ by several orders of magnitude;
• The pH of the equivalence point must not be very high or very low.
Separation of successive equilibrium constants
The effect of the first point is seen by comparing the titration curves of two diprotic acids, sulfurous and succinic. The appearance of only one equivalence point in the latter is a consequence of the closeness of the first and second acid dissociation constants. The pKa's of sulfurous acid (below, left) are sufficiently far apart that its titration curve can be regarded as the superposition of those for two independent monoprotic acids having the corresponding Ka's. This reflects the fact that the two acidic –OH groups are connected to the same central atom, so that the local negative charge that remains when HSO3 is formed acts to suppress the second dissociation step.
*It can be shown that in the limit of large n, the ratio of K1/K2 for a symmetrical dicarboxylic acid HOOC-(CH2)n- COOH converges to a value of 4.
In succinic acid, the two –COOH groups are physically more separated and thus tend to dissociate independently*. Inspection of the species distribution curves for succinic acid (above, right) reveals that the fraction of the ampholyte HA can never exceed 75 percent. That is, there is no pH at which the reaction H2A → HA + H+ can be said to be “complete” while at the same time the second step HA → A2– + H+ has occurred to only a negligible extent. Thus the rise in the pH that would normally be expected as HA is produced will be prevented by consumption of OH in the second step which will be well underway at that point; only when all steps are completed and hydroxide ion is no longer being consumed will the pH rise.
Two other examples of polyprotic acids whose titration curves do not reveal all of the equivalence points are sulfuric and phosphoric acids. Owing to the leveling effect, the apparent Ka1 of H2SO4 is so close to Ka2 = 0.01 that the effect is the same as in succinic acid, so only the second equivalence point is detected.
In phosphoric acid, the third equivalence point (for HPO42) is obscured by H2O-OH buffering as explained previously.
Detection of the equivalence point
Whether or not the equivalence point is revealed by a distinct "break" in the titration curve, it will correspond to a unique hydrogen ion concentration which can be calculated in advance. There are many ways of determining the equivalence point of an acid-base titration.
Indicators
Don't overshoot the equiv point!
The traditional method of detecting the equivalence point has been to employ an indicator dye, which is a second acid-base system in which the protonated and deprotonated forms differ in color, and whose pKa is close to the pH expected at the equivalence point. If the acid being titrated is not a strong one, it is important to keep the indicator concentration as low as possible in order to prevent its own consumption of OH from distorting the titration curve.
The observed color change of an indicator does not take place sharply, but occurs over a range of about 1.5 to 2 pH units. Indicators are therefore only useful in the titration of acids and bases that are sufficiently strong to show a definite break in the titration curve. Some plants contain coloring agents that can act as natural pH indicators. These include cabbage (shown), beets, and hydrangea flowers.
For a strong acid - strong base titration, almost any indicator can be used, although phenolphthalein is most commonly employed. For titrations involving weak acids or bases, as in the acid titration of sodium carbonate solution shown here, the indicator should have a pK close to that of the substance being titrated.
When titrating a polyprotic acid or base, multiple indicators are required if more than one equivalence point is to be seen. The pKas of phenolphthalein and methyl orange are 9.3 and 3.7, respectively.
Potentiometry: Use a pH meter
The pH meter detects the voltage produced when the H+ ions in the solution displace Na+ ions from a thin glass membrane that is dipped into the solution.
A more modern way of finding an equivalence point is to follow the titration by means of a pH meter. Because it involves measuring the electrical potential difference between two electrodes, this method is known as potentiometry. Until around 1980, pH meters were too expensive for regular use in student laboratories, but this has changed; potentiometry is now the standard tool for determining equivalence points.
Improved plotting methods
Plotting the pH after each volume increment of titrant has been added can yield a titration curve as detailed as desired, but there are better ways of locating the equivalence point. The most common of these is to take the first or second derivatives of the plot: d(pH)/dV or d2(pH)/dV2 (of course, for finite increments of pH and volume, these terms would be expressed as Δ(pH)/ΔV and Δ2(pH)/ΔV2 .)
A second-derivative curve locates the inflection point by finding where the rate at which the pH changes is zero.
The differential plot, showing rate-of-change of pH against titrant volume, locates the inflection point which is also the equivalence point
In a standard plot of pH-vs-volume of titrant added, the inflection point is located visually as half-way along the steepest part of the curve.
The idealized plots shown above are unlikely to be seen in practice. When the titration is carried out manually, the titrant is added in increments, so even the simple titration curve must be constructed from points subject to uncertainties in volume measurement and pH (especially if the latter is visually estimated by color change of an indicator.)
If this data is then converted to differential form, these uncertainties add a certain amount of "noise" to the data.
A second-derivative plot uses pH readings on both sides of the equivalence point, making it easier to locate in the presence of noise.
Locating the equivalence point depends very strongly on correct reading of only one or two pH readings near the top of the plot.
A simple curve, plotted from a small number of pH readings, will not always unambiguously locate the equivalence point.
The "noise" in differential plots can usually be minimized by keeping the titrant and analyte concentrations above 10–3 M.
Some other ways of following a titration
Monitoring the pH by means of an indicator or by potentiometry as described above are the standard ways of detecting the equivalence point of a titration. However, we have already seen that in certain cases involving polyprotic acids or bases, some of the equivalence points are obscured by their close proximity to others, or by the buffering that occurs near the extremes of the pH range. Similar problems can arise when the solution to be titrated contains several different acids, as often happens when fluids connected with industrial processes must be monitored.
Take the temperature!
Acid-base neutralization reactions HA + B → A + BH+ are always exothermic; when protons fall from their level in the acid to that in the base, most of the free energy drop gets released as heat. If the acid and base are both strong (i.e., totally dissociated), the enthalpy of neutralization for the reaction
$\ce{H3O+ + OH- → 2 H2O}$
is -68 kJ mol–1.
See this Wikipedia page for more on thermometric titrations, including many examples. Note also the video on this topic in the "Videos" section near the end of this page.
• Thermometric titrations are not limited to acid-base determinations; they can also be used to follow precipitation-, complex formation-, and oxidation-reduction reactions.
• They can be used with polyprotic acids or bases, and with mixtures containing more than one acid or base.
• They are able to follow acid-base titrations that must be carried out in non-aqueous solvents, where other titration methods are not possible.
A typical thermometric titration curve consists of two branches, beginning with a steep rise in temperature as the titrant being added reacts with the analyte, liberating heat. Once the equivalence point is reached, the rise quickly diminishes as heat production stops. Then, as the mixture begins to cool, the plot assumes a negative slope.
Although a rough indication of the equivalence point can be estimated by extrapolating the linear parts of the curve (blue dashed lines), the differential methods described above are generally preferred.
Follow the electrolytic conductance
Acids and bases are electrolytes, meaning that their solutions conduct electric current. The conductivity of such solutions depends on the concentrations of the ions, and to a lesser extent, on the nature of the particular ions. Any chemical reaction in which there is a change in the total quantity of ions in the solution can usually be followed by monitoring the conductance. Acid-base titrations fall into this category. Consider, for example, the titration of hydrochloric acid with sodium hydroxide. This can be described by the equation
$\ce{H^{+} + Cl^{–} + Na^{+} + OH^{–} → H2O + Na^{+} + Cl^{–}} \label{2-1}$
which shows that two of the four species of ions being combined disappear at the equivalence point. During the course of the titration, the conductance of the solution falls as H+ and Cl ions are consumed. At the equivalence point the conductance passes through a minimum, and then rises as continued addition of titrant adds more Na+ and OH ions to the solution.
Each kind of ion makes its own contribution to the solution conductivity. If we could observe the contribution of each ion separately, see that the slopes for H+ and OH are much greater. This reflects the much greater conductivities of these ions owing to their uniquely rapid movement through the solution by hopping across water molecules.
However, because the conductances of individual ions cannot be observed directly, conductance measurements always register the total conductances of all ions in the solution. The change in conductance that is actually observed during the titration of HCl by sodium hydroxide is the sum of the ionic conductances shown above.
For most ordinary acid-base titrations, conductimetry rarely offers any special advantage over regular volumetric analysis using indicators or potentiometry. This is especially true if the acid being titrated is weak; if the pKa is much below 2, the rising salt line (Na+ when titrating with NaOH) will overwhelm the fall in the contribution the small amount of H+ makes to the conductance, thus preventing any minimum in the total conductance curve from being seen.
However, in some special cases such as those illustrated below, conductimetry is the only method capable of yielding useful results.
These examples illustrate two unique capabilities of conductimetric titrations: (left) Titration of a mixture of two acids and (right) Titration of a strong polyprotic acid →
Because pure H2SO2 is a neutral molecule, it makes no contribution to the conductance, which rises to a maximum at the equivalence point.
Automating the process
In four years of college lab sessions, many Chemistry majors will likely carry out fewer than a dozen titrations. However, in the real world, time is money, and long gone are the days when technicians were employed full time just to titrate multiple samples in such enterprises as breweries, food processing (such as blending of canned orange juice), clinical labs, and biochemical research.
Titration calculations
A titration is carried out by adding a sufficient volume \Vo of the titrant solution to a known volume \Vt of the solution being titrated. This addition continues until the end point is reached. The end point is our experimental approximation of the equivalence point at which the acid-base reaction is stoichiometrically complete (ƒ = 1). The quantity we actually measure at the end point is the volume V_ep of titrant delivered to the solution undergoing titration.
The solution being titrated is often referred to as the analyte (the substance being "analyzed") or, less commonly, as the titrand. We shall employ the latter term in what follows.
In a simple titration of a monoprotic acid HA by a base B, the equivalence point corresponds to completion of the reaction
$\ce{HA + B → A^{–} + BH^{+} }\label{3-1}$
where equimolar quantities of HA and B have been combined; that is,
MHA = MB (where M represents the number of moles.)
Recall that the number of moles is given by the product of the volume and concentration:
L × mol L–1 = mol,
mL x mMol ml–1 = mmol.
Because we are measuring the volume of titrant rather than the number of moles, we need to use its concentration to link the two quantities. So if we are titrating the base B with acid HA, the end point is reached when a volume V of HA has been added. The number of moles of HA we have added at the end point is given by the product of its volume and concentration
$M_{HA} = V_{HA} \times C_{HA} \label{3-2}$
And because the reaction \ref{3-1} is now complete, MHA = MB ; thus, at the end point of any monoprotic titration,
$V_{HA} C_{HA} = V_B C_B \label{3-3}$
Equation \ref{3-3} is important! In any titration, both the volume and the concentration of the titrant are known, so the unknown concentration is easily calculated.
In titrations carried out in the laboratory, the titrant is delivered by a burette that is usually calibrated in milliliters, so it is more convenient to express MHA in millimoles and CHA in millimoles/mL (mMol ml–1); note that the latter is numerically the same as moles/L.
Example $1$
50.0 mL of 0.100 M hydrochloric acid is titrated with 0.200 M sodium hydroxide. What volume of NaOH solution will have been added at the equivalence point?
Solution
First, find the number of moles of HCl in the titrand:
\begin{align*} M_{\ce{HCl}} &= C_{\ce{HCl}} \times V_{\ce{HCl}} \[4pt] &= (0.100\, mMol\, mL^{–1}) \times (50.0\, mL) \[4pt] &= 5.0 \,mMol\,\ce{HCl} \end{align*}
This same number of moles of NaOH must be delivered by the burette in order to reach the equivalence point (i.e., MNaOH = 5.0 mMol.)
The volume of NaOH solution MNaOH required is
VNaOH = (MNaOH / CNaOH ) = (5.0 mMol) / (0.200 mMol/mL) = 25 mL | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.05%3A_Acid_Base_Titration.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Describe the major features of the global carbon cycle, and outline the role of acid-base processes in both the slow and fast parts of the cycle.
• Describe the principal acid-base reactions in aerobic respiration.
• Define acid rain, and explain its origins. Outline its major effects on the environment.
Acid-base reactions pervade every aspect of industrial-, physiological-, and environmental chemistry. In this unit we touch on a few highlights that anyone who studies or practices chemical science should be aware of.
Some applications of Buffers
Buffer solutions and the buffering effect they produce are extremely important in many practical applications of chemistry. The reason for this is that many chemical processes are quite sensitive to the pH; the extent of the reaction, its rate, and even the nature of the products can be altered if the pH is allowed to change. Such a change will tend to occur, for example, when the reaction in question, or an unrelated parallel reaction, consumes or releases hydrogen- or hydroxide ions.
Buffers in biochemistry and physiology
Many reactions that take place in living organisms fall into this category. Most biochemical processes are catalyzed by enzymes whose activities are highly dependent on the pH; if the local pH deviates too far from the optimum value, the enzyme may become permanently deactivated.
• For example, the oxidation of glucose to carbon dioxide and water — the fundamental energy-producing process of aerobic metabolism — releases H+ ions. The multiple steps of this process are catalyzed by enzymes whose activity is strongly affected by the pH. If the region of the cell where these reactions take place were not buffered, the very act of living would soon reduce the local pH below the range in which the enzymes are active, halting energy production and killing the cell.
• Blood is strongly buffered (mainly by bicarbonate) to maintain its pH at 7.4±0.3; pH values below 7.0 or above 7.8 cause death within minutes.
• The interiors of most cells are buffered near pH 7.0 (by phosphates and proteins)
Buffers in industry
Buffers are employed in a wide variety of laboratory procedures and industrial processes:
• solutions for the calibration of pH meters
• growth of bacteria in culture media
• fermentation processes, including winemaking and the brewing of beer
• controlling the colors of dyes used in coloring fabrics
• shampoos (maintaining pH at or below 7 to conteract the alkalinity of soaps that can cause irritation)
• other personal care products such as baby lotions (keeping the pH around 6 to inhibit the growth of bacteria)
• pH control in printing inks to assure their proper penetration into the paper
Buffers in the environment
• Natural waters (lakes and streams) able to support aquatic life are buffered by the sediments they are in contact with. This buffering is not always able to offset the effects of acid rain.
• Seawater is buffered mainly by bicarbonate and borates. This allows the oceans to absorb about half of the CO2 emitted to the atmosphere by human activities.
• Soils, in order to remain productive must maintain a near-neutral pH, not only for plants but also for microorganisms which fix- and recycle nitrogen. Soils rendered highly acidic by acid rain can keep essential nutrients such as phosphates in forms such as insoluble Ca3(PO4)2 that prevents their uptake by plants.
Acid-base chemistry in physiology
Acid-base chemistry plays a crucial role in physiology, both at the level of the individual cell and of the total organism. The reasons for this are twofold:
• Many of the major chemical components of an organism can themselves act as acids and/or bases. Thus proteins contain both acidic and basic groups, so that their shapes and their functional activities are highly dependent on the pH.
• Virtually all important metabolic processes involve the uptake or release of hydrogen ions. The very act of being alive tends to change the surrounding pH (usually reducing it); this will eventually kill the organism in the absence of buDering mechanisms.
About two-thirds of the weight of an adult human consists of water. About two-thirds of this water is located within cells, while the remaining third consists of extracellular water, mostly interstial fluid that bathes the cells, and the blood plasma. The latter, amounting to about 5% of body weight (about 5 L in the adult), serves as a supporting fluid for the blood cells and acts as a means of transporting chemicals between cells and the external environment. It is basically a 0.15 M solution of NaCl containing smaller amounts of other electrolytes, the most important of which are HCO3 and protein anions.
Respiration, the most important physiological activity of a cell, is an acid-producing process. Carbohydrate substances are broken down into carbon dioxide, and thus carbonic acid:
$C(H_2O)_{n} + O_2 → CO_2 + n H_2O$
Maintenance of acid-base balance
It is remarkable that the pH of most cellular fluids can be kept within such a narrow range, given the large number of processes that tend to upset it. This is due to the exquisite balance between a large number of interlinked processes operating at many different levels. Acid-base balance in the body is maintained by two general mechanisms: selective excretion of acids or bases, and by the buffering action of weak acid-base systems in body fluids.
• Over a 24-hour period, the adult human eliminates the equivalent of about 20-40 moles of H+ by way of the lungs in the form of CO2. In addition, the kidneys excrete perhaps 5% of this amount of acid, mostly in the form of H2PO4 and
NH4+. Owing to their electric charges, these two species are closely linked to salt balance with ions such as Na+ or K+ and Cl.
• The major buffering system in the body is the carbonate system, which exists mainly in the form of HCO3 at normal physiological pH. Secondary buffering action comes from phosphates, from proteins and other weak organic acids, and (within blood cells), the hemoglobin.
There is a huge industry aimed at the notoriously science-challenged "alternative health" market that promotes worthless nostrums such as "ionized water" that are claimed to restore or preserve "the body's acid-base balance". They commonly point out that most foods are "acidic", (i.e., they are metabolized to H2CO3), but they never explain that most of this acid is almot immediately exhaled in the form of CO2. The implication is that our health is being ruined by the resulting "acidification" of the body; some further imply this can be a cause of cancer and other assorted ailments.
People who fall for these expensive scams are in effect paying a tax on scientific ignorance. Those who, like you, have studied Chemistry, should consider that they have a social duty to counter this kind of predatory and deceptive pseudoscience.
Disturbances of acid-base balance
Deviations of the blood plasma pH from its normal value of 7.4 by more than about ±.1 can be very serious. These conditions are known medically as acidosis and alkalosis. They can be caused by metabolic disturbances such as diabetes and by kidney failure in which excretion of H2PO4 , for example, is inhibited.
Numerous other processes lead to temporary unbalances. Thus hyperventilation, which can result from emotional upset, leads to above-normal loss of CO2, and thus to alkalosis. Similarly, hypoventilation can act as a compensatory mechanism for acidosis. On the other hand, retention of CO2 caused by bronchopneumonia, for example, can give rise to acidosis. Acidosis can also result from diarrhea (loss of alkaline fluid from the intestine), while loss of gastric contents by vomiting promotes alkalosis.
The atmosphere is naturally acidic
The natural, unpolluted atmosphere receives acidic, basic, and neutral substances from natural sources (volcanic emissions, salt spray, windblown dust and microbial metabolism) as well as from pollution (Figure 11.6.X). These react in a kind of gigantic acid-base titration to give a solution in which hydrogen ions must predominate to maintain charge balance (indicated by the identical widths of the bar graphs at the bottom labeled "anions" and "cations").
Carbon dioxide, however, is the major source of acidity in the unpolluted atmosphere. As will be explained farther on, CO2 makes up 0.032 vol-% of dry air, and dissolves in water to form carbonic acid:
$CO_{2(g) }+ H_2O(l) \rightleftharpoons H_2CO_{3(aq)}$
Thus all rain is “acidic” in the sense that it contains dissolved CO2 which will reduce its pH to 5.7. The term acid rain is therefore taken to mean rain whose pH is controlled by pollutants which can lower the pH into the range of 3-4, resulting in severe damage to the environment.
Origin of acid rain
Acid rain originates from emissions of SO2 and various oxides of nitrogen (known collectively as "NOx") that are formed during combustion processes, especially those associated with the burning of coal. Incineration of wastes, industrial operations such as smelting, and forest fires are other combustion-related sources; these also release particulate material into the atmosphere which plays an important role in concentrating and distributing these acid-forming substances.
Origins of NOx
Nitrogen oxides are formed naturally by soil bacteria acting on nitrate ions, an essential plant nutrient and itself a product of natural nitrogen fixation. Ammonia, a product of bacterial decomposition of organic matter, is eventually transformed into NO3 by bacteria into nitrogen oxides.
The quantity of fixed nitrogen produced industrially to support intensive agriculture now exceeds the amount formed naturally, and has become the major source of anthropogenic nitrogen oxides.
• Atmospheric nitrogen can be thermally decomposed to its various oxides NOx when subjected to high temperatures; one minor natural source is lightning. More significantly, these oxides are also formed when "clean" fuels such as natural gas (mainly methane, CH4) is burnt in air.
• NOx can also react with other atmospheric pollutants to produce photochemical smog.
Formation of acids
The gases noted above react with atmospheric oxygen, with each other, and with particulate materials to form sulfuric, nitric, and hydrochloric acids.
• Most of the H2SO4 comes from the photooxidation of SO2 released from the burning of fossil fuels and from industrial operations such as smelting.
• Nitric acid similarly results from photooxidation of nitrogen oxides. Large quantities of these oxides are formed in high-temperature combustion processes in gasoline- and jet engines, and in the turbine engines of fossil-fueled electric power plants.
• Hydrochloric acid is formed naturally through the reaction of sea salt aerosols with atmospheric H2SO4 and HNO3. The salt particles form when ocean waves break the surface and some of the water evaporates before the spray settles back to the surface. On a much smaller scale, fossil fuel combustion and waste incineration are believed to the principal human-caused sources of HCl.
Wet and dry deposition
There are two basic mechanisms by which acidic material is transported through the environment. The acid that dissolves in the water droplets that form clouds and precipitation, and is eventually incorporated into fog and snowflakes, undergoes wet deposition; this is ordinarily what is referred to as "acid rain". Because it is widely dispersed, often high in the atmosphere, it can travel very large distances from the original source.
About 40% of the acidic substances introduced into the atmosphere becomes adsorbed onto particulate matter, which can be soot, smoke, windblown dust, and salt particles formed naturally from sea spray. This spreads less widely (typically around 30 km from the source), and falls onto surfaces by "dry deposition". Once the acid-laden particles land on surfaces, ordinary rain, fog, and dew release the acidic components, often in considerably more concentrated form than occurs through dry deposition alone.
Effects on soils
Because soils support the growth of plants and of the soil microorganisms that are essential agents in the recycling of dead plant materials, acid rain has a indirect but profound effect on soil health and plant growth.
↓ Click on image to enlarge
Regions of poorly-buffered soil in the U.S.
[link]
↓ Click on image to enlarge
Regions of poorly-buffered soil in Canada [gc.ca]
Soils containing alkaline components (most commonly limestone CaCO3 and other insoluble carbonates) can neutralize added acid and mitigate its effects. But soils in high mountain regions tend to be thin and unable to provide adequate buffering capacity. The same is true in almost half of Canada, in which granite rock of the Canadian Shield is very close to the surface; the eastern provinces of the country are strongly impacted by acid rain.
• Most soils also contain clay and humic substances that bind and retain ions such as Ca2+, Mg2+ and K+ which are essential plant nutrients. Added H+ binds even more strongly to these substances, displacing the nutrient ions from the upper layers. The overall effect is to leach these ions to greater depths where they may be inaccessible to plants, or to wash them away into ground water.
• The SO42– component of acid rain converts some nutrient cations into insoluble sulfates, reducing their availability to plants.
• Clays, which are complex aluminosilicates, gradually break down, releasing aluminum ions which normally form insoluble Al(OH)3. Addition of H+ dissolves this hydroxide, raising the concentration of Al(H2O)63+ to levels that can be toxic to plants.
• Many toxic heavy metals such as Pb, Cd and Cr are present in trace amounts in soils but are tied up in the form of insoluble salts. Acid rain can mobilize the ions of these metals, much as happens with aluminum.
Effects on plants
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Leaf damaged by acid rain
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Forest in the U.S Andirondacks damaged by acid rain
[link]
The effects on soils noted above affect plants most strongly. However, direct impingment of acidic rain and fog on leaves has other effects which can be especially serious when air pollutants such as SO2 are present.
• Acid deposition on leaves and needles tends to weaken them by eroding away their protective waxy coatings. This often results in the development of brown spots that interfere with photosynthesis.
• Forests in mountain regions are frequently bathed in clouds and fog which can be even more acidic than normal precipitation, exacerbating the above problem. In addition, the leaves and needles tend to accumulate the fog aerosol into larger droplets. These eventually fall to the forest floor, magnifying the effects on soils described in the previous section.
• Food crops tend to be less affected by acid rain where good farming practices such as the addition of fertilizers to replace depleted nutrients and the addition of limestone to raise the soil pH.
Effects on lakes and aquatic ecosystems
Lakes are subject not only to wet and dry acidic deposition, but also by the water they receive from streams and surface runoff. Any toxic elements released by the action of acid rain on soils and sediments can thus be conveyed to, and concentrated within lakes and the streams that empty them.
↓ Click on image to enlarge
Regions of poorly-buffered lakes in North America in 1980 [link]
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May Lake in the California Sierras - a typical alpine lake
[link]
Lakes in poorly-buffered areas such as are found in alpine regions (western North America, Colorado and much of Switzerland) or on the Canadian Shield and in the Adirondacs and Appalachians) are highly sensitive to acid deposition.
Severely acidified lakes (such as the one depicted at the above right) can be so devoid of life, including algae, that the water appears to be perfectly clear and bright blue in color.
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pH tolerances of some aquatic organisms
[U.S. EPA]
Aquatic organisms are generally adapted to "ordinary" pH conditions of 6-8, but vary greatly in their tolerance of low pH. As pH declines below 6, the diversity of aquatic animals, plants, and microorganisms diminishes.
• Some plant species such as sphagnum mosses and certain filamentous algaes that thrive at very low pH can proliferate in acidified lakes, producing thick mats that seal off oxygen and thus inhibit the decay of litter on the lake floor.
• Acid deposited onto winter snow is released during spring melting and can cause rapid drops of pH in poorly-buffered lakes that receive waters from these sources. This "spring acid shock" can seriously affect viability of organisms such as fish, amphibians, and insects that lay their eggs in the water which hatch in the spring. The hatchlings are often unable to adapt to the rapid change, and end up deformed or killed.
• ↓ Click on image to enlarge
Fish damaged by acid deposition [link]
Even those aquatic species able to survive at in low-pH waters can be indirectly affected if their food supply is pH-limited. For example, mayflies and some other insects, which, which are important food sources for frogs, cannot survive below pH 5.5.
• Fish are severely affected by the aluminum that is released from the action of acid on sediments; it causes a coating of mucus to form on the gills, impeding absorption of oxygen. This has led to the extinction of some species from affected lakes.
• Low pH increases the solubility of calcium salts, making it more difficult for bone to form in developing embryos of fish and amphibians.
• Low pH can also make the eggs of aquatic organism thicker and more difficult for embryos to penetrate, thus delaying hatching. When the embryos continue to grow within the confined space of the egg, their spines can be deformed, interfering with their viability when they finally hatch.
Effects on buildings and statues
Acid rain is not new!
"It has often been observed that the stones and bricks of buildings, especially under projecting parts, crumble more readily in large towns where coal is burnt... I was led to attribute this effect to the slow but constant action of acid rain." Robert Angus Smith, 1856
Acid deposition most strongly affects heritage buildings made of limestone and similar carbonate-containing stone materials.
The principal chemical process involves the reaction of sulfuric acid with calcium carbonate:
$\ce{CO3(s) + H2SO4 → CaSO4(s) + CO2}.$
The acid first erodes and breaks up the surface of the stone. As the hydrated calcium sulfate (gypsum) forms, it picks up iron and other components of the stone and forms an unsightly black coating. Some of this gradually blisters off, exposing yet more stone suface. The gypsum crystals can sometimes grow into the stone, further undermining the surface.
Very old structures such as the Taj Mahal, Notre Dame, the Colosseum and Westminster Abbey have all been affected.
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Stone buildings and monuments are rapidly damaged by acid rain [link]
The exposed upper survaces of the gargoyle are beginning to deteriorate.
[link]
Statues and monuments, including those made from marble, are also susceptible to erosion and damage from acid deposition.
Modern buildings are less affected, although acid rain can erode some painted surfaces.
The Geochemical Carbonate system
The carbonate system, consisting as it does of a soluble gas CO2, soluble ions HCO3 and CO32–, and sparingly soluble carbonate salts, spans all four realms of nature: the atmosphere, hydrosphere (mainly the oceans), the lithosphere, and the biosphere. And owing to the acid-base equilibria that govern the transformations between these carbonate species, carbon is readily transported between these geochemical reservoirs. This chapter presents an overview of carbon geochemistry.
Distribution of carbon on Earth
Carbon is the fourth most abundant element in the universe. Within the Earth's crust it ranks 15th, mostly in the form of carbonates in limestones and dolomites. Kerogens, which are fossilized plant-derived carbon mostly in the form of oil shale, constitute another major repository of terrestrial carbon.
Source relative to atmosphere
carbonate rock
carbonate rock
28,500
fossil carbon
572
10,600
land - organic carbon
0.065
1.22
ocean
3.2
61.8
atmospheric CO2
0.0535
1.0
The geochemical carbon cycle
The carbonate system encompasses virtually all of the environmental compartments– the atmosphere, hydrosphere, biosphere, and major parts of the lithosphere. The complementary processes of photosynthesis and respiration drive a global cycle in which carbon passes slowly between the atmosphere and the lithosphere, and more rapidly between the atmosphere and the oceans. Thus the "carbon cycle" can be divided into "fast" and "slow" parts, operating roughly on annual and geological time scales.
Figure (2\): Image by David Bise, Pennsylvania State University; see here for a full discussion of this diagram.
Another excellent depiction of the carbon cycle, by the U.S. Dept of Energy. (See description):
Carbon Cycle
Carbon dioxide in the atmosphere
CO2 has probably always been present in the atmosphere in the relatively small absolute amounts now observed. Precambrian limestones, possibly formed by reactions with rock-forming silicates, e.g.
$\ce{CaSiO_3 + CO_2 → CaCO_3 + SiO_2} \label{4-1}$
have likely had a moderating influence on the CO2 abundance throughout geological time.
The volume-percent of CO2 in dry air is 0.032%, leading to a partial pressure of 3 × 10–4 (103.5) atm. In a crowded and poorly-ventilated room, PCO2 can be as high as 100 × 10–4 atm. About 54 × 1014 moles per year of CO2 is taken from the atmosphere by photosynthesis, divided about equally between land and sea. Of this, all except 0.05% is returned by respiration (mostly by microorganisms); the remainder leaks into the slow, sedimentary part of the geochemical cycle where it can remain for thousands to millions of years. Trends in the growth of atmospheric CO2 for the past five years can be seen on this US NOAA page.
Since the advent of large-scale industrialization around 1860, the amount of CO2 in the atmosphere has been increasing. Most of this has been due to fossil-fuel combustion; in 1966, about 3.6 × 1015 g of C was released to the atmosphere; this is about 12 times greater than the estimated natural removal of carbon into sediments. The large-scale destruction of tropical forests, which has accelerated greatly in recent years, is believed to exacerbate this effect by removing a temporary sink for CO2.
About 30-50% of the CO2 released into the atmosphere by combustion remains there; the remainder enters the hydrosphere and biosphere. The oceans have a large absorptive capacity for CO2 by virtue of its transformation into bicarbonate and carbonate in a slightly alkaline aqueous medium, and they contain about 60 times as much inorganic carbon as is in the atmosphere. However, effcient transfer takes place only into the topmost
(100 m) wind-mixed layer, which contains only about one atmosphere equivalent of CO2; mixing time into the deeper parts of the ocean is of the order of 1000 years. For this reason, only about ten percent of added CO2 is taken up by the oceans.
Most of the carbon in the oceans is in the form of bicarbonate, as would be expected from the pH which ranges between 7.8 and 8.2. In addition to atmospheric CO2, there is a carbonate input to the ocean from streams. This is mostly in the form of HCO3 which derives from the weathering of rocks and terrestrial carbonate sediments, and the acid-base reaction
$H_2CO_3 + CO_3^{2–} → 2 HCO_3^– \label{4-2}$
which can be considered to be the source of bicarbonate in seawater. In this sense, the ocean can be considered the site of a gigantic acid-base titration in which atmospheric acids (mainly CO2 but also SO2, HCl, and other acids of volcanic origin) react with bases that originate from rocks and are introduced through carbonate-bearing streams or in windblown dust.
Carbon in the hydrosphere
Dissolution of CO2 in water
Carbon dioxide is slightly soluble in water:
°C
0
4
10
20
mol L–1
0.077
0.066
0.054
0.039
Henry's law is followed up to a CO2 pressure of about 5 atm:
$[\ce{CO2 (aq)}] = 0.032 P_{\ce{CO_2}} \label{4-3}$
“Dissolved carbon dioxide” consists mostly of the hydrated oxide CO2(aq) together with a small proportion of carbonic acid:
$[\ce{CO2 (aq)}] = 650 [\ce{H2CO3}] \label{4-4}$
The acid dissociation constant $K_{a1}$ that is commonly quoted for "H2CO3" is really a composite equilibrium constant that includes the above equilibrium. This means that "pure" H2CO3 (which cannot be isolated) is a considerably stronger acid than is usually appreciated.
Distribution of carbonate species in aqueous solutions
Water exposed to the atmosphere with PCO2 = 103.5 atm will take up carbon dioxide, which becomes distributed between the three carbonate species CO2, HCO3 and CO32– in proportions that depend on K1 and K2 and on the pH. The "total dissolved carbon" is given by the mass balance
$C_t = [H_2CO_3] + [HCO_3^–] + [CO_3^{2–}] \label{4-5}$
The distribution of these species as a function of pH can best be seen by constructing a log C-pH diagram for Ct = 10–5 M.
This Sillén plot is drawn for two different concentrations of the carbonate system. The lower one, in heavier lines, is for a 10–5 M solution, corresponding roughly to atmospheric CO2 in equilibrium with pure water. The upper plot, for a 10–3 M solution, is representative of many natural waters such as lakes and streams where the water is in contact with sediments.
It is important to note that this diagram applies only to a system in which Ct is constant. In a solution that is open to the atmosphere, this will not be the case at high pH values where the concentration of CO2 is appreciable. Under these conditions this ion will react with H2CO3 and the solution will absorb CO2 from the atmosphere, eventually resulting in the formation of a solid carbonate precipitate.
The lower of the two above plots can be used to predict the pH of 10^{5} M solutions of carbon dioxide, sodium bicarbonate, and sodium carbonate in pure water. The pH values are estimated by using the mass and charge balance conditions for each solution as noted below.
The reasoning leading to these calculations is explained in the discussion of the 10–3 M carbonate system in a previous chapter.
Solution of CO2 or H2CO3
[H+] = [OH] + [HCO3] + 2 [CO32–](4-6)
which, since the solution will be acidic, can be simplified to
[H+] = [HCO3] (point )(4-7)
Solution of NaHCO3
[H+] + [H2CO3] = [CO32–] + [OH](4-8)
or
[H+] = [HCO3] (point ) (4-9)
Solution of Na2CO3
[H+] + 2 [H2CO3] + [CO32–] = [OH](4-10)
or
[HCO3] = [OH] (point ) (4-11)
Carbon in the oceans
Natural waters aquire carbon from sediments they are in contact with, and of course also from the atmosphere. The pH is an important factor here; CO2 and solid carbonates are more soluble at high pH, and pH also controls the distribution of carbon species, as is seen in the plot just above.
Carbon content and pH of typical natural waters
rain river/lake water sea water
ppm of carbon
1.2
35
140
pH (unpolluted)
5.6
6.5 - 8.5
7.5 - 8.4
At the slightly alkaline pH of most bodies of water (of which the oceans constitute 97% of the earth's surface waters), bicarbonate is the principal dissolved carbon species. The quantity of organic carbon is fairly small.
Carbon content of the oceans , moles × 1018
CO2(aq) HCO3 CO32– dead org. living org.
0.18
2.6
0.33
0.23
0.05 | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.06%3A_Applications_of_Acid-Base_Equilibria.txt |
Learning Objectives
• Understand the exact equations that are involves in complex acid-base equilibria in aqueous solutions
The methods for dealing with acid-base equilibria that we developed in the earlier units of this series are widely used in ordinary practice. Although many of these involve approximations of various kinds, the results are usually good enough for most purposes. Sometimes, however — for example, in problems involving very dilute solutions, the approximations break down, often because they ignore the small quantities of H+ and OH ions always present in pure water. In this unit, we look at exact, or "comprehensive" treatment of some of the more common kinds of acid-base equilibria problems.
Strong acid-base systems
The usual definition of a “strong” acid or base is one that is completely dissociated in aqueous solution. Hydrochloric acid is a common example of a strong acid. When HCl gas is dissolved in water, the resulting solution contains the ions H3O+, OH, and Cl, but except in very concentrated solutions, the concentration of HCl is negligible; for all practical purposes, molecules of “hydrochloric acid”, HCl, do not exist in dilute aqueous solutions. To specify the concentrations of the three species present in an aqueous solution of HCl, we need three independent relations between them.
Limiting Conditions
These relations are obtained by observing that certain conditions must always hold for aqueous solutions:
1. The dissociation equilibrium of water must always be satisfied: $[H^+][OH^–] = K_w \label{1-1}$
2. The undissociated acid and its conjugate base must be in mass balance. The actual concentrations of the acid and its conjugate base can depend on a number of factors, but their sum must be constant, and equal to the “nominal concentration”, which we designate here as $C_a$. For a solution of HCl, this equation would be $[HCl] + [Cl^–] = C_a \label{1-2}$ Because a strong acid is by definition completely dissociated, we can neglect the first term and write the mass balance condition as $[Cl^–] = C_a \label{1-3}$
3. In any ionic solution, the sum of the positive and negative electric charges must be zero; in other words, all solutions are electrically neutral. This is known as the electroneutrality principle. $[H^+] = [OH^–] + [Cl^–] \label{1-4}$
The next step is to combine these three limiting conditions into a single expression that relates the hydronium ion concentration to $C_a$. This is best done by starting with an equation that relates several quantities and substituting the terms that we want to eliminate. Thus we can get rid of the $[Cl^–]$ term by substituting Equation $\ref{1-3}$ into Equation $\ref{1-4}$ :
$[H^+] = [OH^–] + C_a \label{1-5}$
The $[OH^–]$ term can be eliminated by the use of Equation $\ref{1-1}$:
$[H^+] = C_a + \dfrac{K_w}{[H^+]} \label{1-6}$
This equation tells us that the hydronium ion concentration will be the same as the nominal concentration of a strong acid as long as the solution is not very dilute. As the acid concentration falls below about 10–6 M, however, the second term predominates; $[H^+]$ approaches $\sqrt{K_w}$ or $10^{–7} M$ at 25 °C. The hydronium ion concentration can of course never fall below this value; no amount of dilution can make the solution alkaline!
No amount of dilution can make the solution of a strong acid alkaline!
Notice that Equation $\ref{1-6}$ is a quadratic equation; in regular polynomial form it would be rewritten as
$[H^+]^2 – C_a[H^+] – K_w = 0 \label{1-7}$
Most practical problems involving strong acids are concerned with more concentrated solutions in which the second term of Equation $\ref{1-7}$ can be dropped, yielding the simple relation
$[H^+] \approx [A^–]$
Activities and Concentrated Solutions of Strong Acids
In more concentrated solutions, interactions between ions cause their “effective” concentrations, known as their activities, to deviate from their “analytical” concentrations. Thus in a solution prepared by adding 0.5 mole of the very strong acid HClO4 to sufficient water to make the volume 1 liter, freezing-point depression measurements indicate that the concentrations of hydronium and perchlorate ions are only about 0.4 M. This does not mean that the acid is only 80% dissociated; there is no evidence of HClO4 molecules in the solution. What has happened is that about 20% of the H3O+ and ClO4 ions have formed ion-pair complexes in which the oppositely-charged species are loosely bound by electrostatic forces. Similarly, in a 0.10 M solution of hydrochloric acid, the activity of H+ is 0.81, or only 81% of its concentration. (See the green box below for more on this.)
Activities are important because only these work properly in equilibrium calculations. The relation between the concentration of a species and its activity is expressed by the activity coefficient $\gamma$:
$a = \gamma C \label{1-8}$
As a solution becomes more dilute, $\gamma$ approaches unity. At ionic concentrations below about 0.001 M, concentrations can generally be used in place of activities with negligible error. Recall that pH is defined as the negative logarithm of the hydrogen ion activity, not its concentration.
Activities of single ions cannot be determined, so activity coefficients in ionic solutions are always the average, or mean, of those for all ionic species present. This quantity is denoted as $\gamma_{\pm}$.
At very high concentrations, activities can depart wildly from concentrations. This is a practical consideration when dealing with strong mineral acids which are available at concentrations of 10 M or greater. In a 12 M solution of hydrochloric acid, for example, the mean ionic activity coefficient* is 207. This means that under these conditions with [H+] = 12, the activity {H+} = 2500, corresponding to a pH of about –3.4, instead of –1.1 as might be predicted if concentrations were being used. These very high activity coefficients also explain another phenomenon: why you can detect the odor of HCl over a concentrated hydrochloric acid solution even though this acid is supposedly "100% dissociated".
At these high concentrations, a pair of "dissociated" ions $H^+$ and $Cl^–$ will occasionally find themselves so close together that they may momentarily act as an HCl unit; some of these may escape as $HCl(g)$ before thermal motions break them up again. Under these conditions, “dissociation” begins to lose its meaning so that in effect, dissociation is no longer complete. Although the concentration of $HCl(aq)$ will always be very small, its own activity coefficient can be as great as 2000, which means that its escaping tendency from the solution is extremely high, so that the presence of even a tiny amount is very noticeable.
Weak monoprotic acids and bases
Most acids are weak; there are hundreds of thousands of them, whereas there are no more than a few dozen strong acids. We can treat weak acid solutions in exactly the same general way as we did for strong acids. The only difference is that we must now include the equilibrium expression for the acid. We will start with the simple case of the pure acid in water, and then go from there to the more general one in which strong cations are present. In this exposition, we will refer to “hydrogen ions” and $[H^+]$ for brevity, and will assume that the acid $HA$ dissociates into $H^+$ and its conjugate base $A^-$.
Pure acid in water
In addition to the species H+, OH, and A− which we had in the strong-acid case, we now have the undissociated acid HA; four variables, requiring four equations.
• Equilibria
$[H^+][OH^–] = K_w \label{2-1}$
$K_a = \dfrac{[H^+][A^–]}{[HA]} \label{2-2}$
• Mass balance
$C_a = [HA] + [A^–] \label{2-3}$
• Charge balance
$[H^+] = [OH^–] + [HA^–] \label{2-4}$
To eliminate [HA] from Equation $\ref{2-2}$, we solve Equation $\ref{2-4}$ for this term, and substitute the resulting expression into the numerator:
$K_a =\dfrac{[H^+]([H^+] - [OH^-])}{C_a-([H^+] - [OH^-]) } \label{2-5}$
The latter equation is simplified by multiplying out and replacing [H+][OH] with Kw. We then get rid of the [OH] term by replacing it with Kw/[H+]
$[H^+] C_b + [H^+]^2 – [H^+][OH^–] = K_a C_a – K_a [H^+] + K_a [OH^–]$
$[H^+]^2 C_b + [H^+]^3 – [H^+] K_w = K_a C_a – K_a [H^+] + \dfrac{K_a K_w}{[H^+]}$
Rearranged into standard polynomial form, this becomes
$[H^+]^3 + K_a[H^+]^2 – (K_w + C_aK_a) [H^+] – K_a K_w = 0 \label{2-5a}$
For most practical applications, we can make approximations that eliminate the need to solve a cubic equation.
Approximation 1: Neglecting Hydroxide Population
Unless the acid is extremely weak or the solution is very dilute, the concentration of OH can be neglected in comparison to that of [H+]. If we assume that [OH] ≪ [H+], then Equation $\ref{2-5a}$ can be simplified to
$K_a \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{2-6}$
which is a quadratic equation:
$[H^+]^2 +K_a[H^+]– K_aC_a \approx 0 \label{2-7}$
and thus, from the quadratic formula,
$[H^+] \approx \dfrac{K_a + \sqrt{K_a + 4K_aC_a}}{2} \label{2-8}$
Example $1$: Acetic Acid
Calculate the pH of a 0.0010 M solution of acetic acid, $K_a = 1.74 \times 10^{–5}$.
Solution
First approximation:
$[H^+] = \sqrt{(1.0 \times 10^{–3}) × (1.74 \times 10^{–5}} = \sqrt{1.74 \times 10^{–8}} = 1.3 \times 10^{–4}\; M. \nonumber$
Applying the "5% test",
$\dfrac{1.3 \times 10^{–4}}{1.0 \times 10^{–3}} = 0.13\nonumber$
This exceeds 0.05, so we must explicitly solve the quadratic Equation $\ref{2-7}$ to obtain two roots: $+1.2 \times 10^{–4}$ and $–1.4 \times 10^{-4}$. Taking the positive root, we have
$pH = –\log (1.2 \times 10^{–4}) = 3.9 \nonumber$
Approximation 2: Very Concentrated Acids
If the acid is fairly concentrated (usually more than 10–3 M), a further simplification can frequently be achieved by making the assumption that $[H^+] \ll C_a$. This is justified when most of the acid remains in its protonated form [HA], so that relatively little H+ is produced. In this event, Equation $\ref{2-6}$ reduces to
$K_a \approx \dfrac{[H^+]^2}{C_a} \label{2-9}$
or
$[H^+] \approx \sqrt{K_aC_a} \label{2-10}$
Example $2$
Calculate the pH and percent ionization of 0.10 M acetic acid "HAc" (CH3COOH), $K_a = 1.74 \times 10^{–5}$.
Solution:
It is usually best to start by using Equation $\ref{2-9}$ as a first approximation:
$[H^+] = \sqrt{(0.10)(1.74 \times 10^{–5})} = \sqrt{1.74 \times 10^{–6}} = 1.3 \times 10^{–3}\; M\nonumber$
This approximation is generally considered valid if [H+] is less than 5% of Ca; in this case, [H+]/Ca = 0.013, which is smaller than 0.05 and thus within the limit. This same quantity also corresponds to the ionization fraction, so the percent ionization is 1.3%. The pH of the solution is
$pH = –\log 1.2 \times 10^{-3} = 2.9\nonumber$
Approximation 3: Very Weak and Acidic
If the acid is very weak or its concentration is very low, the $H^+$ produced by its dissociation may be little greater than that due to the ionization of water. However, if the solution is still acidic, it may still be possible to avoid solving the cubic equation $\ref{2-5a}$ by assuming that the term $([H^+] - [OH^–]) \ll C_a$ in Equation $\ref{2-5}$:
$K_a = \dfrac{[H^+]^2}{C_a - [H^+]} \label{2-11}$
This can be rearranged into standard quadratic form
$[H^+]^2 + K_a [H^+] – K_a C_a = 0 \label{2-12}$
For dilute solutions of weak acids, an exact treatment may be required. With the aid of a computer or graphic calculator, solving a cubic polynomial is now far less formidable than it used to be. However, round-off errors can cause these computerized cubic solvers to blow up; it is generally safer to use a quadratic approximation.
Example $3$: Boric Acid
Boric acid, B(OH)3 ("H3BO3") is a weak acid found in the ocean and in some natural waters. As with many boron compounds, there is some question about its true nature, but for most practical purposes it can be considered to be monoprotic with $K_a = 7.3 \times 10^{–10}$:
$Bi(OH)_3 + 2 H_2O \rightleftharpoons Bi(OH)_4^– + H_3O^+\nonumber$
Find the [H+] and pH of a 0.00050 M solution of boric acid in pure water.
Solution
Because this acid is quite weak and its concentration low, we will use the quadratic form Equation $\ref{2-7}$, which yields the positive root $6.12 \times 10^{–7}$, corresponding to pH = 6.21. Notice that this is only six times the concentration of $H^+$ present in pure water!
It is instructive to compare this result with what the quadratic approximation would yield, which yield $[H^+] = 6.04 \times 10^{–7}$ so $pH = 6.22$.
Weak bases
The weak bases most commonly encountered are:
• anions A of weak acids:
$A^– + H_2O \rightleftharpoons HA + OH^–$
$CO_3^{2–} + H_2O \rightleftharpoons HCO_3^– + OH^–$
• ammonia
$NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^–$
• amines, e.g. methylamine
$CH_3NH_2 + H_2O \rightleftharpoons CH_3NH_3^++ H_2O$
Solution of an anion of a weak acid
Note that, in order to maintain electroneutrality, anions must be accompanied by sufficient cations to balance their charges. Thus for a Cb M solution of the salt NaA in water, we have the following conditions:
• Species
Na+, A, HA, H2O, H+, OH
• Equilibria
$[H^+][OH^–] = K_w \label{2-13}$
$K_b =\dfrac{[HA][OH^-]}{[A^-]} \label{2-14}$
• Mass balance
$C_b = [Na^+] = [HA] + [A^–] \label{2-15}$
• Charge balance
$[Na^+] + [H^+] = [OH^–] + [A^–] \label{2-16}$
Replacing the [Na+] term in Equation $\ref{2-15}$ by $C_b$ and combining with $K_w$ and the mass balance, a relation is obtained that is analogous to that of Equation $\ref{2-5}$ for weak acids:
$K_b =\dfrac{[OH^-] ([OH^-] - [H^+])}{C_b - ([OH^-] - [H^+])} \label{2-17}$
The approximations
$K_b \approx \dfrac{[OH^-]^2}{C_b - [OH^-]} \label{2-18}$
and
$[OH^–] \approx \sqrt{K_b C_b} \label{2-19}$
can be derived in a similar manner.
Example $4$: Methylamine
Calculate the pH and the concentrations of all species in a 0.01 M solution of methylamine, CH3NH2 ($K_b = 4.2 \times 10^{–4}$).
Solution
We begin by using the simplest approximation Equation $\ref{2-14}$:
$[OH^–] = \sqrt{(K_b C_b}- = \sqrt{(4.2 \times 10^{-4})(10^{–2})} = 2.1 \times 10^{–3}\nonumber$
To see if this approximation is justified, we apply a criterion similar to what we used for a weak acid: [OH] must not exceed 5% of Cb. In this case,
$\dfrac{[OH^–]}{ C_b} = \dfrac{(2.1 \times 10^{-3}} { 10^{–2}} = 0.21\nonumber$
so we must use the quadratic form Equation $\ref{2-12}$ that yields the positive root $1.9 \times 10^{–3}$ which corresponds to $[OH^–]$
$[H^+] = \dfrac{K_w}{[OH^–} = \dfrac{1 \times 10^{-14}}{1.9 \times 10^{–3}} = 5.3 \times 10^{-12}\nonumber$
and
$pH = –\log 5.3 \times 10^{–12} = 11.3.\nonumber$
From the charge balance equation, solve for
$[CH_3NH_2] = [OH^–] – [H^+] \approx [OH^–] = 5.3 \times 10^{–12}\; M. \nonumber$
For the concentration of the acid form (methylaminium ion CH3NH3+), use the mass balance equation:
$[CH_3NH_3^+] = C_b – [CH_3NH_2] = 0.01 – 0.0019 =0.0081\; M.\nonumber$
Mixtures of Acids
Many practical problems relating to environmental and physiological chemistry involve solutions containing more than one acid. In this section, we will restrict ourselves to a much simpler case of two acids, with a view toward showing the general method of approaching such problems by starting with charge- and mass-balance equations and making simplifying assumptions when justified. In general, the hydrogen ions produced by the stronger acid will tend to suppress dissociation of the weaker one, and both will tend to suppress the dissociation of water, thus reducing the sources of H+ that must be dealt with.
Consider a mixture of two weak acids HX and HY; their respective nominal concentrations and equilibrium constants are denoted by Cx, Cy, Kx and Ky,
Starting with the charge balance expression
$[H^+] = [X^–] + [Y^–] + [OH^–] \label{3-1}$
We use the equilibrium constants to replace the conjugate base concentrations with expressions of the form
$[X^-] = K_x \dfrac{[HX]}{[H^+]} \label{3-2}$
to yield
$[H^+] = \dfrac{[HX]}{K_x} + \dfrac{[HY]}{K_y} + K_w \label{3-3}$
If neither acid is very strong or very dilute, we can replace equilibrium concentrations with nominal concentrations:
$[H^+] \approx \sqrt{C_cK_x + C_yK_y K_w} \label{3-4}$
Example $5$: Acetic Acid and Formic Acid
Estimate the pH of a solution that is 0.10M in acetic acid ($K_a = 1.8 \times 10^{–5}$) and 0.01M in formic acid ($K_a = 1.7 \times 10^{–4}$).
Solution
Because Kw is negligible compared to the CaKa products, we can simplify \Equation $ref{3-4}$:
$[H^+] = \sqrt{1.8 \times 10^{–6} + 1.7 \times 10^{-6}} = 0.0019\nonumber$
Which corresponds to a pH of $–\log 0.0019 = 2.7$
Note that the pH of each acid separately at its specified concentration would be around 2.8. However, if 0.001 M chloroacetic acid (Ka= 0.0014) is used in place of formic acid, the above expression becomes
$[H^+] \approx \sqrt{ 1.4 \times 10^{-6} + 1.75 \times 10^{-14}} = 0.00188 \label{3-5}$
which exceeds the concentration of the stronger acid; because the acetic acid makes a negligible contribution to [H+] here, the simple approximation given above \Equation $\ref{3-3}$ is clearly invalid. We now use the mass balance expression for the stronger acid
$[HX] + [X^–] = C_x \label{3-6}$
to solve for [X] which is combined with the equilibrium constant Kx to yield
$[X^-] = C_x - \dfrac{[H^+][X^]}{K_x} \label{3-7}$
Solving this for [X] gives
$[X^-] = \dfrac{C_xK_x}{K_x + [H^+]} \label{3-8}$
The approximation for the weaker acetic acid (HY) is still valid, so we retain it in the substituted electronegativity expression:
$[H^+] \dfrac{C_xK_x}{K_x+[H^+]} + \dfrac{C_yK_y}{[H^+]} \label{3-9}$
which is a cubic equation that can be solved by approximation.
Several methods have been published for calculating the hydrogen ion concentration in solutions containing an arbitrary number of acids and bases. These generally involve iterative calculations carried out by a computer. See, for example, J. Chem. Education 67(6) 501-503 (1990) and 67(12) 1036-1037 (1990).
Equilibria of Polyprotic Acids
Owing to the large number of species involved, exact solutions of problems involving polyprotic acids can become very complicated. Thus for phosphoric acid H3PO4, the three "dissociation" steps yield three conjugate bases:
Fortunately, it is usually possible to make simplifying assumptions in most practical applications. In the section that follows, we will show how this is done for the less-complicated case of a diprotic acid. A diprotic acid HA can donate its protons in two steps, yielding first a monoprotonated species HA and then the completely deprotonated form A2–.
Since there are five unknowns (the concentrations of the acid, of the two conjugate bases and of H+ and OH), we need five equations to define the relations between these quantities. These are
• Equilibria
$[H^+][OH^–] = K_w \label{4-1}$
$K_1 = \dfrac{[H^+][HA^-]}{[H_2A]} \label{4-2}$
$K_1 = \dfrac{[H^+][HA^{2-}]}{[HA^-]} \label{4-3}$
• Mass balance
$C_a = [H_2A] + [HA^–] + [A^{2–}] \label{4-4}$
• Charge balance
$[H^+] = [OH^–] + [HA^–] + 2 [A^{2–}] \label{4-5}$
(It takes 2 moles of $H^+$ to balance the charge of 1 mole of $A^{2–}$)
Solving these five equations simultaneously for $K_1$ yields the rather intimidating expression
$K_1 = \dfrac{[H^+] \left( [H^+] - [OH^-] \dfrac{2K_2[H^+] - [OH^-]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] - [OH^-] \dfrac{K_2 [H^+] -[OH^-]}{[H^+] + 2K_2} \right)} \label{4-6}$
which is of little practical use except insofar as it provides the starting point for various simplifying approximations.
If the solution is even slightly acidic, then ([H+] – [OH]) ≈ [H+] and
$K_1 = \dfrac{[H^+] \left( [H^+] \dfrac{2K_2[H^+]}{[H^+ + 2K_2} \right)}{C_a - \left( [H^+] \dfrac{K_2 [H^+]}{[H^+] + 2K_2} \right)} \label{4-7}$
For any of the common diprotic acids, $K_2$ is much smaller than $K_1$. If the solution is sufficiently acidic that $K_2 \ll [H^+]$, then a further simplification can be made that removes $K_2$ from Equation $\ref{4-7}$; this is the starting point for most practical calculations.
$K_1 \approx \dfrac{[H^+]^2}{C_a-[H^+]} \label{4-8}$
Finally, if the solution is sufficiently concentrated and $K_1$ sufficiently small so that $[H^+] \ll C_a$, then Equation $\ref{4-8}$ reduces to:
$[H^+] \approx \sqrt{K_a C_a}$
Acid with conjugate base: Buffer solutions
Solutions containing a weak acid together with a salt of the acid are collectively known as buffers. When they are employed to control the pH of a solution (such as in a microbial growth medium), a sodium or potassium salt is commonly used and the concentrations are usually high enough for the Henderson-Hasselbalch equation to yield adequate results.
Exact solution
In this section, we will develop an exact analytical treatment of weak acid-salt solutions, and show how the HH equation arises as an approximation. A typical buffer system is formed by adding a quantity of strong base such as sodium hydroxide to a solution of a weak acid HA. Alternatively, the same system can be made by combining appropriate amounts of a weak acid and its salt NaA. A system of this kind can be treated in much the same way as a weak acid, but now with the parameter Cb in addition to Ca.
• Species
Na+, A, HA, H2O, H+, OH
• Equilibria
$[H^+][OH^–] = K_w \label{5-1}$
$K_a = \dfrac{[H^+][A^-]}{[HA]} \label{5-2}$
• Mass balance
$C_a + C-b = [HA] + [A^–] \label{5-3}$
$C_b = [Na^+] \label{5-4}$
• Charge balance
$[Na^+] + [H^+] = [OH^–] + [A^–] \label{5-5}$
Substituting Equation $\ref{5-4}$ into Equation $\ref{5-5}$ yields an expression for [A]:
$[A^–] = C_b + [H^+] – [OH^–] \label{5-6}$
Inserting this into Equation $\ref{5-3}$ and solving for [HA] yields
$[HA] = C_b + [H^+] – [OH^–] \label{5-7)}$
Finally, we substitute these last two expressions into the equilibrium constant (Equation $\ref{5-2}$):
$[H^+] = K_a \dfrac{C_a - [H^+] + [OH^-]}{C_b + [H^+] - [OH^-]} \label{5-8}$
which becomes cubic in [H+] when [OH] is replaced by (Kw / [H+]).
$[H^+]^3 +(C_b +K_a)[H^+]^2 – (K_w + C_aK_a) [H^+] – K_aK_w = 0 \label{5-8a}$
Approximations
In almost all practical cases it is possible to make simplifying assumptions. Thus if the solution is known to be acidic or alkaline, then the [OH] or [H+] terms in Equation $\ref{5-8}$ can be neglected. In acidic solutions, for example, Equation $\ref{5-8}$ becomes
$[H^+] = K_a \dfrac{C_a - [H^+]}{C_b + [H^+]} \label{5-9}$
which can be rearranged into a quadratic in standard polynomial form:
$[H^+]^2 + (C_b + C_a) [H^+] – K_a C_a = 0 \label{5-10}$
If the concentrations Ca and Cb are sufficiently large, it may be possible to neglect the [H+] terms entirely, leading to the commonly-seen Henderson-Hasselbalch Approximation.
$\color{red} [H^+] \approx K_a \dfrac{C_a}{C_b} \label{5-11}$
It's important to bear in mind that the Henderson-Hasselbalch Approximation is an "approximation of an approximation" that is generally valid only for combinations of Ka and concentrations that fall within the colored portion of this plot.
Most buffer solutions tend to be fairly concentrated, with Ca and Cb typically around 0.01 - 0.1 M. For more dilute buffers and larger Ka's that bring you near the boundary of the colored area, it is safer to start with Equation $\ref{5-9}$.
Example $6$: Chlorous Acid Buffer
Chlorous acid HClO2 has a pKa of 1.94. Calculate the pH of a solution made by adding 0.01 M/L of sodium hydroxide to a -.02 M/L solution of chloric acid.
Solution
In the resulting solution, Ca = Cb = 0.01M. On the plots shown above, the intersection of the log Ca = –2 line with the plot for pKa = 2 falls near the left boundary of the colored area, so we will use the quadratic form $\ref{5-10}$.
Substitution in Equation $\ref{5-10}$ yields
$H^+ + 0.02 H^+ – (10^{–1.9} x 10^{–2}) = 0 \nonumber$
which yields a positive root 0.0047 = [H+] that corresponds to pH = 2.3.
Note: Using the Henderson-Hassalbach Approximateion (Equation $\ref{5-11}$) would give pH = pKa = 1.9. | textbooks/chem/General_Chemistry/Chem1_(Lower)/13%3A_Acid-Base_Equilibria/13.07%3A_Exact_Calculations_and_Approximations.txt |
All chemical changes are accompanied by the absorption or release of heat. The intimate connection between matter and energy has been a source of wonder and speculation from the most primitive times; it is no accident that fire was considered one of the four basic elements (along with earth, air, and water) as early as the fifth century BCE. In this set of lessons, we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes. This will build the groundwork for a more ambitious goal: to predict the direction and extent of change itself.
• 14.1: Energy, Heat and Work
All chemical changes are accompanied by the absorption or release of heat. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes.
• 14.2: The First Law of Thermodynamics
"Energy cannot be created or destroyed"— this fundamental law of nature, more properly known as conservation of energy, is familiar to anyone who has studied science. Under its more formal name of the First Law of Thermodynamics, it governs all aspects of energy in science and engineering .
• 14.3: Molecules as Energy Carriers and Converters
All molecules at temperatures above absolue zero possess thermal energy— the randomized kinetic energy associated with the various motions the molecules as a whole, and also the atoms within them, can undergo. Polyatomic molecules also possess potential energy in the form of chemical bonds. Molecules are thus both vehicles for storing and transporting energy, and the means of converting it from one form to another is accompanied by the uptake or release of heat.
• 14.4: Thermochemistry and Calorimetry
The heat that flows across the boundaries of a system undergoing a change is a fundamental property that characterizes the process. It is easily measured, and if the process is a chemical reaction carried out at constant pressure, it can also be predicted from the difference between the enthalpies of the products and reactants. The quantitative study and measurement of heat and enthalpy changes is known as thermochemistry.
• 14.5: Calorimetry
Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.
• 14.6: Applications of Thermochemistry
Virtually all chemical processes involve the absorption or release of heat, and thus changes in the internal energy of the system. In this section, we survey some of the more common chemistry-related applications of enthalpy and the First Law. While the first two sections relate mainly to chemistry, the remaining ones impact the everyday lives of everyone.
• 14.E: Thermochemistry (Exercises)
14: Thermochemistry
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• The potential energy of an object relates to its location, but there is one additional requirement that must be satisfied for potential energy be present. Explain and give an example.
• Distinguish between the nature of kinetic energy that is associated with macroscopic bodies and that is found in microscopic objects such as atoms and molecules.
• Describe the meaning and origins of "chemical" energy.
• Define the calorie.
• Heat and work are both expressed in energy units, but they differ from "plain" energy in a fundamental way. Explain.
• ... and state the distinction between heat and work.
All chemical changes are accompanied by the absorption or release of heat. The intimate connection between matter and energy has been a source of wonder and speculation from the most primitive times; it is no accident that fire was considered one of the four basic elements (along with earth, air, and water) as early as the fifth century BCE. In this unit we will review some of the fundamental concepts of energy and heat and the relation between them. We will begin the study of thermodynamics, which treats the energetic aspects of change in general, and we will finally apply this specifically to chemical change. Our purpose will be to provide you with the tools to predict the energy changes associated with chemical processes. This will build the groundwork for a more ambitious goal: to predict the direction and extent of change itself.
What is Energy?
Energy is one of the most fundamental and universal concepts of physical science, but one that is remarkably difficult to define in way that is meaningful to most people. This perhaps reflects the fact that energy is not a “thing” that exists by itself, but is rather an attribute of matter (and also of electromagnetic radiation) that can manifest itself in different ways. It can be observed and measured only indirectly through its effects on matter that acquires, loses, or possesses it.
The concept that we call energy was very slow to develop; it took more than a hundred years just to get people to agree on the definitions of many of the terms we use to describe energy and the interconversion between its various forms. But even now, most people have some difficulty in explaining what it is; somehow, the definition we all learned in elementary science ("the capacity to do work") seems less than adequate to convey its meaning. Although the term "energy" was not used in science prior to 1802, it had long been suggested that certain properties related to the motions of objects exhibit an endurance which is incorporated into the modern concept of "conservation of energy". René Descartes (1596-1650) stated it explicitly:
When God created the world, He "caused some of its parts to push others and to transfer their motions to others..." and thus "He conserves motion".*
In the 17th Century, the great mathematician Gottfried Leibniz (1646-1716) suggested the distinction between vis viva ("live force") and vis mortua ("dead force"), which later became known as kinetic energy (1829) and potential energy (1853).
Kinetic Energy and Potential Energy
Whatever energy may be, there are basically two kinds. Kinetic energy is associated with the motion of an object, and its direct consequences are part of everyone's daily experience; the faster the ball you catch in your hand, and the heavier it is, the more you feel it. Quantitatively, a body with a mass m and moving at a velocity v possesses the kinetic energy mv2/2.
Example \(1\)
A rifle shoots a 4.25 g bullet at a velocity of 965 m s–1. What is its kinetic energy?
Solution
The only additional information you need here is that
1 J = 1 kg m2 s–2:
KE = ½ × (.00425 kg) (965 m s–1)2 = 1980 J
Potential energy is energy a body has by virtue of its location. But there is more: the body must be subject to a "restoring force" of some kind that tends to move it to a location of lower potential energy. Think of an arrow that is subjected to the force from a stretched bowstring; the more tightly the arrow is pulled back against the string, the more potential energy it has. More generally, the restoring force comes from what we call a force field— a gravitational, electrostatic, or magnetic field. We observe the consequences of gravitational potential energy all the time, such as when we walk, but seldom give it any thought.
If an object of mass m is raised off the floor to a height h, its potential energy increases by mgh, where g is a proportionality constant known as the acceleration of gravity; its value at the earth's surface is 9.8 m s–2.
Example \(2\)
Find the change in potential energy of a 2.6 kg textbook that falls from the 66-cm height of a table top onto the floor.
Solution
PE = m g h = (2.6 kg)(9.8 m s–2)(0.66 m) = 16.8 kg m2 s–2 = 16.8 J
Similarly, the potential energy of a particle having an electric charge q depends on its location in an electrostatic field.
"Chemical energy"
Electrostatic potential energy plays a major role in chemistry; the potential energies of electrons in the force field created by atomic nuclei lie at the heart of the chemical behavior of atoms and molecules.
"Chemical energy" usually refers to the energy that is stored in the chemical bonds of molecules. These bonds form when electrons are able to respond to the force fields created by two or more atomic nuclei, so they can be regarded as manifestations of electrostatic potential energy.
In an exothermic chemical reaction, the electrons and nuclei within the reactants undergo rearrangment into products possessing lower energies, and the difference is released to the environment in the form of heat.
Interconversion of potential and kinetic energy
Transitions between potential and kinetic energy are such an intimate part of our daily lives that we hardly give them a thought. It happens in walking as the body moves up and down. Our bodies utilize the chemical energy in glucose to keep us warm and to move our muscles. In fact, life itself depends on the conversion of chemical energy to other forms.
Energy is conserved: it can neither be created nor destroyed. So when you go uphill, your kinetic energy is transformed into potential energy, which gets changed back into kinetic energy as you coast down the other side. And where did the kinetic energy you expended in peddling uphill come from? By conversion of some of the chemical potential energy in your breakfast cereal.
• When drop a book, its potential energy is transformed into kinetic energy. When it strikes the floor, this transformation is complete. What happens to the energy then? The kinetic energy that at the moment of impact was formerly situated exclusively in the moving book, now becomes shared between the book and the floor, and in the form of randomized thermal motions of the molecular units of which they are made; we can observe this effect as a rise in temperature.
• ← Much of the potential energy of falling water can be captured by a water wheel or other device that transforms the kinetic energy of the exit water into kinetic energy. The output of a hydroelectric power is directly proportional to its height above the level of the generator turbines in the valley below. At this point, the kinetic energy of the exit water is transferred to that of the turbine, most of which (up to 90 percent in the largest installations) is then converted into electrical energy.
• Will the temperature of the water at the bottom of a water fall be greater than that at the top? James Joule himself predicted that it would be. It has been calculated that at Niagra falls, that complete conversion of the potential energy of 1 kg of water at the top into kinetic energy when it hits the plunge pool 58 meters below will result in a temperature increase of about 0.14 C°. (But there are lots of complications. For example, some of the water breaks up into tiny droplets as it falls, and water evaporates from droplets quite rapidly, producing a cooling effect.)
• Chemical energy can also be converted, at least partially, into electrical energy: this is what happens in a battery. If a highly exothermic reaction also produces gaseous products, the latter may expand so rapidly that the result is an explosion — a net conversion of chemical energy into kinetic energy (including sound).
Thermal energy
Kinetic energy is associated with motion, but in two different ways. For a macroscopic object such as a book or a ball, or a parcel of flowing water, it is simply given by ½ mv2.
But as we mentioned above, when an object is dropped onto the floor, or when an exothermic chemical reaction heats surrounding matter, the kinetic energy gets dispersed into the molecular units in the environment. This "microscopic" form of kinetic energy, unlike that of a speeding bullet, is completely random in the kinds of motions it exhibits and in its direction. We refer to this as "thermalized" kinetic energy, or more commonly simply as thermal energy. We observe the effects of this as a rise in the temperature of the surroundings. The temperature of a body is direct measure of the quantity of thermal energy is contains.
Thermal energy is never completely recoverable
Once kinetic energy is thermalized, only a portion of it can be converted back into potential energy. The remainder simply gets dispersed and diluted into the environment, and is effectively lost.
To summarize, then:
• Potential energy can be converted entirely into kinetic energy..
• Potential energy can also be converted, with varying degrees of efficiency,into electrical energy.
• The kinetic energy of macroscopic objects can be transferred between objects (barring the effects of friction).
• Once kinetic energy becomes thermalized, only a portion of it can be converted back into either potential energy or be concentrated back into the kinetic energy of a macroscopic. This limitation, which has nothing to do with technology but is a fundamental property of nature, is the subject of the second law of thermodynamics.
• A device that is intended to accomplish the partial transformation of thermal energy into organized kinetic energy is known as a heat engine.
Energy scales and units
You might at first think that a book sitting on the table has zero kinetic energy since it is not moving. But if you think about it, the earth itself is moving; it is spinning on its axis, it is orbiting the sun, and the sun itself is moving away from the other stars in the general expansion of the universe. Since these motions are normally of no interest to us, we are free to adopt an arbitrary scale in which the velocity of the book is measured with respect to the table; on this so-called laboratory coordinate system, the kinetic energy of the book can be considered zero.
We do the same thing with potential energy. If the book is on the table, its potential energy with respect to the surface of the table will be zero. If we adopt this as our zero of potential energy, and then push the book off the table, its potential energy will be negative after it reaches the floor.
Energy is measured in terms of its ability to perform work or to transfer heat. Mechanical work is done when a force f displaces an object by a distance d: w = f × d.
The basic unit of energy is the joule. One joule is the amount of work done when a force of 1 newton acts over a distance of 1 m; thus 1 J = 1 N-m. The newton is the amount of force required to accelerate a 1-kg mass by 1 m/sec2, so the basic dimensions of the joule are kg m2 s–2. The other two units in wide use. the calorie and the BTU (British thermal unit) are defined in terms of the heating effect on water. Because of the many forms that energy can take, there are a correspondingly large number of units in which it can be expressed, a few of which are summarized below.
1 calorie will raise the temperature of 1 g of water by 1 C°. The “dietary” calorie is actually 1 kcal. An average young adult expends about 1800 kcal per day just to stay alive.
(you should know this definition)
1 cal = 4.184 J
1 BTU (British Thermal Unit) will raise the temperature of 1 lb of water by 1F°. 1 BTU = 1055 J
The erg is the c.g.s. unit of energy and a very small one; the work done when a 1-dyne force acts over a distance of 1 cm.
1 J = 107 ergs
1 erg = 1 d-cm = 1 g cm2 s–2
The electron-volt is even tinier: 1 e-v is the work required to move a unit electric charge (1 C) through a potential difference of 1 volt. 1 J = 6.24 × 1018 e-v
The watt is a unit of power, which measures the rate of energy flow in J sec–1. Thus the watt-hour is a unit of energy. An average human consumes energy at a rate of about 100 watts; the brain alone runs at about 5 watts.
1 J = 2.78 × 10–4 watt-hr
1 w-h = 3.6 kJ
The liter-atmosphere is a variant of force-displacement work associated with volume changes in gases. 1 L-atm = 101.325 J
The huge quantities of energy consumed by cities and countries are expressed in quads; the therm is a similar but smaller unit. 1 quad = 1015 Btu = 1.05 × 1018 J
If the object is to obliterate cities or countries with nuclear weapons, the energy unit of choice is the ton of TNT equivalent. 1 ton of TNT = 4.184 GJ
(by definition)
In terms of fossil fuels, we have barrel-of-oil equivalent, cubic-meter-of-natural gas equivalent, and ton-of-coal equivalent.
1 bboe = 6.1 GJ
1 cmge = 37-39 mJ
1 toce = 29 GJ
Heat and work
Heat and work are both measured in energy units, so they must both represent energy. How do they differ from each other, and from just plain “energy” itself? In our daily language, we often say that "this object contains a lot of heat", but this kind of talk is a no-no in thermodynamics! It's ok to say that the object is "hot", meaning that its temperature is high. The term "heat" has a special meaning in thermodynamics: it is a process in which a body (the contents of a tea kettle, for example) acquires or loses energy as a direct consequence of its having a different temperature than its surroundings (the rest of the world).
Heat and work are processes and cannot be stored
Thermal energy can only flow from a higher temperature to a lower temperature. It is this flow that constitutes "heat". Use of the term "flow" of heat recalls the 18th-century notion that heat is an actual substance called “caloric” that could flow like a liquid.
Heat is transferred by conduction or radiation
Transfer of thermal energy can be accomplished by bringing two bodies into physical contact (the kettle on top of the stove, or through an electric heating element inside the kettle). Another mechanism of thermal energy transfer is by radiation; a hot object will convey energy to any body in sight of it via electromagnetic radiation in the infrared part of the spectrum. In many cases, a combination of modes will be active. Thus when you place a can of beer in the refrigerator, both processes are operative: the can radiates heat to the cold surfaces around it, and absorbs it by direct conduction from the ambient air.
So what is work?
Work refers to the transfer of energy some means that does not depend on temperature difference.
Work, like energy, can take various forms, the most familiar being mechanical and electrical. Mechanical work arises when an object moves a distance Δx against an opposing force f:
\[w = f Δx N^{-m}\]
with \(1\, N^{-m}\) = 1\, J.\)
Electrical work is done when a body having a charge q moves through a potential difference ΔV.
Work, like heat, exists only when energy is being transferred.When two bodies are placed in thermal contact and energy flows from the warmer body to the cooler one,we call the process “heat”. A transfer of energy to or from a system by any means other than heat is called “work”.
Interconvertability of heat and work
Work can be completely converted into heat (by friction, for example), but heat can only be partially converted to work. Conversion of heat into work is accomplished by means of a heat engine, the most common example of which is an ordinary gasoline engine. The science of thermodynamics developed out of the need to understand the limitations of steam-driven heat engines at the beginning of the Industrial Age. A fundamental law of Nature, the Second Law of Thermodynamics, states that the complete conversion of heat into work is impossible. Something to think about when you purchase fuel for your car! | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.01%3A_Energy_Heat_and_Work.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Define: system, surroundings, state properties, change of state (how the latter is calculated).
• Write out the equation that defines the First Law, and explain its physical meaning.
• Describe the meaning of a pathway in thermodynamics, and how it relates to state functions.
• Comment on the meaning of pressure-volume work: how is it calculated? What kinds of chemical reactions involve P-V work?
• Define isothermal and adiabatic processes, and give examples of each.
• Using the expansion of a gas as an example, state the fundamental distinction between reversible and irreversible changes in terms of the system + surroundings.
• Explain how enthalpy change relates to internal energy, and how it can be observed experimentally.
• Define heat capacity; explain its physical significance, and why there is a difference between Cp and Cv.
"Energy cannot be created or destroyed"— this fundamental law of nature, more properly known as conservation of energy, is familiar to anyone who has studied science. Under its more formal name of the First Law of Thermodynamics, it governs all aspects of energy in science and engineering applications. It's special importance in Chemistry arises from the fact that virtually all chemical reactions are accompanied by the uptake or release of energy. One of the interesting things about thermodynamics is that although it deals with matter, it makes no assumptions about the microscopic nature of that matter. Thermodynamics deals with matter in a macroscopic sense; it would be valid even if the atomic theory of matter were wrong. This is an important quality, because it means that reasoning based on thermodynamics is unlikely to require alteration as new facts about atomic structure and atomic interactions come to light.
A Thermodynamic view of the world
In thermodynamics, we must be very precise in our use of certain words. The two most important of these are system and surroundings. A thermodynamic system is that part of the world to which we are directing our attention. Everything that is not a part of the system constitutes the surroundings. The system and surroundings are separated by a boundary.
If our system is one mole of a gas in a container, then the boundary is simply the inner wall of the container itself. The boundary need not be a physical barrier; for example, if our system is a factory or a forest, then the boundary can be wherever we wish to define it. We can even focus our attention on the dissolved ions in an aqueous solution of a salt, leaving the water molecules as part of the surroundings. The single property that the boundary must have is that it be clearly defined, so we can unambiguously say whether a given part of the world is in our system or in the surroundings.
If matter is not able to pass across the boundary, then the system is said to be closed; otherwise, it is open. A closed system may still exchange energy with the surroundings unless the system is an isolated one, in which case neither matter nor energy can pass across the boundary. The tea in a closed Thermos bottle approximates a closed system over a short time interval.
The properties of a system are those quantities such as the pressure, volume, temperature, and its composition, which are in principle measurable and capable of assuming definite values. There are of course many properties other than those mentioned above; the density and thermal conductivity are two examples. However, the pressure, volume, and temperature have special significance because they determine the values of all the other properties; they are therefore known as state properties because if their values are known then the system is in a definite state.
In dealing with thermodynamics, we must be able to unambiguously define the change in the state of a system when it undergoes some process. This is done by specifying changes in the values of the different state properties using the symbol Δ (delta) as illustrated here for a change in the volume:
$ΔV = V_{final} – V_{initial} \label{1-1}$
We can compute similar delta-values for changes in P, V, ni (the number of moles of component i), and the other state properties we will meet later.
Internal energy is simply the totality of all forms of kinetic and potential energy of the system. Thermodynamics makes no distinction between these two forms of energy and it does not assume the existence of atoms and molecules. But since we are studying thermodynamics in the context of chemistry, we can allow ourselves to depart from “pure” thermodynamics enough to point out that the internal energy is the sum of the kinetic energy of motion of the molecules, and the potential energy represented by the chemical bonds between the atoms and any other intermolecular forces that may be operative.
How can we know how much internal energy a system possesses? The answer is that we cannot, at least not on an absolute basis; all scales of energy are arbitrary. The best we can do is measure changes in energy. However, we are perfectly free to define zero energy as the energy of the system in some arbitrary reference state, and then say that the internal energy of the system in any other state is the difference between the energies of the system in these two different states.
The First Law of Thermodynamics
This law is one of the most fundamental principles of the physical world. Also known as the Law of Conservation of Energy, it states that energy can not be created or destroyed; it can only be redistributed or changed from one form to another. A way of expressing this law that is generally more useful in Chemistry is that any change in the internal energy of a system is given by the sum of the heat q that flows across its boundaries and the work w done on the system by the surroundings.
$ΔU = q + w \label{2-1}$
This says that there are two kinds of processes, heat and work, that can lead to a change in the internal energy of a system. Since both heat and work can be measured and quantified, this is the same as saying that any change in the energy of a system must result in a corresponding change in the energy of the world outside the system- in other words, energy cannot be created or destroyed.
There is an important sign convention for heat and work that you are expected to know. If heat flows into a system or the surroundings to do work on it, the internal energy increases and the sign of q or w is positive. Conversely, heat flow out of the system or work done by the system will be at the expense of the internal energy, and will therefore be negative. (Note that this is the opposite of the sign convention that was commonly used in much of the pre-1970 literature.) The full significance of Equation $\ref{2-1}$ cannot be grasped without understanding that U is a state function. This means that a given change in internal energy ΔU can follow an infinite variety of pathways corresponding to all the possible combinations of q and w that can add up to a given value of ΔU.
As a simple example of how this principle can simplify our understanding of change, consider two identical containers of water initially at the same temperature. We place a flame under one until its temperature has risen by 1°C. The water in the other container is stirred vigorously until its temperature has increased by the same amount. There is now no physical test by which you could determine which sample of water was warmed by performing work on it, by allowing heat to flow into it, or by some combination of the two processes. In other words, there is no basis for saying that one sample of water now contains more “work”, and the other more “heat”. The only thing we can know for certain is that both samples have undergone identical increases in internal energy, and we can determine the value of simply by measuring the increase in the temperature of the water.
Pressure-volume work
The kind of work most frequently associated with chemical change occurs when the volume of the system changes owing to the disappearance or formation of gaseous substances. This is sometimes called expansion work or PV-work, and it can most easily be understood by reference to the simplest form of matter we can deal with, the hypothetical ideal gas.
The figure shows a quantity of gas confined in a cylinder by means of a moveable piston. Weights placed on top of the piston exert a force f over the cross-section area A, producing a pressure P = f / A which is exactly countered by the pressure of the gas, so that the piston remains stationary. Now suppose that we heat the gas slightly; according to Charles’ law, this will cause the gas to expand, so the piston will be forced upward by a distance Δx. Since this motion is opposed by the force f, a quantity of work f Δx will be done by the gas on the piston. By convention, work done by the system (in this case, the gas) on the surroundings is negative, so the work is given by
$w = – f Δx \label{3-1}$
When dealing with a gas, it is convenient to think in terms of the more relevant quantities pressure and volume rather than force and distance. We can accomplish this by multiplying the second term by A/A which of course leaves it unchanged:
$w = -f Δx \dfrac{A}{A} \label{3-2}$
By grouping the terms differently, but still not changing anything, we obtain
$w = -\dfrac{f}{A} Δx A \label{3-3}$
Since pressure is force per unit area and the product of the length A and the area has the dimensions of volume, this expression becomes
$w = –P ΔV \label{3-4}$
It is important to note that although $P$ and $V$ are state functions, the work is not (that's why we denote it by a lower-case w.) As is shown farther below, the quantity of work done will depend on whether the same net volume change is realized in a single step (by setting the external pressure to the final pressure P), or in multiple stages by adjusting the restraining pressure on the gas to successively smaller values approaching the final value of $P$.
Example $1$
Find the amount of work done on the surroundings when 1 liter of an ideal gas, initially at a pressure of 10 atm, is allowed to expand at constant temperature to 10 liters by:
1. reducing the external pressure to 1 atm in a single step,
2. reducing $P$ first to 5 atm, and then to 1 atm,
3. allowing the gas to expand into an evacuated space so its total volume is 10 liters.
Solution
First, note that $ΔV$, which is a state function, is the same for each path:
$V_2 = (10/1) × (1 L) = 10 L$
so
$ΔV = 9\; L$
For path (a)
w = –(1 atm)× (9 L) = –9 L-atm.
For path (b), the work is calculated for each stage separately:
w = –(5 atm) × (2–1 L) – (1 atm) × (10–2 L) = –13 L-atm
For path (c) the process would be carried out by removing all weights from the piston in Figure $1$ so that the gas expands to 10 L against zero external pressure. In this case w = (0 atm) × 9 L = 0; that is, no work is done because there is no force to oppose the expansion.
Adiabatic and Isothermal Processes
When a gas expands, it does work on the surroundings; compression of a gas to a smaller volume similarly requires that the surroundings perform work on the gas. If the gas is thermally isolated from the surroundings, then the process is said to occur adiabatically. In an adiabatic change, q = 0, so the First Law becomes ΔU = 0 + w. Since the temperature of the gas changes with its internal energy, it follows that adiabatic compression of a gas will cause it to warm up, while adiabatic expansion will result in cooling.
In contrast to this, consider a gas that is allowed to slowly escape from a container immersed in a constant-temperature bath. As the gas expands, it does work on the surroundings and therefore tends to cool, but the thermal gradient that results causes heat to pass into the gas from the surroundings to exactly compensate for this change. This is called an isothermal expansion. In an isothermal process the internal energy remains constant and we can write the First Law (Equation $Ref{2-1}$) as
$0 = q + w$
or
$q = –w$
This illustrates that the heat flow and work done exactly balance each other.
Because no thermal insulation is perfect, truly adiabatic processes do not occur. However, heat flow does take time, so a compression or expansion that occurs more rapidly than thermal equilibration can be considered adiabatic for practical purposes. If you have ever used a hand pump to inflate a bicycle tire, you may have noticed that the bottom of the pump barrel can get quite warm. Although a small part of this warming may be due to friction, it is mostly a result of the work you (the surroundings) are doing on the system (the gas.)
Adiabatic expansion and contractions are especially important in understanding the behavior of the atmosphere. Although we commonly think of the atmosphere as homogeneous, it is really not, due largely to uneven heating and cooling over localized areas. Because mixing and heat transfer between adjoining parcels of air does not occur rapidly, many common atmospheric phenomena can be considered at least quasi-adiabatic.
Reversible processes
From Example $1$ we see that when a gas expands into a vacuum ($P_{external} = 0$ the work done is zero. This is the minimum work the gas can do; what is the maximum work the gas can perform on the surroundings? To answer this, notice that more work is done when the process is carried out in two stages than in one stage; a simple calculation will show that even more work can be obtained by increasing the number of stages— that is, by allowing the gas to expand against a series of successively lower external pressures. In order to extract the maximum possible work from the process, the expansion would have to be carried out in an infinite sequence of infinitesimal steps. Each step yields an increment of work P ΔV which can be expressed as (RT/V) dV and integrated:
\begin{align} w &= \int_{V_1}^{V_2} \dfrac{RT}{V} dv \[4pt] &= RT \ln \dfrac{V_2}{V_1} \label{3-5} \end{align}
Although such a path (which corresponds to what is called a reversible process) cannot be realized in practice, it can be approximated as closely as desired.
Even though no real process can take place reversibly (it would take an infinitely long time!), reversible processes play an essential role in thermodynamics. The main reason for this is that qrev and wrev are state functions which are important and are easily calculated. Moreover, many real processes take place sufficiently gradually that they can be treated as approximately reversible processes for easier calculation.
Each expansion-compression cycle leaves the gas unchanged, but in all but the one in the bottom row, the surroundings are forever altered, having expended more work in compressing the gas than was performed on it when the gas expanded. Only when the processes are carried out in an infinite number of steps will the system and the surroundings be restored to their initial states— this is the meaning of thermodynamic reversibility.
Heat changes at constant pressure: the Enthalpy
For a chemical reaction that performs no work on the surroundings, the heat absorbed is the same as the change in internal energy: q = ΔU. But many chemical processes do involve work in one form or another:
• If the total volume of the reaction products exceeds that of the reactants, then the process performs work on the surroundings in the amount PΔV, in which P is the pressure exerted by the surroundings (usually the atmosphere) on the system.
• A reaction that drives an electrical current through an external circuit performs electrical work on the surroundings.
For an isothermal process, pressure-volume work affects the heat q
We will consider only pressure-volume work in this lesson. If the process takes place at a constant pressure, then the work is given by PΔV and the change in internal energy will be
$ΔU = q – PΔV \label{4-1}$
Bear in mind why $q$ is so important: the heat flow into or out of the system is directly measurable. ΔU, being "internal" to the system, is not directly observable. Thus the amount of heat that passes between the system and the surroundings is given by
$q = ΔU + PΔV \label{4-2}$
This means that if an exothermic reaction is accompanied by a net increase in volume under conditons of constant pressure, some heat additional to ΔU must be absorbed in order to supply the energy expended as work done on the surroundings if the temperature is to remain unchanged (isothermal process.)
For most practical purposes, changes in the volume of the system are only significant if the reaction is accompanied by a difference in the moles of gaseous reactants and products.
For example, in the reaction H2(g) + O2(g) → ½H2O(l), the total volume of the system decreases from that correponding to 2 moles of gaseous reactants to 0.5 mol of liquid water which occupies only 9 mL — a volume so small in comparison to that of the reactants that it can be neglected without significant error. So all we are really concerned with is the difference in the number of moles of gas Δng:
Δng = (0 – 2) mol = –2 mol
This corresponds to a net contraction (negative expansion) of the system, meaning that the surroundings perform work on the system.
The molar volume of an ideal gas at 25° C and 1 atm is
(298/273) × (22.4 L mol–1) = 24.5 L mol–1
Remember the sign convention: a flow of heat or performance of work that supplies energy is positive; if it consumes energy, it is negative. Thus work performed by the surroundings diminishes the energy of the surroundings (wsurr < 0) and increases the energy of the system (wsys > 0).
and the work done (by the surroundings on the system) is
(1 atm) (–2 mol)(24.5 L mol–1) = –49.0 L-atm.
Using the conversion factor 1 J = 101.3 J, and bearing in mind that work performed on the system supplies energy to the system, the work associated solely with the volume change of the system increases its energy by
(101.3 J/L-atm)(–49.0 L-atm) = 4964 J = 4.06 kJ
Example $2$
The above reaction H2(g) + ½ O2(g) → H2O(l) is carried out at a constant pressure of 1 atm and a constant temperature of 25° C. What quantity of heat q will cross the system boundary (and in which direction?) For this reaction, the change in internal energy is ΔU = –281.8 kJ/mol.
Solution
281.8 kJ of heat. The work performed by the surroundings supplies an additional energy of P ΔV = 4.06 kJ to the system. The total energy change of the system is thus
q = (–281.8 + 4.06) k J = –285.8 kJ
(Eq. 4.2 above.) In order to maintain the constant 25° temperature, an equivalent quantity of heat must pass from the system to the surroundings.
This terminology can be somewhat misleading unless you bear in mind that the conditions ΔP and ΔT refer to the differences between the inital and final states of the system — that is, before and after the reaction. During the time the reaction is in progress, the temperature of the mixture will rise or fall, depending on whether the process is exothermic or endothermic. But because ΔT is a state function, its value is independent of what happens "in between" the initial state (reactants) and final state (products). The same is true of ΔV.
Enthalpy hides work and saves it too!
Because most chemical changes we deal with take place at constant pressure, it would be tedious to have to explicitly deal with the pressure-volume work details that were described above. Fortunately, chemists have found a way around this; they have simply defined a new state function that incorporates and thus hides within itself any terms relating to incidental kinds of work (P-V, electrical, etc.). Since both ΔP and ΔV in Equation $\ref{4-2}$ are state functions, then $q_P$, the heat that is absorbed or released when a process takes place at constant pressure, must also be a state function and is known as the enthalpy change ΔH
$ΔH ≡ q_P = ΔU + PΔV \label{4-3}$
Since most processes that occur in the laboratory, on the surface of the earth, and in organisms do so under a constant pressure of one atmosphere, Equation $\ref{4-3}$ is the form of the First Law that is of greatest interest to most of us most of the time.
Example $3$
Hydrogen chloride gas readily dissolves in water, releasing 75.3 kJ/mol of heat in the process. If one mole of HCl at 298 K and 1 atm pressure occupies 24.5 liters, find ΔU for the system when one mole of HCl dissolves in water under these conditions.
Solution
In this process the volume of liquid remains practically unchanged, so ΔV = –24.5 L. The work done is
\begin{align*} w &= –PΔV \[4pt] &= –(1\; atm)(–24.5\; L) \[4pt] &= 24.6 \;L-atm \end{align*}
(The work is positive because it is being done on the system as its volume decreases due to the dissolution of the gas into the much smaller volume of the solution.) Using the conversion factor 1 L-atm = 101.33 J mol–1 and substituting in Equation \ref{4-3} we obtain
\begin{align*} ΔU &= q +PΔV \[4pt] = –(75,300\; J) + [101.33\; J/L-atm) (24.5\; L-atm)] \[4pt] &= –72.82\; kJ \end{align*}
In other words, if the gaseous HCl could dissolve without volume change, the heat released by the process (75.3 kJ) would cause the system’s internal energy to diminish by 75.3 kJ. But the disappearance of the gaseous phase reduces the volume of the system. This is equivalent to compression of the system by the pressure of the atmosphere performing work on it and consuming part of the energy that would otherwise be liberated, reducing the net value of ΔU to –72.82 kJ.
The Heat Capacity
For systems in which no change in composition (chemical reaction) occurs, things are even simpler: to a very good approximation, the enthalpy depends only on the temperature. This means that the temperature of such a system can serve as a direct measure of its enthalpy. The functional relation between the internal energy and the temperature is given by the heat capacity measured at constant pressure:
$c_p =\dfrac{dH}{dT} \label{5-1}$
(or ΔHT over a finite duration) An analogous quantity relates the heat capacity at constant volume to the internal energy:
$c_v =\dfrac{dU}{dT} \label{5-2}$
The greater the heat capacity of a substance, the smaller will be the effect of a given absorption or loss of heat on its temperature. Heat capacity can be expressed in joules or calories per mole per degree (molar heat capacity), or in joules or calories per gram per degree; the latter is called the specific heat capacity or just the specific heat. The difference between $c_p$ and $c_v$ is of importance only when the volume of the system changes significantly— that is, when different numbers of moles of gases appear on either side of the chemical equation. For reactions involving only liquids and solids, $c_p$ and $c_v$ are for all practical purposes identical. | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.02%3A_The_First_Law_of_Thermodynamics.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Describe the sources of potential energy and kinetic energy contained in a molecule.
• Describe the nature of "thermal" energy, and how it relates to other forms of kinetic energy and to temperature.
• Explain why the simplest molecules (monatomic and diatomic) have smaller heat capacities than polyatomic molecules.
• Similarly, explain why the dependence of heat capacity on the temperature is different for monatomic and polyatomic molecules.
All molecules at temperatures above absolue zero possess thermal energy— the randomized kinetic energy associated with the various motions the molecules as a whole, and also the atoms within them, can undergo. Polyatomic molecules also possess potential energy in the form of chemical bonds. Molecules are thus both vehicles for storing and transporting energy, and the means of converting it from one form to another when the formation, breaking, or rearrangement of the chemical bonds within them is accompanied by the uptake or release of heat.
Chemical Energy: Potential + Kinetic
When you buy a liter of gasoline for your car, a cubic meter of natural gas to heat your home, or a small battery for your flashlight, you are purchasing energy in a chemical form. In each case, some kind of a chemical change will have to occur before this energy can be released and utilized: the fuel must be burned in the presence of oxygen, or the two poles of the battery must be connected through an external circuit (thereby initiating a chemical reaction inside the battery.) And eventually, when each of these reactions is complete, our source of energy will be exhausted; the fuel will be used up, or the battery will be “dead”.
Where did the energy go? It could have gone to raise the temperature of the products, to perform work in expanding any gaseous products or to push electrons through a circuit. The remainder will reside in the chemical potential energy associated with the products of the reaction.
Chemical substances are made of atoms, or more generally, of positively charged nuclei surrounded by negatively charged electrons. A molecule such as dihydrogen, H2, is held together by electrostatic attractions mediated by the electrons shared between the two nuclei. The total potential energy of the molecule is the sum of the repulsions between like charges and the attractions between electrons and nuclei:
$PE_{total} = PE_{electron-electron} + PE_{nucleus-nucleus} + PE_{nucleus-electron} \label{1-1}$
In other words, the potential energy of a molecule depends on the time-averaged relative locations of its constituent nuclei and electrons. This dependence is expressed by the familiar potential energy curve which serves as an important description of the chemical bond between two atoms.
Translation refers to movement of an object as a complete unit. Translational motions of molecules in solids or liquids are restricted to very short distances comparable to the dimensions of the molecules themselves, whereas in gases the molecules typically travel hundreds of molecular diameters between collisions.
In gaseous hydrogen, for example, the molecules will be moving freely from one location to another; this is called translational motion, and the molecules therefore possess translational kinetic energy
$KE_{trans} = \dfrac{mv^2}{2}$
in which $v$ stands for the average velocity of the molecules; you may recall from your study of gases that $v$, and therefore $KE_{trans}$, depends on the temperature.
In addition to translation, molecules composed of two or more atoms can possess other kinds of motion. Because a chemical bond acts as a kind of spring, the two atoms in H2 will have a natural vibrational frequency. In more complicated molecules, many different modes of vibration become possible, and these all contribute a vibrational term KEvib to the total kinetic energy. Finally, a molecule can undergo rotational motions which give rise to a third term $KE_{rot}$. Thus the total kinetic energy of a molecule is the sum
$KE_{total} = KE_{trans} + KE_{vib} + KE_{rot} \label{1-2}$
The total energy of the molecule (its internal energy U) is just the sum
$U = KE_{total} + PE_{total} \label{1-3}$
Although this formula is simple and straightforward, it cannot take us very far in understanding and predicting the behavior of even one molecule, let alone a large number of them. The reason, of course, is the chaotic and unpredictable nature of molecular motion. Fortunately, the behavior of a large collection of molecules, like that of a large population of people, can be described by statistical methods.
How Molecules Store Thermal Energy
As noted above, the heat capacity of a substance is a measure of how sensitively its temperature is affected by a change in heat content; the greater the heat capacity, the less effect a given flow of heat q will have on the temperature.
Thermal energy is randomized kinetic energy
We also pointed out that temperature is a measure of the average kinetic energy due to translational motions of molecules. If vibrational or rotational motions are also active, these will also accept thermal energy and reduce the amount that goes into translational motions. Because the temperature depends only on the latter, the effect of the other kinds of motions will be to reduce the dependence of the internal energy on the temperature, thus raising the heat capacity of a substance.
Table $1$: Molar heat capacities (kJ mol–1 K–1) of some gaseous substances at constant pressure.
monatomic diatomic triatomic
He 20.5 CO 29.3 H2O 33.5
Ne 20.5 N2 29.5 D2O 34.3
Ar 20.5 F2 31.4 CO2 37.2
Kr 20.5 Cl2 33.9 CS2 45.6
Whereas monatomic molecules can only possess translational thermal energy, two additional kinds of motions become possible in polyatomic molecules.
A linear molecule has an axis that defines two perpendicular directions in which rotations can occur; each represents an additional degree of freedom, so the two together contribute a total of ½ R to the heat capacity.
Vibrational and rotational motions are not possible for monatomic species such as the noble gas elements, so these substances have the lowest heat capacities. Moreover, as you can see in the leftmost column of Table 1, their heat capacities are all the same. This reflects the fact that translational motions are the same for all particles; all such motions can be resolved into three directions in space, each contributing one degree of freedom to the molecule and ½ R to its heat capacity. (R is the gas constant, 8.314 J K1).
Think of a "degree of freedom" as a kind of motion that adds kinetic energy to a molecule.
For a non-linear molecule, rotations are possible along all three directions of space, so these molecules have a rotational heat capacity of 3/2 R. Finally, the individual atoms within a molecule can move relative to each other, producing a vibrational motion. A molecule consisting of N atoms can vibrate in 3N –6 different ways or modes. Each vibrational mode contributes R (rather than ½ R) to the total heat capacity. (These results come from advanced mechanics and will not be proven here.)
Table $2$: Contribution of molecular motions to heat capacity
type of motion → translation
rotation
vibration
monatomic 3/2R 0 0
diatomic 3/2 R R R
polyatomic 3/2 R 3/2 R 3N – 6
separation between adjacent levels, (kJ mol–1) 6.0 × 10–17 J (O2) 373 J (HCl) 373 J (HCl)
Monatomic molecules have the smallest heat capacities
Now we are in a position to understand why more complicated molecules have higher heat capacities. The total kinetic energy of a molecule is the sum of those due to the various kinds of motions:
$KE_{total} = KE_{trans} + KE_{rot} + KE_{vib} \label{2-1}$
When a monatomic gas absorbs heat, all of the energy ends up in translational motion, and thus goes to increase its temperature. In a polyatomic gas, by contrast, the absorbed energy is partitioned among the other kinds of motions; since only the translational motions contribute to the temperature, the temperature rise is smaller, and thus the heat capacity is larger. There is one very significant complication, however: classical mechanics predicts that the energy is always partitioned equally between all degrees of freedom. Experiments, however, show that this is observed only at quite high temperatures. The reason is that these motions are all quantized. This means that only certain increments of energy are possible for each mode of motion, and unless a certain minimum amount of energy is available, a given mode will not be active at all and will contribute nothing to the heat capacity.
Translational energy levels are effectively a continuum
The shading indicates the average thermal energy available at 300 K. Only those levels within this range will have significant occupancy as indicated by the thickness of the lines in the two rightmost columns. At 300 K, only the lowest vibrational state and the first few rotational states will be active. Most of the thermal energy will be confined to the translational levels whose minute spacing (10–17 J) causes them to appear as a continuum.
Heat capacity of dihydrogen as a function of temperature. This plot is typical of those for other polyatomic molecules, and shows the practical consequences of the spacings of the various forms of thermal energy. Thus translational motions are available at virtually all temperatures, but contributions to heat acapacity by rotational or vibrational motions can only develop at temperatures sufficiently large to excite these motions.
It turns out that translational energy levels are spaced so closely that they these motions are active almost down to absolute zero, so all gases possess a heat capacity of at least 3/2 R at all temperatures. Rotational motions do not get started until intermediate temperatures, typically 300-500K, so within this range heat capacities begin to increase with temperature. Finally, at very high temperatures, vibrations begin to make a significant contribution to the heat capacity
The strong intermolecular forces of liquids and many solids allow heat to be channeled into vibrational motions involving more than a single molecule, further increasing heat capacities. One of the well known “anomalous” properties of liquid water is its high heat capacity (75 J mol–1 K–1) due to intermolecular hydrogen bonding, which is directly responsible for the moderating influence of large bodies of water on coastal climates.
Heat capacities of metals
Metallic solids are a rather special case. In metals, the atoms oscillate about their equilibrium positions in a rather uniform way which is essentially the same for all metals, so they should all have about the same heat capacity. That this is indeed the case is embodied in the Law of Dulong and Petit. In the 19th century these workers discovered that the molar heat capacities of all the metallic elements they studied were around to 25 J mol–1 K–1, which is close to what classical physics predicts for crystalline metals. This observation played an important role in characterizing new elements, for it provided a means of estimating their molar masses by a simple heat capacity measurement.
Standard enthalpy change
Under the special conditions in which the pressure is 1 atm and the reactants and products are at a temperature of 298 K, ΔH becomes the standard enthalpy change Δ. Chemists usually refer to the "enthalpy change of a reaction" as simply the "enthalpy of reaction", or even more simply as the "heat of reaction". But students are allowed to employ this latter shortcut only if they are able to prove that they know the meaning of enthalpy.
Since most changes that occur in the laboratory, on the surface of the earth, and in organisms are subjected to an approximately constant pressure of "one atmosphere" and reasonably salubrious temperatures, most reaction heats quoted in the literature refer to Δ. But the high pressures and extreme temperatures frequently encountered by chemical engineers, geochemists, and practicioners of chemical oceanography, often preclude the convenience of the "standard" values.
The rearrangement of atoms that occurs in a chemical reaction is virtually always accompanied by the liberation or absorption of heat. If the purpose of the reaction is to serve as a source of heat, such as in the combustion of a fuel, then these heat effects are of direct and obvious interest. We will soon see, however, that a study of the energetics of chemical reactions in general can lead us to a deeper understanding of chemical equilibrium and the basis of chemical change itself.
In chemical thermodynamics, we define the zero of the enthalpy and internal energy as that of the elements as they exist in their stable forms at 298K and 1 atm pressure. Thus the enthalpies H of Xe(g), O2(g) and C(diamond) are all zero, as are those of H2 and Cl2 in the reaction
$H_{2(g)} + Cl_{2(g)} → 2 HCl_{(g)}$
The enthalpy of two moles of HCl is smaller than that of the reactants, so the difference is released as heat. Such a reaction is said to be exothermic. The reverse of this reaction would absorb the same quantity of heat from the surroundings and be endothermic. In comparing the internal energies and enthalpies of different substances as we have been doing here, it is important to compare equal numbers of moles, because energy is an extensive property of matter. However, heats of reaction are commonly expressed on a molar basis and treated as intensive properties.
Changes in enthalpy and internal energy
We can characterize any chemical reaction by the change in the internal energy or enthalpy:
$ΔH = H_{final} – H_{initial} \label{3-1}$
The significance of this can hardly be exaggerated because ΔH, being a state function, is entirely independent of how the system gets from the initial state to the final state. In other words, the value of ΔH or ΔU for a given change in state is independent of the pathway of the process.
Consider, for example, the oxidation of a lump of sugar to carbon dioxide and water:
$\ce{C12H22O11 + 12 O2(g) → 12 CO2(g) + 11 H2O(l)}$
This process can be carried out in many ways, for example by burning the sugar in air, or by eating the sugar and letting your body carry out the oxidation. Although the mechanisms of the transformation are completely different for these two pathways, the overall change in the enthalpy of the system (the atoms of carbon, hydrogen and oxygen that were originally in the sugar) will be identical, and can be calculated simply by looking up the standard enthalpies of the reactants and products and calculating the difference
$ΔH = [12 \times H(\ce{CO2})] + [11 \times H(\ce{H2O})] – H(\ce{C12H22O11}) = –5606\, kJ$
The same quantity of heat is released whether the sugar is burnt in the air or oxidized in a series of enzyme-catalyzed steps in your body.
Enthalpy increases with temperature
When the temperature of a substance is raised, it absorbs heat. The enthalpy of a system increases with the temperature by the amount $ΔH = C_p ΔT$. The defining relation
$ΔH = ΔU + P ΔV$
tells us that this change is dominated by the internal energy, subject to a slight correction for the work associated with volume change. Heating a substance causes it to expand, making ΔV positive and causing the enthalpy to increase slightly more than the internal energy. Physically, what this means is that if the temperature is increased while holding the pressure constant, some extra energy must be expended to push back the external atmosphere while the system expands. The difference between the dependence of U and H on temperature is only really significant for gases, since the coefficients of thermal expansion of liquids and solids are very small.
Enthalpy of phase changes
A plot of the enthalpy of a system as a function of its temperature is called an enthalpy diagram. The slope of the line is given by Cp. The enthalpy diagram of a pure substance such as water shows that this plot is not uniform, but is interrupted by sharp breaks at which the value of Cp is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature the water boiling in a kettle can never exceed 100 until all the liquid has evaporated, at which point the temperature of the steam will rise as more heat flows into the system.
A plot of the enthalpy of carbon tetrachloride as a function of its temperature provides a concise view of its thermal behavior. The slope of the line is given by the heat capacity Cp. All H-vs.-C plots show sharp breaks at which the value of Cp is apparently infinite, meaning that the substance can absorb or lose heat without undergoing any change in temperature at all. This, of course, is exactly what happens when a substance undergoes a phase change; you already know that the temperature of the water boiling in a kettle can never exceed 100°C until all the liquid has evaporated, at which point the temperature (of the steam) will rise as more heat flows into the system.
The lowest-temperature discontinuity on the CCl4 diagram corresponds to a solid-solid phase transition associated with a rearrangement of molecules in the crystalline solid. | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.03%3A_Molecules_as_Energy_Carriers_and_Converters.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concepts:
• Write a balanced thermochemical equation that expresses the enthalpy change for a given chemical reaction. Be sure to correctly specificy the physical state (and, if necessary, the concentration) of each component.
• Write an equation that defines the standard enthalpy of formation of a given chemical species.
• Define the standard enthalpy change of a chemical reaction. Use a table of standard enthalpies of formation to evaluate ΔHf°.
• State the principle on which Hess' law depends, and explain why this law is so useful.
• Describe a simple calorimeter and explain how it is employed and how its heat capacity is determined.
The heat that flows across the boundaries of a system undergoing a change is a fundamental property that characterizes the process. It is easily measured, and if the process is a chemical reaction carried out at constant pressure, it can also be predicted from the difference between the enthalpies of the products and reactants. The quantitative study and measurement of heat and enthalpy changes is known as thermochemistry.
Thermochemical Equations and Standard States
In order to define the thermochemical properties of a process, it is first necessary to write a thermochemical equation that defines the actual change taking place, both in terms of the formulas of the substances involved and their physical states (temperature, pressure, and whether solid, liquid, or gaseous.
To take a very simple example, here is the complete thermochemical equation for the vaporization of water at its normal boiling point:
$\ce{H2O(\ell,\, 373 \,K, \,1 \,atm) → H2O(g, \,373 \,K, \,1 \,atm)} \,\,\,ΔH = 40.7\, kJ\, mol^{-1} \nonumber$
The quantity 40.7 is known as the enthalpy of vaporization (often referred to as “heat of vaporization”) of liquid water. The following points should be kept in mind when writing thermochemical equations:
Any thermodynamic quantity such as $ΔH$ that is associated with a thermochemical equation always refers to the number of moles of substances explicitly shown in the equation. Thus for the synthesis of water we can write
$\ce{2 H2(g) + O2(g)→ 2 H2O(l) }\,\,\,ΔH = -572 \,kJ \nonumber$
or
$\ce{H2(g) + 1/2 O2(g)→ H2O(l)} \,\,\, ΔH = -286 \,kJ \nonumber$
The thermochemical equations for reactions taking place in solution must also specify the concentrations of the dissolved species. For example, the enthalpy of neutralization of a strong acid by a strong base is given by
$\ce{H^{+}(aq,\, 1M,\, 298\, K,\, 1\, atm) + OH^{-} (aq,\, 1M,\, 298 \,K,\, 1 \,atm) → H2O(\ell, 373\, K,\, 1 \,atm) } \,\,\, ΔH = -56.9\, kJ\, mol^{-1} \nonumber$
in which the abbreviation aq refers to the hydrated ions as they exist in aqueous solution. Since most thermochemical equations are written for the standard conditions of 298 K and 1 atm pressure, we can leave these quantities out if these conditions apply both before and after the reaction. If, under these same conditions, the substance is in its preferred (most stable) physical state, then the substance is said to be in its standard state.
Thus the standard state of water at 1 atm is the solid below 0°C, and the gas above 100°C. A thermochemical quantity such as $ΔH$ that refers to reactants and products in their standard states is denoted by $ΔH^°$.
Concentrations vs. Effective Concentrations (Activities)
In the case of dissolved substances, the standard state of a solute is that in which the “effective concentration”, known as the activity, is unity. For non-ionic solutes the activity and molarity are usually about the same for concentrations up to about 1M, but for an ionic solute this approximation is generally valid only for solutions more dilute than 0.001-0.01M, depending on electric charge and size of the particular ion.
Standard Enthalpy of Formation
The enthalpy change for a chemical reaction is the difference
$ΔH = H_{products} - H_{reactants}$
If the reaction in question represents the formation of one mole of the compound from its elements in their standard states, as in
$\ce{H2(g) + 1/2O2(g) -> H2O(l)} \;\;\; ΔH = -286\; kJ$
then we can arbitrarily set the enthalpy of the elements to zero and write
\begin{align*} H_f^o &= \sum H_f^o (products) - \sum H_f^o (reactants) \[4pt] &= -286\; kJ - 0 \[4pt] &= -268\; kJ \,mol^{-1} \end{align*}
which defines the standard enthalpy of formation of water at 298 K. The value Hf ° = -268 kJ tells us that when hydrogen and oxygen, each at a pressure of 1 atm and at 298 K (25° C) react to form 1 mole of liquid water also at 25°C and 1 atm pressure, 268 kJ will have passed from the system (the reaction mixture) into the surroundings. The negative sign indicates that the reaction is exothermic: the enthalpy of the product is smaller than that of the reactants. The standard enthalpy of formation is a fundamental property of any compound. Table T1 list Hf ° values (usually alongside values of other thermodynamic properties) in their appendices.
The standard enthalpy of formation of a compound is defined as the heat associated with the formation of one mole of the compound from its elements in their standard states.
In general, the standard enthalpy change for a reaction is given by the expression
$ΔH_f^o= \sum H_f^o (products) - \sum H_f^o (reactants) \label{2-1}$
in which the $\sum H_f^o$ terms indicate the sums of the standard enthalpies of formations of all products and reactants. The above definition is one of the most important in chemistry because it allows us to predict the enthalpy change of any reaction without knowing any more than the standard enthalpies of formation of the products and reactants, which are widely available in tables.
The following examples illustrate some important aspects of the standard enthalpy of formation of substances.
Molarity
The thermochemical equation defining $H_f^o$ is always written in terms of one mole of the substance in question> For example, the relavante thermodynamic equation for the heat of formation of ammonia ($\ce{NH3}$) is:
$\ce{ 1/2 N2(g) + 3/2 H2(g)→ NH3(g)}\,\,\, ΔH^o = -46.1\, kJ \,(\text{per mole of } \ce{NH3}) \nonumber$
Allotropes
The standard heat of formation of a compound is always taken in reference to the forms of the elements that are most stable at 25°C and 1 atm pressure.
A number of elements, of which sulfur and carbon are common examples, can exist in more then one solid crystalline form (called allotropes). For carbon dixoide ($ce{CO2}$, one can construct two thermodynamics equations:
$\ce{C(graphite) + O2(g) → CO2(g)}\,\,\, ΔH^o ≡ H_f^o = -393.5\, kJ\, mol^{-1} \nonumber$
$\ce{C(diamond) + O2(g) → CO2(g)}\,\,\, ΔH^{o} = -395.8\, kJ\, mol^{-1} \nonumber$
However for carbon, the graphite form is the more stable form and the correct thermodynamic equations for the heat of formation.
Heats of Vaporization
The physical state of the product of the formation reaction must be indicated explicitly if it is not the most stable one at 25°C and 1 atm pressure:
$\ce{H2(g) + 1/2 O2(g) → H2O(aq)}\,\,\, ΔH^o ≡ H_f^o = -285.8\, kJ\, mol^{-1} \nonumber$
$\ce{H2(g) + 1/2 O2(g) → H2O(g)} \,\,\, ΔH^o = -241.8\, kJ\, mol^{-1} \nonumber$
Notice that the difference between these two $ΔH^o$ values is just the heat of vaporization of water.
Endothermic Heats of Formation
Although the formation of most molecules from their elements is an exothermic process, the formation of some compounds is mildly endothermic:
$\ce{1/2 N2(g) + O2(g) -> NO2(g)}\,\,\, ΔH^o ≡ H_f^° = +33.2\, kJ\, mol^{-1} \nonumber$
A positive heat of formation is frequently associated with instability— the tendency of a molecule to decompose into its elements, although it is not in itself a sufficient cause. In many cases, however, the rate of this decomposition is essentially zero, so it is still possible for the substance to exist. In this connection, it is worth noting that all molecules will become unstable at higher temperatures.
Non-Existing Compounds
The thermochemical reactions that define the heats of formation of most compounds cannot actually take place. For example, the direct synthesis of methane from its elements
$\ce{C(graphite) + 2 H2(g) → CH4(g)} \nonumber$
cannot be observed directly owing to the large number of other possible reactions between these two elements. However, the standard enthalpy change for such a reaction be found indirectly from other data, as explained in the next section.
Heats of Atomization
The standard enthalpy of formation of gaseous atoms from the element is known as the heat of atomization.
Heats of atomization are always positive, and are important in the calculation of bond energies.
$\ce{Fe(s) → Fe(g)}\,\, ΔH° = 417 \,kJ\, mol^{-1} \nonumber$
Ions in Solution
The standard enthalpy of formation of an ion dissolved in water is expressed on a separate scale in which that of $\ce{H^{+}(aq)}$ is defined as zero.
The standard heat of formation of a dissolved ion such as $\ce{Cl^{-}(aq)}$ (that is, formation of the ion from the element) cannot be measured because it is impossible to have a solution containing a single kind of ion. For this reason, ionic enthalpies are expressed on a separate scale on which $\ce{H_f^o}$ of the hydrogen ion at unit activity (1 M effective concentration) is defined as zero:
$\ce{1/2 H2(g) → H^{+}(aq)}\,\,\, ΔH^o ≡ H_f^o = 0\, kJ\, mol^{-1} \nonumber$
Other ionic enthalpies (as they are commonly known) are found by combining appropriate thermochemical equations (as explained in Section 3 below). For example, Hf° of HCl(aq) is found from the enthalpies of formation and solution of HCl(g), yielding
$\ce{1/2 H2(g) + 1/2 Cl2(g) → HCl(aq)}\,\,\,\ ΔH^o ≡ H_f^o = -167\, kJ\, mol^{-1} \nonumber$
Because Hf° for H+(aq) is zero, this value establishes the standard enthalpy of the chloride ion. The standard enthalpy of formation of Ca2+(aq), given by
$\ce{Ca(s) + Cl2(g) -> CaCl2(aq)} \nonumber$
could then be calculated by combining other measurable quantities such as the enthalpies of formation and solution of $\ce{CaCl2(s)}$ to find Hf° for CaCl2(aq), from which Hf° of $\ce{Ca^{2+}(aq)}$ is found by difference from that of $\ce{Cl^{-}(aq)}$. Tables of the resulting ionic enthalpies are widely available (see here).
Hess’ Law and Thermochemical Calculations
Two or more chemical equations can be combined algebraically to give a new equation. Even before the science of thermodynamics developed in the late nineteenth century, it was observed that the heats associated with chemical reactions can be combined in the same way to yield the heat of another reaction. For example, the standard enthalpy changes for the oxidation of graphite and diamond can be combined to obtain ΔH° for the transformation between these two forms of solid carbon, a reaction that cannot be studied experimentally.
$\ce{C(graphite) + O2(g)→ CO2(g)}\,\,\, ΔH^o = -393.51\, kJ\, mol^{-1}$
$\ce{C(diamond) + O2(g)→ CO2(g)} \,\,\, ΔH^o = -395.40\, kJ\, mol^{-1}$
Subtraction of the second reaction from the first (i.e., writing the second equation in reverse and adding it to the first one) yields
$\ce{C(graphite) → C(diamond)}\,\,\, ΔH^{°} = 1.89\, kJ\, mol^{-1}$
This principle, known as Hess’ law of independent heat summation is a direct consequence of the enthalpy being a state function. Hess’ law is one of the most powerful tools of chemistry, for it allows the change in the enthalpy (and in other thermodynamic functions) of huge numbers of chemical reactions to be predicted from a relatively small base of experimental data.
Germain Henri Hess
Germain Henri Hess (1802-1850) was a Swiss-born professor of chemistry at St. Petersburg, Russia. He formulated his famous law, which he discovered empirically, in 1840. Very little appears to be known about his other work in chemistry.
Standard Enthalpies of Combustion
Because most substances cannot be prepared directly from their elements, heats of formation of compounds are seldom determined by direct measurement. Instead, Hess’ law is employed to calculate enthalpies of formation from more accessible data. The most important of these are the standard enthalpies of combustion. Most elements and compounds combine with oxygen, and many of these oxidations are highly exothermic, making the measurement of their heats relatively easy. For example, by combining the heats of combustion of carbon, hydrogen, and methane, we can obtain the standard enthalpy of formation of methane, which as we noted above, cannot be determined directly (Example $1$).
Example $1$: Combustion of Methane
Use the following heat of formation/combustion information to estimate the standard heat of formation of methane $\ce{CH4}$.
\begin{aligned} \ce{C(graphite)} + \ce{O2(g)} &→ \ce{CO2(g)} \quad &ΔH° = -393\, kJ\, mol^{-1} \label{P1-1} \[4pt] \ce{H2(g)} + \ce{1/2 O2(g)} &→ \ce{H2O(g)} \quad &ΔH° = -286 kJ mol^{-1} \label{P1-2} \[4pt] \ce{CH4(g)} + \ce{2O2(g)} &→ \ce{CO2(g)} + \ce{2H2O(g)} \quad &ΔH^o = -890\, kJ\, mol^{-1} \label{P1-3} \end{aligned}
Solution
The standard heat of formation of methane is defined by the reaction
$\ce{C(graphite) + 2H2(g) → CH4(g)} \quad ΔH^o = ? \label{P1-4}$
Our task is thus to combine the top three equations in such a way that they add up to (4).
Step 1
Begin by noting that (3), the combustion of methane, is the only equation that contains the CH4 term, so we need to write it in reverse (not forgetting to reverse the sign of ΔH°!) so that CH4 appears as the product.
$\ce{CO2(g) + 2H2O(g) → CH4(g) + 2O2(g)} \quad ΔH^o = +890\, kJ\, mol^{-1} \label{P1-3Rev}$
Step 2
Since H2O does not appear in the net reaction (4), add two times (2) to cancel these out. Notice that this also cancels one of the oxygens in (3Rev):
CO2(g) + 2H2O(g)CH4(g) + 2O2(g) ΔH° = +890 kJ mol-1(P1-3Rev)
2H2O(g) ΔH° = -484 kJ mol-1(P1-2)
Step 3
Finally, get rid of the remaining O2 and CO2 by adding (1); this also adds a needed C:
CH4(g) + 2O2(g) ΔH° = +890 kJ mol-1(P1-3Rev)
2H2O(g) ΔH° = -572 kJ mol-1(P1-2)
CO2(g) ΔH° = -393 kJ mol-1(P1-1)
Step 4
So our creative cancelling has eliminated all except the substances that appear in (4). Just add up the enthalpy changes and we are done:
$\ce{C(graphite) + 2H2(g) → CH4(g)}$
(The tablulated value is -74.6 kJ mol-1)
Calorimetry: Measuring ΔH in the laboratory
How are enthalpy changes determined experimentally? First, you must understand that the only thermal quantity that can be observed directly is the heat q that flows into or out of a reaction vessel, and that q is numerically equal to ΔH° only under the special condition of constant pressure. Moreover, q is equal to the standard enthalpy change only when the reactants and products are both at the same temperature, normally 25°C. The measurement of q is generally known as calorimetry.
The most common types of calorimeters contain a known quantity of water which absorbs the heat released by the reaction. Because the specific heat capacity of water (4.184 J g-1 K-1) is known to high precision, a measurement of its temperature rise due to the reaction enables one to calculate the quantity of heat released.
In all but the very simplest calorimeters, some of the heat released by the reaction is absorbed by the components of the calorimeter itself. It is therefore necessary to "calibrate" the calorimeter by measuring the temperature change that results from the introduction of a known quantity of heat. The resulting calorimeter constant, expressed in J K-1, can be regarded as the “heat capacity of the calorimeter”. The known source of heat is usually produced by passing a known quantity of electric current through a resistor within the calorimeter, but it can be measured by other means as described in the following problem example.
Example $2$: Heat Capacity of a Calorimeter
In determining the heat capacity of a calorimeter, a student mixes 100.0 g of water at 57.0 °C with 100.0 g of water, already in the calorimeter, at 24.2°C. (yhe specific heat of water is 4.184 J g-1 K-1). After mixing and thermal equilibration with the calorimeter, the temperature of the water stabilizes at 38.7°C. Calculate the heat capacity of the calorimeter in J/K.
Solution
The hot water loses heat, the cold water gains heat, and the calorimter itself gains heat, so this is essentially a thermal balance problem. Conservation of energy requires that
$q_{hot} + q_{cold} + q_{cal} = 0 \nonumber$
We can evaluate the first two terms from the observed temperature changes:
$q_{hot} = (100\; g) (38.7\;K - 57.0\;K) (4.184\; J \;g^{-1}K^{-1}) = -7,657\; J \nonumber$
$q_{cold} = (100\; g) (38.7 \;K - 24.2 \;K) (4.184\; J\; g^{-1} K^{-1}) = 6,067\; J \nonumber$
So
$q_{cal} = 7,657 \;J - ,6067\; J = 1,590;\ J \nonumber$
The calorimeter constant is
$\dfrac{1590\; J}{38.7\; K - 24.2\; K} = 110\; J\; K^{-1} \nonumber$
Note: Strictly speaking, there is a fourth thermal balance term that must be considered in a highly accurate calculation: the water in the calorimeter expands as it is heated, performing work on the atmosphere.
For reactions that can be initiated by combining two solutions, the temperature rise of the solution itself can provide an approximate value of the reaction enthalpy if we assume that the heat capacity of the solution is close to that of the pure water — which will be nearly true if the solutions are dilute.
For example, a very simple calorimetric determination of the standard enthalpy of the reaction
$\ce{H^{+}(aq) + OH^{-}(aq) → H2O(\ell)} \nonumber$
could be carried out by combining equal volumes of 0.1M solutions of HCl and of NaOH initially at 25°C. Since this reaction is exothermic, a quantity of heat q will be released into the solution. From the temperature rise and the specific heat of water, we obtain the number of joules of heat released into each gram of the solution, and q can then be calculated from the mass of the solution. Since the entire process is carried out at constant pressure, we have ΔH° = q.
For reactions that cannot be carried out in dilute aqueous solution, the reaction vessel is commonly placed within a larger insulated container of water. During the reaction, heat passes between the inner and outer containers until their temperatures become identical. Again, the temperature change of the water is observed, but in this case we need to know the value of the calorimeter constant described above.
The Bomb Calorimeter
$\Delta H_{combustion}$, since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, and this accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:
• Steel bomb which contains the reactants
• Water bath in which the bomb is submerged
• Thermometer
• A motorized stirrer
• Wire for ignition
Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as bomb calorimetry. The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.
Another consequence of the constant-volume condition is that the heat released corresponds to qv , and thus to the internal energy change ΔU rather than to ΔH. The enthalpy change is calculated according to the formula
$ΔH = q_v + Δn_gRT$
• $Δn_g$ is the change in the number of moles of gases in the reaction.
Example $3$: Combustion of Biphenyl
A sample of biphenyl (C6H5)2 weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid C6H5COOH weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant pressure is known to be 3226 kJ mol-1 (that is, ΔU = -3226 kJ mol-1.) Use this information to determine the standard enthalpy of combustion of biphenyl.
Solution
Begin by working out the calorimeter constant:
• Moles of benzoic acid:
$\dfrac{(0.825 g}{122.1 \;g/mol} = 0.00676\; mol\nonumber$
• Heat released to calorimeter:
$(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ\nonumber$
• Calorimeter constant:
$\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K\nonumber$
Now determine $ΔU_{combustion}$ of the biphenyl ("BP"):
• moles of biphenyl:
$\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol\nonumber$
• heat released to calorimeter:
$(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ\nonumber$
• heat released per mole of biphenyl:
$\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol\nonumber$
$ΔU_{combustion} (BP) = -6,293\; kJ/mol\nonumber$
(This is the heat change at constant volume, $q_v$; the negative sign indicates that the reaction is exothermic, as all combustion reactions are.)
From the reaction equation
$(C_6H_5)_{2(s)} + \frac{19}{2} O_{2(g)} \rightarrow 12 CO_{2(g)} + 5 H_2O_{(l)} \nonumber$
we have
$Δn_g = 12 - \frac{19}{2} = \frac{-5}{2} \nonumber$
Thus the volume of the system decreases when the reaction takes place. Converting to ΔH, we can write the following equation. Additionally, recall that at constant volume, $ΔU = q_V$.
\begin{align*} ΔH &= q_V + Δn_gRT \[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \[4pt] &=(-6,293 \; kJ/mol)-(6,194\; J/mol)=(-6,293-6.2)\;kJ/mol= -6299 \; kJ/mol \end{align*}
A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in $ΔH$ compared to $ΔU$ reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation.
Determining the Heat of Reaction
The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the heat of reaction. The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents ($q_{calorimeter}$) through the combustion reaction.
$q_{rxn} = -q_{calorimeter} \label{2A}$
where
$q_{calorimeter} = q_{bomb} + q_{water} \label{3A}$
If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula:
$q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}$
Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate $q_{rxn}$ from $q_{calorimeter}$ calculated by Equation 2. The heat capacity of the calorimeter can be determined by conducting an experiment.
Example $4$: Heat of Combustion
1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42°C to 27.64°C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, $C_{12}H_{22}O_{11}$, in kJ per mole of $C_{12}H_{22}O_{11}$.
Given:
• mass of $C_{12}H_{22}O_{11}$: 1.150 g
• $T_{initial}$: 23.42°C
• $T_{final}$:27.64°C
• Heat Capacity of Calorimeter: 4.90 kJ/°C
Solution
Using Equation 4 to calculate $q_{calorimeter}$:
$q_{calorimeter} = (4.90\; kJ/°C) \times (27.64 - 23.42)°C = (4.90 \times 4.22) \;kJ = 20.7\; kJ$
Plug into Equation 2:
$q_{rxn} = -q_{calorimeter} = -20.7 \; kJ \;$
But the question asks for kJ/mol $C_{12}H_{22}O_{11}$, so this needs to be converted:
$q_{rxn} = \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} = \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}}$
Per Mole $C_{12}H_{22}O_{11}$:
$q_{rxn} = \dfrac{-18.0 \; kJ}{g \; C_{12}H_{22}O_{11}} \times \dfrac{342.3 \; g \; C_{12}H_{22}O_{11}}{1 \; mol \; C_{12}H_{22}O_{11}} = \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; C_{12}H_{22}O_{11}}$
Calorimeters
Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms:
The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.
The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C. | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.04%3A_Thermochemistry_and_Calorimetry.txt |
Learning Objectives
Make sure you thoroughly understand the following essential concept:
• Describe a simple calorimeter and explain how it is employed and how its heat capacity is determined.
Constant Volume Calorimetry, also know as bomb calorimetry, is used to measure the heat of a reaction while holding volume constant and resisting large amounts of pressure. Although these two aspects of bomb calorimetry make for accurate results, they also contribute to the difficulty of bomb calorimetry. In this module, the basic assembly of a bomb calorimeter will be addressed, as well as how bomb calorimetry relates to the heat of reaction and heat capacity and the calculations involved in regards to these two topics.
Introduction
Calorimetry is used to measure quantities of heat, and can be used to determine the heat of a reaction through experiments. Usually a coffee-cup calorimeter is used since it is simpler than a bomb calorimeter, but to measure the heat evolved in a combustion reaction, constant volume or bomb calorimetry is ideal. A constant volume calorimeter is also more accurate than a coffee-cup calorimeter, but it is more difficult to use since it requires a well-built reaction container that is able to withstand large amounts of pressure changes that happen in many chemical reactions.
Most serious calorimetry carried out in research laboratories involves the determination of heats of combustion $\Delta H_{combustion}$, since these are essential to the determination of standard enthalpies of formation of the thousands of new compounds that are prepared and characterized each month. In a constant volume calorimeter, the system is sealed or isolated from its surroundings, which accounts for why its volume is fixed and there is no volume-pressure work done. A bomb calorimeter structure consists of the following:
• Steel bomb which contains the reactants
• Water bath in which the bomb is submerged
• Thermometer
• A motorized stirrer
• Wire for ignition
Since the process takes place at constant volume, the reaction vessel must be constructed to withstand the high pressure resulting from the combustion process, which amounts to a confined explosion. The vessel is usually called a “bomb”, and the technique is known as bomb calorimetry. The reaction is initiated by discharging a capacitor through a thin wire which ignites the mixture.
Another consequence of the constant-volume condition is that the heat released corresponds to $q_v$, and thus to the internal energy change $ΔU$ rather than to $ΔH$. The enthalpy change is calculated according to the formula
$ΔH = q_v + Δn_gRT$
where $Δn_g$ is the change in the number of moles of gases in the reaction.
Example $1$: Combustion of Biphenyl
A sample of biphenyl ($\ce{(C6H5)2}$) weighing 0.526 g was ignited in a bomb calorimeter initially at 25°C, producing a temperature rise of 1.91 K. In a separate calibration experiment, a sample of benzoic acid ($\ce{C6H5COOH}$) weighing 0.825 g was ignited under identical conditions and produced a temperature rise of 1.94 K. For benzoic acid, the heat of combustion at constant volume is known to be 3,226 kJ mol–1 (that is, ΔU = –3,226 kJ mol–1.) Use this information to determine the standard enthalpy of combustion of biphenyl.
Solution
Begin by working out the calorimeter constant:
• Moles of benzoic acid:
$\dfrac{0.825 g}{122.1 \;g/mol} = 0.00676\; mol \nonumber$
• Heat released to calorimeter:
$(0.00676\; mol) \times (3226\; kJ/mol) = 21.80\; kJ \nonumber$
• Calorimeter constant:
$\dfrac{21.80\; kJ}{1.94\; K} = 11.24\; kJ/K \nonumber$
Now determine $ΔU_{combustion}$ of the biphenyl ("BP"):
• moles of biphenyl:
$\dfrac{0.526\; g}{154.12\; g/mol} = 0.00341 \; mol \nonumber$
• heat released to calorimeter:
$(1.91\; K) \times (11.24\; kJ/K) = 21.46\; kJ \nonumber$
• heat released per mole of biphenyl:
$\dfrac{21.46\; kJ}{0.00341\; mol} = 6,293\; kJ/mol \nonumber$
$ΔU_{combustion} (BP) = –6,293\; kJ/mol \nonumber$
This is the heat change at constant volume, $q_v$; the negative sign indicates that the reaction is exothermic, as all combustion reactions are.
From the balanced reaction equation
$\ce{(C6H5)2(s) + 29/2 O2(g) \rightarrow 12 CO2(g) + 5 H2O(l)} \nonumber$
we can calculate the change in the moles of gasses for this reaction
$Δn_g = 12 - \frac{29}{2} = \frac{-5}{2} \nonumber$
Thus the volume of the system decreases when the reaction takes place. Converting to $ΔH$, we can write the following equation. Additionally, recall that at constant volume, $ΔU = q_V$.
\begin{align*} ΔH &= q_V + Δn_gRT \[4pt] &= ΔU -\left( \dfrac{5}{2}\right) (8.314\; J\; mol^{-1}\; K^{-1}) (298 \;K) \[4pt] &= (-6,293 \; kJ/mol)–(6,194\; J/mol) \[4pt] &= (-6,293-6.2)\;kJ/mol \[4pt] &= -6299 \; kJ/mol \end{align*}
A common mistake here is to forget that the subtracted term is in J, not kJ. Note that the additional 6.2 kJ in $ΔH$ compared to $ΔU$ reflects the work that the surroundings do on the system as the volume of gases decreases according to the reaction equation.
Determining the Heat of Reaction
The amount of heat that the system gives up to its surroundings so that it can return to its initial temperature is the heat of reaction. The heat of reaction is just the negative of the thermal energy gained by the calorimeter and its contents ($q_{calorimeter}$) through the combustion reaction.
$q_{rxn} = -q_{calorimeter} \label{2A}$
where
$q_{calorimeter} = q_{bomb} + q_{water} \label{3A}$
If the constant volume calorimeter is set up the same way as before, (same steel bomb, same amount of water, etc.) then the heat capacity of the calorimeter can be measured using the following formula:
$q_{calorimeter} = \text{( heat capacity of calorimeter)} \times \Delta{T} \label{4A}$
Heat capacity is defined as the amount of heat needed to increase the temperature of the entire calorimeter by 1 °C. The equation above can also be used to calculate $q_{rxn}$ from $q_{calorimeter}$ calculated by Equation \ref{2A}. The heat capacity of the calorimeter can be determined by conducting an experiment.
Example $4$: Heat of Combustion
1.150 g of sucrose goes through combustion in a bomb calorimeter. If the temperature rose from 23.42 °C to 27.64 °C and the heat capacity of the calorimeter is 4.90 kJ/°C, then determine the heat of combustion of sucrose, $\ce{C12H22O11}$ (in kJ per mole of $\ce{C12H22O11}$).
Solution
Given:
• mass of $C_{12}H_{22}O_{11}$: 1.150 g
• $T_{initial}$: 23.42°C
• $T_{final}$:27.64°C
• Heat Capacity of Calorimeter: 4.90 kJ/°C
Using Equation \ref{4A} to calculate $q_{calorimeter}$:
\begin{align*} q_{calorimeter} &= (4.90\; kJ/°C) \times (27.64 - 23.42)°C \[4pt] &= (4.90 \times 4.22) \;kJ = 20.7\; kJ \end{align*}
Plug into Equation \ref{2A}:
\begin{align*} q_{rxn} &= -q_{calorimeter} \[4pt] &= -20.7 \; kJ \; \end{align*}
But the question asks for kJ/mol $\ce{C12H22O11}$, so this needs to be converted:
\begin{align*}q_{rxn} &= \dfrac{-20.7 \; kJ}{1.150 \; g \; C_{12}H_{22}O_{11}} \[4pt] &= \dfrac{-18.0 \; kJ}{g\; C_{12}H_{22}O_{11}} \end{align*}
Convert to per Mole $\ce{C12H22O11}$:
\begin{align*}q_{rxn} &= \dfrac{-18.0 \; kJ}{\cancel{g \; \ce{C12H22O11}}} \times \dfrac{342.3 \; \cancel{ g \; \ce{C12H22O11}}}{1 \; mol \; \ce{C12H22O11}} \[4pt] &= \dfrac{-6.16 \times 10^3 \; kJ \;}{mol \; \ce{C12H22O11}} \end{align*}
"Ice Calorimeter"
Although calorimetry is simple in principle, its practice is a highly exacting art, especially when applied to processes that take place slowly or involve very small heat changes, such as the germination of seeds. Calorimeters can be as simple as a foam plastic coffee cup, which is often used in student laboratories. Research-grade calorimeters, able to detect minute temperature changes, are more likely to occupy table tops, or even entire rooms:
The ice calorimeter is an important tool for measuring the heat capacities of liquids and solids, as well as the heats of certain reactions. This simple yet ingenious apparatus is essentially a device for measuring the change in volume due to melting of ice. To measure a heat capacity, a warm sample is placed in the inner compartment, which is surrounded by a mixture of ice and water.
The heat withdrawn from the sample as it cools causes some of the ice to melt. Since ice is less dense than water, the volume of water in the insulated chamber decreases. This causes an equivalent volume of mercury to be sucked into the inner reservoir from the outside container. The loss in weight of this container gives the decrease in volume of the water, and thus the mass of ice melted. This, combined with the heat of fusion of ice, gives the quantity of heat lost by the sample as it cools to 0°C. | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.05%3A_Calorimetry.txt |
Virtually all chemical processes involve the absorption or release of heat, and thus changes in the internal energy of the system. In this section, we survey some of the more common chemistry-related applications of enthalpy and the First Law. While the first two sections relate mainly to chemistry, the remaining ones impact the everyday lives of everyone.
Enthalpy diagrams and their uses
Comparison and interpretation of enthalpy changes is materially aided by a graphical construction in which the relative enthalpies of various substances are represented by horizontal lines on a vertical energy scale. The zero of the scale can be placed anywhere, since energies are always arbitrary; it is generally most useful to locate the elements at zero energy, which reflects the convention that their standard enthlapies of formation are zero.
This very simple enthalpy diagram for carbon and oxygen and its two stable oxides (Figure \(1\)) shows the changes in enthalpy associated with the various reactions this system can undergo. Notice how Hess’s law is implicit in this diagram; we can calculate the enthalpy change for the combustion of carbon monoxide to carbon dioxide, for example, by subtraction of the appropriate arrow lengths without writing out the thermochemical equations in a formal way.
The zero-enthalpy reference states refer to graphite, the most stable form of carbon, and gaseous oxygen. All temperatures are 298 K.
This enthalpy diagram for the hydrogen-oxygen system (Figure \(2\)) shows the known stable configurations of these two elements. Reaction of gaseous H2 and O2 to yield one mole of liquid water releases 285 kJ of heat . If the H2O is formed in the gaseous state, the energy release will be smaller. Notice also that...
• The heat of vaporization of water (an endothermic process) is clearly found from the diagram.
• Hydrogen peroxide H2O2, which spontaneously decomposes into O2 and H2O, releases some heat in this process. Ordinarily this reaction is so slow that the heat is not noticed. But the use of an appropriate catalyst can make the reaction so fast that it has been used to fuel a racing car.
Why you cannot run your car on water
You may have heard the venerable urban legend, probably by now over 80 years old, that some obscure inventor discovered a process to do this, but the invention was secretly bought up by the oil companies in order to preserve their monopoly. The enthalpy diagram for the hydrogen-oxygen system shows why this cannot be true — there is simply no known compound of H and O that resides at a lower enthalpy level.
Enthalpy diagrams are especially useful for comparing groups of substances having some common feature. This one shows the molar enthalpies of species relating to two hydrogen halides, with respect to those of the elements. From this diagram we can see at a glance that the formation of HF from the elements is considerably more exothermic than the corresponding formation of HCl. The upper part of this diagram shows the gaseous atoms at positive enthalpies with respect to the elements. The endothermic processes in which the H2 and the dihalogen are dissociated into atoms can be imagined as taking place in two stages, also shown. From the enthalpy change associated with the dissociation of H2 (218 kJ mol–1), the dissociation enthalpies of F2 and Cl2 can be calculated and placed on the diagram.
Bond Enthalpies vs. Bond Energies
The enthalpy change associated with the reaction
\[\ce{HI(g) → H(g) + I(g)}\]
is the enthalpy of dissociation of the \(\ce{HI}\) molecule; it is also the bond energy of the hydrogen-iodine bond in this molecule. Under the usual standard conditions, it would be expressed either as the bond enthalpy H°(HI,298K) or internal energy (HI,298); in this case the two quantities differ from each other by ΔPV = RT. Since this reaction cannot be studied directly, the H–I bond enthalpy is calculated from the appropriate standard enthalpies of formation:
½ H2(g)→ H(g) + 218 kJ
½ I2(g)→ I(g) +107 kJ
½ H2(g) + 1/2 I2(g)→ HI(g) –36 kJ
HI(g) → H(g) + I(g) +299 kJ
Bond energies and enthalpies are important properties of chemical bonds, and it is very important to be able to estimate their values from other thermochemical data. The total bond enthalpy of a more complex molecule such as ethane can be found from the following combination of reactions:
C2H6(g)→ 2C(graphite) + 3 H2(g) 84.7 kJ
3 H2(g)→ 6 H(g) 1308 kJ
2 C(graphite) → 2 C(g) 1430 kJ
C2H6(g)→ 2 C(g) + 6H(g) 2823 kJ
When a molecule in its ordinary state is broken up into gaseous atoms, the process is known as atomization (Figure \(4\)). The standard enthalpy of atomization refers to the transformation of an element into gaseous atoms:
\[\ce{ C_{(graphite)} → C(g)} \;\;\;\; ΔH^o = 716.7\; kJ\]
Atomization is always an endothermic process. Heats of atomization are most commonly used for calculating bond energies. They are usually measured spectroscopically.
Pauling’s rule and Average Bond Energy
Pauling’s Rule
The total bond energy of a molecule can be thought of as the sum of the energies of the individual bonds.
Pauling’s Rule is only an approximation, because the energy of a given type of bond is not really a constant, but depends somewhat on the particular chemical environment of the two atoms. In other words, all we can really talk about is the average energy of a particular kind of bond, such as C–O, for example, the average being taken over a representative sample of compounds containing this type of bond, such as CO, CO2, COCl2, (CH3)2CO, CH3COOH, etc.
Despite the lack of strict additivity of bond energies, Pauling’s Rule is extremely useful because it allows one to estimate the heats of formation of compounds that have not been studied, or have not even been prepared. Thus in the foregoing example, if we know the enthalpies of the C–C and C–H bonds from other data, we could estimate the total bond enthalpy of ethane, and then work back to get some other quantity of interest, such as ethane’s enthalpy of formation. By assembling a large amount of experimental information of this kind, a consistent set of average bond energies can be obtained. The energies of double bonds are greater than those of single bonds, and those of triple bonds are higher still (Table \(1\)).
Table \(1\): Average energies of some single bonds (KJ/mol)
H C N O F Cl Br I Si
H 436 415 390 464 569 432 370 295 395
C 345 290 350 439 330 275 240 360
N 160 200 270 200 270
O 140 185 205 185 200 370
F 160 255 160 280 540
Cl 243 220 210 359
Br 190 180 290
I 150 210
Si 230
Energy Content of Fuels
A fuel is any substance capable of providing useful amounts of energy through a process that can be carried out in a controlled manner at economical cost. For most practical fuels, the process is combustion in air (in which the oxidizing agent O2 is available at zero cost.) The enthalpy of combustion is obviously an important criterion for a substance’s suitability as a fuel, but it is not the only one; a useful fuel must also be easily ignited, and in the case of a fuel intended for self-powered vehicles, its energy density in terms of both mass (kJ kg–1) and volume (kJ m–3) must be reasonably large. Thus substances such as methane and propane which are gases at 1 atm must be stored as pressurized liquids for transportation and portable applications.
Table \(2\): Energy densities of some common fuels
fuel
MJ kg–1
wood (dry) 15
coal (poor) 15
coal (premium) 27
ethanola 30
petroleum-derived products 45
methane, liquified natural gas 54
hydrogenb 140
Notes on the above table
a Ethanol is being strongly promoted as a motor fuel by the U.S. agricultural industry. Note, however, that according to some estimates, it takes 46 MJ of energy to produce 1 kg of ethanol from corn. Some other analyses, which take into account optimal farming practices and the use of by-products arrive at different conclusions; see, for example, this summary with links to several reports.
b Owing to its low molar mass and high heat of combustion, hydrogen possesses an extraordinarily high energy density, and would be an ideal fuel if its critical temperature (33 K, the temperature above which it cannot exist as a liquid) were not so low. The potential benefits of using hydrogen as a fuel have motivated a great deal of research into other methods of getting a large amount of H2 into a small volume of space. Simply compressing the gas to a very high pressure is not practical because the weight of the heavy-walled steel vessel required to withstand the pressure would increase the effective weight of the fuel to an unacceptably large value. One scheme that has shown some promise exploits the ability of H2 to “dissolve” in certain transition metals. The hydrogen can be recovered from the resulting solid solution (actually a loosely-bound compound) by heating.
Energy content of foods
What, exactly, is meant by the statement that a particular food “contains 1200 calories” per serving? This simply refers to the standard enthalpy of combustion of the foodstuff, as measured in a bomb calorimeter. Note, however, that in nutritional usage, the calorie is really a kilocalorie (sometimes called “large calorie”), that is, 4184 J. Although this unit is still employed in the popular literature, the SI unit is now commonly used in the scientific and clinical literature, in which energy contents of foods are usually quoted in kJ per unit of weight.
Although the mechanisms of oxidation of a carbohydrate such as glucose to carbon dioxide and water in a bomb calorimeter and in the body are complex and entirely different, the net reaction involves the same initial and final states, and must be the same for any possible pathway:
\[C_6H_{12}O_6 + 6 O_2 → 6 CO_2 + 6 H_2O \;\;\;\; ΔH^o = – 20.8\; kJ \;mol^{–1}\]
Glucose is a sugar, a breakdown product of starch, and is the most important energy source at the cellular level; fats, proteins, and other sugars are readily converted to glucose. By writing balanced equations for the combustion of sugars, fats, and proteins, a comparison of their relative energy contents can be made. The stoichiometry of each reaction gives the amounts of oxygen taken up and released when a given amount of each kind of food is oxidized; these gas volumes are often taken as indirect measures of energy consumption and metabolic activity; a commonly-accepted value that seems to apply to a variety of food sources is 20.1 J (4.8 kcal) per liter of O2 consumed.
For some components of food, particularly proteins, oxidation may not always be complete in the body, so the energy that is actually available will be smaller than that given by the heat of combustion. Mammals, for example, are unable to break down cellulose (a polymer of sugar) at all; animals that derive a major part of their nutrition from grass and leaves must rely on the action of symbiotic bacteria which colonize their digestive tracts. The amount of energy available from a food can be found by measuring the heat of combustion of the waste products excreted by an organism that has been restricted to a controlled diet, and subtracting this from the heat of combustion of the food (Table \(3\)).
Table \(3\): Energy content and availability of the major food components
type of food food
ΔH° (kJ g–1)
percent availability
Protein meat 22.4 92
egg 23.4
Fat butter 38.2
animal fat 39.2 95
Carbohydrate starch 17.2
glucose (sugar) 15.5 99
ethanol 29.7 100
The amount of energy an animal requires depends on the age, sex, surface area of the body, and of course on the amount of physical activity. The rate at which energy is expended is expressed in watts: 1 W = 1 J sec–1. For humans, this value varies from about 200-800 W. This translates into daily food intakes having energy equivalents of about 10-15 MJ for most working adults. In order to just maintain weight in the absence of any physical activity, about 6 MJ per day is required.
Table \(4\): Metabolic rates of some animals
animal kJ hr–1 kJ kg–1hr–1
mouse 82 17
cat 34 6.8
dog 78 3.3
sheep 193 2.2
human 300 2.1
horse 1430 1.1
elephant 5380 0.7
The above table is instructive in that although larger animals consume more energy, the energy consumption per unit of body weight decreases with size. This reflects the fact the rate of heat loss to the environment depends largely on the surface area of an animal, which increases with mass at a greater rate than does an animal's volume ("size").
Thermodynamics and the weather
Hydrogen bonds at work
It is common knowledge that large bodies of water have a “moderating” effect on the local weather, reducing the extremes of temperature that occur in other areas. Water temperatures change much more slowly than do those of soil, rock, and vegetation, and this effect tends to affect nearby land masses. This is largely due to the high heat capacity of water in relation to that of land surfaces— and thus ultimately to the effects of hydrogen bonding. The lower efficiency of water as an absorber and radiator of infrared energy also plays a role.
The specific heat capacity of water is about four times greater than that of soil. This has a direct consequence to anyone who lives near the ocean and is familiar with the daily variations in the direction of the winds between the land and the water. Even large lakes can exert a moderating influence on the local weather due to water's relative insensitivity to temperature change.
During the daytime the land and sea receive approximately equal amounts of heat from the Sun, but the much smaller heat capacity of the land causes its temperature to rise more rapidly. This causes the air above the land to heat, reducing its density and causing it to rise. Cooler oceanic air is drawn in to fill the void, thus giving rise to the daytime sea breeze. In the evening, both land and ocean lose heat by radiation to the sky, but the temperature of the water drops less than that of the land, continuing to supply heat to the oceanic air and causing it to rise, thus reversing the direction of air flow and producing the evening land breeze.
Why it gets colder as you go higher: the adiabatic lapse rate
The air receives its heat by absorbing far-infrared radiation from the earth, which of course receives its heat from the sun. The amount of heat radiated to the air immediately above the surface varies with what's on it (forest, fields, water, buildings) and of course on the time and season. When a parcel of air above a particular location happens to be warmed more than the air immediately surrounding it, this air expands and becomes less dense. It therefore rises up through the surrounding air and undergoes further expansion as it encounters lower pressures at greater altitudes.
Whenever a gas expands against an opposing pressure, it does work on the surroundings. According to the First Law ΔU = q + w, if this work is not accompanied by a compensating flow of heat into the system, its internal energy will fall, and so, therefore, will its temperature. It turns out that heat flow and mixing are rather slow processes in the atmosphere in comparison to the convective motion we are describing, so the First Law can be written as ΔU = w (recall that w is negative when a gas expands.) Thus as air rises above the surface of the earth it undergoes adiabatic expansion and cools. The actual rate of temperature decrease with altitude depends on the composition of the air (the main variable being its moisture content) and on its heat capacity. For dry air, this results in an adiabatic lapse rate of 9.8 C° per km of altitude.
Santa Anas and Chinooks
Just the opposite happens when winds develop in high-altitude areas and head downhill. As the air descends, it undergoes compression from the pressure of the air above it. The surroundings are now doing work on the system, and because the process occurs too rapidly for the increased internal energy to be removed as heat, the compression is approximately adiabatic.
The resulting winds are warm (and therefore dry) and are often very irritating to mucous membranes. These are known generically as Föhn winds (which is the name given to those that originate in the Alps). In North America they are often called chinooks (or, in winter, "snow melters") when they originate along the Rocky Mountains. Among the most notorious are the Santa Ana winds of Southern California which pick up extra heat (and dust) as they pass over the Mohave Desert before plunging down into the Los Angeles basin (Figure 13.5.X). Their dryness and high velocities feed many of the disastrous wildfires that afflict the region.
14.0E: 14.E: Thermochemistry (Exercises)
13.5: Calorimetry
Q13.5.1
After going through combustion in a bomb calorimeter a sample gives off 5,435 cal. The calorimeter experiences an increase of 4.27°C in its temperature. Using this information, determine the heat capacity of the calorimeter in kJ/°C.
Q13.5.2
Referring to the example given above about the heat of combustion, calculate the temperature change that would occur in the combustion of 1.732 g \(C_{12}H_{22}O_{12}\) in a bomb calorimeter that had the heat capacity of 3.87 kJ/°C.
Q13.5.3
Given the following data calculate the heat of combustion in kJ/mol of xylose,\(C_{5}H_{10}O_{5}\)(s), used in a bomb calorimetry experiment: mass of \(C_{5}H_{10}O_{5}\)(s) = 1.250 g, heat capacity of calorimeter = 4.728 kJ/°C, Initial Temperature of the calorimeter = 24.37°C, Final Temperature of calorimeter = 28.29°C.
Q13.5.4
Determine the heat capacity of the bomb calorimeter if 1.714 g of naphthalene, \(C_{10}H_{8}\)(s), experiences an 8.44°C increase in temperature after going through combustion. The heat of combustion of naphthalene is -5156 kJ/mol \(C_{10}H_{8}\).
Q13.5.5
What is the heat capacity of the bomb calorimeter if a 1.232 g sample of benzoic acid causes the temperature to increase by 5.14°C? The heat of combustion of benzoic acid is -26.42 kJ/g.
S13.5.1
Use equation 4 to calculate the heat of capacity:
\(q_{calorimeter} = \; heat \; capicity \; of \; calorimeter \; x \; \Delta{T}\)
5435 cal = heat capacity of calorimeter x 4.27°C
Heat capacity of calorimeter = (5435 cal/ 4.27°C) x (4.184 J/1 cal) x (1kJ/1000J) = 5.32 kJ/°C
S13.5.2
The temperature should increase since bomb calorimetry releases heat in an exothermic combustion reaction.
Change in Temp = (1.732 g \(C_{12}H_{22}O_{11}\)) x (1 mol \(C_{12}H_{22}O_{11}\)/342.3 g \(C_{12}H_{22}O_{11}\)) x (6.61 x 10³ kJ/ 1 mol \(C_{12}H_{22}O_{11}\)) x (1°C/3.87kJ) = 8.64°C
S13.5.3
[(Heat Capacity x Change in Temperature)/mass] =[ ((4.728 kJ/°C) x(28.29 °C – 24.37 °C))/1.250 g] = 14.8 kJ/g xylose
\(q_{rxn}\) = (-14.8 kJ/g xylose) x (150.13 g xylose/ 1 mol xylose) = -2.22x10³ kJ/mol xylose
S13.5.4
Heat Capacity = [(1.714 g \(C_{10}H_{8}\)) x (1 mol \(C_{10}H_{8}\)/128.2 g \(C_{10}H_{8}\)) x (5.156x10³ kJ/1 mol \(C_{10}H_{8}\))]/8.44°C = 8.17 kJ/ °C
S13.5.5
Heat Capacity = [(1.232 g benzoic acid) x (26.42 kJ/1 g benzoic acid)]/5.14°C = 6.31 kJ/ °C | textbooks/chem/General_Chemistry/Chem1_(Lower)/14%3A_Thermochemistry/14.06%3A_Applications_of_Thermochemistry.txt |
Trying to introduce chemical thermodynamics to beginning students is always problematic; to do it "properly" requires a degree of rigor that rarely succeeds for more than a small fraction of the class. Although a full formal development is rarely appropriate at this level, I believe that the value of developing students' understanding of the fundamental concepts is generally underappreciated. This requires some understanding of the ways thermal energy is dispersed in matter— something that is not a part of classical thermodynamics and is not supported by most textbooks, but which is in keeping with the molecular focus of modern chemical science.
The equilibrium value for a reversible reaction is an important quantity that characterizes a chemical reaction, but what factors govern its value? In particular, is there any way that we can predict the value of the equilibrium constant of a reaction solely from information about the products and reactants themselves, without any knowledge at all about the mechanism or other details of the reaction? The answer is yes, and this turns out to be the central purpose of chemical thermodynamics:
The purpose of thermodynamics is to predict the equilibrium composition of a system from the properties of its components.
Don’t let the significance of this pass you by; it means that we can say with complete certainty whether or not a given change is possible, and if it is possible, to what extent it will occur— without the need to study the particular reaction in question. To a large extent, this is what makes chemistry a science, rather than a mere cataloging of facts.
• 15.1: Energy Spreading Drives Spontaneous Change
Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as natural processor spontaneous changes. In many cases our everyday life experiences teach us the direction in which spontaneous change can occur, and anything that runs counter to these expectations is immediately sensed as weird.
• 15.2: Entropy Rules
Previously, we explained how the tendency of thermal energy to disperse as widely as possible is what drives all spontaneous processes, including, of course chemical reactions. We now need to understand how the direction and extent of the spreading and sharing of energy can be related to measurable thermodynamic properties of substances— that is, of reactants and products.
• 15.3: The Second Law of Thermodynamics
The First Law of thermodynamics, expressed as ΔU = q + w, is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process.
• 15.4: Free Energy and the Gibbs Function
In this unit we introduce a new thermodynamic function, the free energy, which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. As we will explain near the bottom of this page, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as the "Gibbs function" or "Gibbs energy."
• 15.5: Thermodynamics of Mixing and Dilution
This lesson goes somewhat beyond what is covered in most first-year courses, and can usually be skipped by students in non-honors beginners' courses. The concepts presented here are not especially complicated, but they don't really become essential until one gets into more advanced courses in chemistry, physiology, and similar subjects.
• 15.6: Free Energy and Equilibrium
Under conditions of constant temperature and pressure, chemical change will tend to occur in whatever direction leads to a decrease in the value of the Gibbs free energy. In this lesson we will see how G varies with the composition of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The equilibrium composition of the mixture is determined by ΔG° which also defines the equilibrium constant K.
• 15.7: Some Applications of Entropy and Free Energy
Thermodynamics may appear at first to be a rather esoteric subject, but when you think about it, almost every chemical (and biological) process is governed by changes in entropy and free energy. Examples such as those given below should help you connect these concepts with the real world.
• 15.8: Quantum states, Microstates, and Energy spreading in Reactions
Entropy (S) is a state function whose value increases with an increase in the number of available microstates.For a given system, the greater the number of microstates, the higher the entropy. During a spontaneous process, the entropy of the universe increases.
15: Thermodynamics of Chemical Equilibria
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in bold.
• Spontaneous change is that which, once initiated, proceeds on its own until some state of equilibrium (mechanical, thermal, chemical, etc.) is attained.
• All natural processes (those that occur in the world) ultimately involve spontaneous change.
• Thermal energy is kinetic energy associated with the random motions of atoms and molecules. These states of motion are quantized into energy states which can be realized in huge numbers of different ways that we call microstates.
• will be thermally accessible and likely to be populated.
• Spontaneous change is driven by the tendency of thermal energy to spread into as many microstates of the system and surroundings as are thermally accessible.
Drop a teabag into a pot of hot water, and you will see the tea diffuse into the water until it is uniformly distributed throughout the water. What you will never see is the reverse of this process, in which the tea would be sucked up and re-absorbed by the teabag. The making of tea, like all changes that take place in the world, possesses a “natural” direction.
What is a spontaneous process?
Processes that proceed in a definite direction when left to themselves and in the absence of any attempt to drive them in reverse — are known as natural processor spontaneous changes. In many cases our everyday life experiences teach us the direction in which spontaneous change can occur, and anything that runs counter to these expectations is immediately sensed as weird. In other cases, including that of most chemical change, we often have no obvious guidelines, and must learn how to apply the laws of thermodynamics which ultimately govern all spontaneous change.
In order to answer the above question, let's begin by thinking about the outcomes of the following four experiments, each of which illustrates a natural process that proceeds spontaneously only in a single direction.
Experiment $1$
A stack of one hundred coins is thrown into the air. After they have come to rest on the floor, the numbers that land “heads up” and “tails up” are noted.
Net change: ordered coins → randomized coins (roughly equal numbers of heads and tails.)
Energetics: no relevant net change in energy
Why it does not go in reverse: Simple statistics shows that the probability that the coins will land in one particular arrangement of the huge number that are possible is vanishingly small.
Example $2$
Two identical blocks of copper, one at 200°C and the other at 100°C, are brought into contact in a thermally-insulated environment. Eventually the temperatures of both blocks reach 150°C.
Net change: block1 (200°) + block2 (100°) → combined blocks (150°)
Energetics: Heat (randomized molecular kinetic energy) flows from the warmer block to the cooler one until their temperatures are identical.
Why it does not go in reverse: Dispersal of kinetic energy amongst the copper atoms is a random process; the chances that such a process would lead to a non-uniform sharing of the energy are even smaller than in the case of the 100 coins because of the much greater number (around 1022) of particles involved.
Experiment $3$
A book or some other solid object is held above a table top, and is then allowed to fall.
Net change: book in air book on table top; potential energy organized kinetic energy thermal energy.
Energetics: At the instant just before the end of its fall, the potential energy the object acquired when it was raised will exist entirely as kinetic energy mv2/2 in which m is the mass of the object and v is its velocity. Each atom of which the object is composed will of course possess a proportionate fraction of this energy, again with its principal velocity component pointing down. Superimposed on this, however, will be minute thermal displacements that vary randomly in magnitude and direction from one instant to the next. The sum total of these constitutes the thermal energy contained in the object.
When the object strikes the table top, its motion ceases and we say its kinetic energy is zero. Energy is supposed to be conserved, so where did it disappear to? The shock of impact has resulted in its dispersal into greatly augmented thermal motions of the atoms, both of the object itself and of the area of the table top where the impact occurred. In other words, the kinetic energy of organized motion the object had just before its motion stopped has been transformed into kinetic energy of random or disorganized motion (thermal energy) which spreads rapidly away from the point of impact.
Why it does not go in reverse: Once the kinetic energy of the book has been dispersed amongst the molecules of the book and the table top, the probability of these randomized motions reappearing at the surface where the two objects are in contact and then acting in concert to propel the object back into the air) is negligible.
Experiment $4$
One mole of gas, initially at 300 K and 2 atm pressure, is allowed to expand to double its volume, keeping the temperature constant.
Net change: Increase in volume of gas.
Energetics: No change in energy if the gas behaves ideally.
Why it does not go in reverse: Simple statistics: the probability that N randomly moving objects (flies in a bottle, for example,) will at any time all be located in one half of the container is (1/2)N. For chemically-significant values of N (1020, say) this probability is indistinguishable from zero.
All of the changes described above take place spontaneously, meaning that
• Once they are allowed to commence, they will proceed to the finish without any outside intervention.
• It would be inconceivable that any of these changes could occur in the reverse direction (that is, be undone) without changing the conditions or actively disturbing the system in some way.
What determines the direction in which spontaneous change will occur? It is clearly not a fall in the energy, since in most cases cited above the energy of the system did not change. Even in the case of the falling book, in which the potential energy of the system (the book) falls, energy is conserved overall; if there is no net loss of energy when these processes operate in the forward or natural direction, it would not require any expenditure of energy for them to operate in reverse. In other words, contrary to what all too many people appear to believe, the First Law of Thermodynamics cannot predict the direction of a natural process.
The direction of a spontaneous process is not governed by the energy change and thus the First Law of Thermodynamics cannot predict the direction of a natural process
Direction through disorder
In our examination of the processes described above, we saw that although the total energy of the system and the surroundings (and thus, of the world) is unchanged, there is something about the world that has changed, and this is its degree of disorder.
After coins have been tossed or cards shuffled, the final state is invariably one of greater disorder. Similarly, the molecules of a gas can occupy a larger number of possible positions in space if the volume is larger, so the expansion of a gas is similarly accompanied by an increase in randomness
How can we express disorder quantitatively? From the example of the coins, you can probably see that simple statistics plays a role: the probability of obtaining three heads and seven tails after tossing ten coins is just the ratio of the number of ways that ten different coins can be arranged in this way, to the number of all possible arrangements of ten coins.
Using the language of molecular statistics, we say that a collection of coins in which a given fraction of its members are heads-up constitutes a macroscopic state of the system. Since we do not care which coins are heads-up, there are clearly numerous configurations of the individual coins which can result in this “macrostate”. Each of these configurations specifies a microscopic state of the system. The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate. To see what this means, consider the possible outcomes of a toss of four coins (Table $1$ ):
macrostate ways probability microstates
Table $1$: Coin Toss Results
0 heads 1 1/16 TTTT
1 head 4 4/16 = 1/4 HTTT THTT TTHT TTTH
2 heads 6 6/16 = 3/8 HHTT HTHT HTTH THHT TTHH THTH
3 heads 4 4/16 = 1/4 HHHT HTHH HHTH THHH
4 heads 1 1/16 HHHH
A toss of four coins will yield one of the five outcomes (macrostates) listed in the leftmost column of the table. The second column gives the number of “ways”— that is, the number of head/tail configurations of the set of coins (the number of microstates)— that can result in the macrostate. The probability of a toss resulting in a particular macrostate is proportional to the number of microstates corresponding to the macrostate, and is equal to this number, divided by the total number of possible microstates (in this example, 24 =16). An important assumption here is that all microstates are equally probable; that is, the toss is a “fair” one in which the many factors that determine the trajectory of each coin operate in an entirely random way.
The greater the number of microstates that correspond to a given macrostate, the greater the probability of that macrostate.
Disorder is more probable than order because there are so many more ways of achieving it. Thus coins and cards tend to assume random configurations when tossed or shuffled, and socks and books tend to become more scattered about a teenager’s room during the course of daily living. But there are some important differences between these large-scale mechanical, or macro systems, and the collections of sub-microscopic particles that constitute the stuff of chemistry, and which we will refer to here generically as molecules. Molecules, unlike macro objects, are capable of accepting, storing, and giving up energy in tiny amounts (quanta), and act as highly efficient carriers and spreaders of thermal energy as they move around. Thus, in chemical systems,
1. We are dealing with huge numbers of particles. This is important because statistical predictions are always more accurate for larger samples. Thus although for the four tosses there is a good chance (62%) that the H/T ratio will fall outside the range of 0.45 - 0.55, this probability becomes almost zero for 1000 tosses. To express this in a different way, the chances that 1000 gas molecules moving about randomly in a container would at any instant be distributed in a sufficiently non-uniform manner to produce a detectable pressure difference between the two halves of a container will be extremely small. If we increase the number of molecules to a chemically significant number (around 1020, say), then the same probability becomes indistinguishable from zero.
2. Once the change begins, it proceeds spontaneously. That is, no external agent (a tosser, shuffler, or teenager) is needed to keep the process going. Gases will spontaneously expand if they are allowed to, and reactions, one started, will proceed toward equilibrium.
3. Thermal energy is continually being exchanged between the particles of the system, and between the system and the surroundings. Collisions between molecules result in exchanges of momentum (and thus of kinetic energy) amongst the particles of the system, and (through collisions with the walls of a container, for example) with the surroundings.
4. Thermal energy spreads rapidly and randomly throughout the various energetically accessible microstates of the system. The direction of spontaneous change is that which results in the maximum possible spreading and sharing of thermal energy.
The importance of these last two points is far greater than you might at first think, but to fully appreciate this, you must recall the various ways in which thermal energy is stored in molecules— hence the following brief review.
The Spreading of Energy
Thermal energy is the portion of a molecule's energy that is proportional to its temperature, and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation. Since there are three directions in space, all molecules possess three modes of translational motion.
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation; a linear molecule such as CO2 in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes.
Finally, molecules consisting of two or more atoms can undergo internal vibrations. For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at a given temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. It is readily apparent that virtually all the thermal energy resides in translational states.
Notice the greatly different spacing of the three kinds of energy levels. This is extremely important because it determines the number of energy quanta that a molecule can accept, and, as the following illustration shows, the number of different ways this energy can be distributed amongst the molecules.
Accessible energy states. The more closely spaced the quantized energy states of a molecule, the greater will be the number of ways in which a given quantity of thermal energy can be shared amongst a collection of these molecules.
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states. This is a feature of the particle-in-a-box model, which predicts that the separation of the energy states of a gas confined within a box depends on the inverse square of the box length (and on the inverse of the particle mass as well.)
Quantum states, microstates, and energy spreading
At the atomic and molecular level, all energy is quantized; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute (roughly 10–30 J) that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations. Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10600— a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line: any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
• Addition of energy quanta (higher temperature),
• Increase in the number of molecules (resulting from dissociation, for example).
• the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Increasing temperature increases the number of microstates in a state and hence the entropy of the state.
Energy-spreading changes the world
Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery.
The profundity of this conclusion was recognized around 1900, when it was first described as the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever change. Not a happy thought!
Gases Spontaneously Expand and never contract
Everybody knows that a gas, if left to itself, will tend to expand so as to fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of energy states (and thus microstates) its thermal energy can occupy. Since all such microstates within the thermally accessible range of energies are equally probable, the expansion of the gas can viewed as a consequence of the tendency of energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
The same can in fact be said for even other highly probable distributions, such as having 49.999% of the molecules in the left half of the container and 50.001% in the right half. Even though the number of possible configurations that would yield this distribution of molecules is inconceivably great, it is essentially negligible compared to the number that would correspond to an exact 50-percent distribution.
The illustration represents the allowed thermal energy states of an ideal gas. The larger the volume in which the gas is confined, the more closely-spaced are these states, resulting in a huge increase in the number of microstates into which the available thermal energy can reside; this can be considered the origin of the thermodynamic "driving force" for the spontaneous expansion of a gas.
Heat Spontaneously flows from Hot to Cold
Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler always operates in the direction “warmer-to-cooler” because this allows thermal energy to occupy a larger number of energy states as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”.
When two bodies at different temperatures are placed in thermal contact, thermal energy flow from the warmer into the cooler one until they reach the same temperature.
1. Schematic depiction of the thermal energy states in two separated identical bodies at different temperatures (indicated by shading.)
2. When the bodies are brought into thermal contact, thermal energy flows from the higher occupied levels in the warmer object into the unoccupied levels of the cooler one until equal numbers are occupied .
As you might expect, the increase in the amount of energy spreading and sharing is proportional to amount of heat transferred q, but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T, the degree of dilution of the thermal energy is given by q /T
To understand why we have to divide by the temperature, consider the effect of very large and very small values of T in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are occupied, so the amount of energy spreading can be very great. Conversely, if the temperature is initially large, the number of new thermal energy states that become occupied will be negligible compared to the number already active.
Energy spreading and sharing in chemical reactions
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
1. The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H2 → 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H2. Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and H2 would not exist.
2. must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in H, as indicated by the vertical displacement of the right half in each of the four panels below.
Energy levels in H2 and 2 H. The number of thermally accessible energy states (indicated by the shading in the diagram below) increases with temperature, but because 2 moles of H possess twice as many translational states as one mole of H2, dissociation becomes increasingly favored at higher temperatures as more of these H states become thermally accessible.
This argument can be generalized to all molecules, illustrating an important principle:
All molecules spontaneously absorb heat and dissociate at high temperatures.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
• As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T1, the number of populated states of H2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
• At some temperature T2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H2 and “2H” in equal amounts; that is, the mole ratio of H2/H will be 1:2.
• As the temperature rises to T3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant.
The result is exactly what the Le Chatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures. This is all very well for helping you understand the direct connection between energy spreading when a chemical reaction occurs, but it is of little help in achieving our goal of predicting the direction and extent of chemical change. For this, we need to incorporate the concept of energy spreading into thermodynamics. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.01%3A_Energy_Spreading_Drives_Spontaneous_Change.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in bold.
• A reversible process is one carried out in infinitessimal steps after which, when undone, both the system and surroundings (that is, the world) remain unchanged (see the example of gas expansion-compression below). Although true reversible change cannot be realized in practice, it can always be approximated.
• ((in which a process is carried out.
• As a process is carried out in a more reversible manner, the value of w approaches its maximum possible value, and q approaches its minimum possible value.
• Although q is not a state function, the quotient qrev/T is, and is known as the entropy.
• energy within a system.
• The entropy of a substance increases with its molecular weight and complexity and with temperature. The entropy also increases as the pressure or concentration becomes smaller. Entropies of gases are much larger than those of condensed phases.
• The absolute entropy of a pure substance at a given temperature is the sum of all the entropy it would acquire on warming from absolute zero (where S=0) to the particular temperature.
Entropy is one of the most fundamental concepts of physical science, with far-reaching consequences ranging from cosmology to chemistry. It is also widely mis-represented as a measure of "disorder", as we discuss below. The German physicist Rudolf Clausius originated the concept as "energy gone to waste" in the early 1850s, and its definition went through a number of more precise definitions over the next 15 years.
Previously, we explained how the tendency of thermal energy to disperse as widely as possible is what drives all spontaneous processes, including, of course chemical reactions. We now need to understand how the direction and extent of the spreading and sharing of energy can be related to measurable thermodynamic properties of substances— that is, of reactants and products.
You will recall that when a quantity of heat q flows from a warmer body to a cooler one, permitting the available thermal energy to spread into and populate more microstates, that the ratio q/T measures the extent of this energy spreading. It turns out that we can generalize this to other processes as well, but there is a difficulty with using q because it is not a state function; that is, its value is dependent on the pathway or manner in which a process is carried out. This means, of course, that the quotient q/T cannot be a state function either, so we are unable to use it to get differences between reactants and products as we do with the other state functions. The way around this is to restrict our consideration to a special class of pathways that are described as reversible.
Reversible and irreversible changes
A change is said to occur reversibly when it can be carried out in a series of infinitesimal steps, each one of which can be undone by making a similarly minute change to the conditions that bring the change about. For example, the reversible expansion of a gas can be achieved by reducing the external pressure in a series of infinitesimal steps; reversing any step will restore the system and the surroundings to their previous state. Similarly, heat can be transferred reversibly between two bodies by changing the temperature difference between them in infinitesimal steps each of which can be undone by reversing the temperature difference.
The most widely cited example of an irreversible change is the free expansion of a gas into a vacuum. Although the system can always be restored to its original state by recompressing the gas, this would require that the surroundings perform work on the gas. Since the gas does no work on the surrounding in a free expansion (the external pressure is zero, so PΔV = 0,) there will be a permanent change in the surroundings. Another example of irreversible change is the conversion of mechanical work into frictional heat; there is no way, by reversing the motion of a weight along a surface, that the heat released due to friction can be restored to the system.
These diagrams show the same expansion and compression ±ΔV carried out in different numbers of steps ranging from a single step at the top to an "infinite" number of steps at the bottom. As the number of steps increases, the processes become less irreversible; that is, the difference between the work done in expansion and that required to re-compress the gas diminishes. In the limit of an ”infinite” number of steps (bottom), these work terms are identical, and both the system and surroundings (the “world”) are unchanged by the expansion-compression cycle. In all other cases the system (the gas) is restored to its initial state, but the surroundings are forever changed.
Definition: Reversible Changes
A reversible change is one carried out in such as way that, when undone, both the system and surroundings (that is, the world) remain unchanged.
It should go without saying, of course, that any process that proceeds in infinitesimal steps would take infinitely long to occur, so thermodynamic reversibility is an idealization that is never achieved in real processes, except when the system is already at equilibrium, in which case no change will occur anyway! So why is the concept of a reversible process so important?
The answer can be seen by recalling that the change in the internal energy that characterizes any process can be distributed in an infinity of ways between heat flow across the boundaries of the system and work done on or by the system, as expressed by the First Law ΔU = q + w. Each combination of q and w represents a different pathway between the initial and final states. It can be shown that as a process such as the expansion of a gas is carried out in successively longer series of smaller steps, the absolute value of q approaches a minimum, and that of w approaches a maximum that is characteristic of the particular process.
Thus when a process is carried out reversibly, the w-term in the First Law expression has its greatest possible value, and the q-term is at its smallest. These special quantities wmax and qmin (which we denote as qrev and pronounce “q-reversible”) have unique values for any given process and are therefore state functions.
Work and reversibility
For a process that reversibly exchanges a quantity of heat qrev with the surroundings, the entropy change is defined as
$\Delta S = \dfrac{q_{rev}}{T} \label{23.2.1}$
This is the basic way of evaluating ΔS for constant-temperature processes such as phase changes, or the isothermal expansion of a gas. For processes in which the temperature is not constant such as heating or cooling of a substance, the equation must be integrated over the required temperature range, as discussed below.
If no real process can take place reversibly, what use is an expression involving qrev? This is a rather fine point that you should understand: although transfer of heat between the system and surroundings is impossible to achieve in a truly reversible manner, this idealized pathway is only crucial for the definition of ΔS; by virtue of its being a state function, the same value of ΔS will apply when the system undergoes the same net change via any pathway. For example, the entropy change a gas undergoes when its volume is doubled at constant temperature will be the same regardless of whether the expansion is carried out in 1000 tiny steps (as reversible as patience is likely to allow) or by a single-step (as irreversible a pathway as you can get!) expansion into a vacuum.
The physical meaning of entropy
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system. This “spreading and sharing” can be spreading of the thermal energy into a larger volume of space or its sharing amongst previously inaccessible microstates of the system. The following table shows how this concept applies to a number of common processes.
Table $1$: Observations and Explanations in terms of Entropy changes
system and process source of entropy increase of system
A deck of cards is shuffled, or 100 coins, initially heads up, are randomly tossed. This has nothing to do with entropy because macro objects are unable to exchange thermal energy with the surroundings within the time scale of the process
Two identical blocks of copper, one at 20°C and the other at 40°C, are placed in contact. The cooler block contains more unoccupied microstates, so heat flows from the warmer block until equal numbers of microstates are populated in the two blocks.
A gas expands isothermally to twice its initial volume. A constant amount of thermal energy spreads over a larger volume of space
1 mole of water is heated by 1C°. The increased thermal energy makes additional microstates accessible. (The increase is by a factor of about
1020,000,000,000,000, 000,000,000.)
Equal volumes of two gases are allowed to mix. The effect is the same as allowing each gas to expand to twice its volume; the thermal energy in each is now spread over a larger volume.
One mole of dihydrogen, H2, is placed in a container and heated to 3000K. Some of the H2 dissociates to H because at this temperature there are more thermally accessible microstates in the 2 moles of H.
The above reaction mixture is cooled to 300K. The composition shifts back to virtually all H2because this molecule contains more thermally accessible microstates at low temperatures.
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Entropy and "disorder"
Entropy is still described, particularly in older textbooks, as a measure of disorder. In a narrow technical sense this is correct, since the spreading and sharing of thermal energy does have the effect of randomizing the disposition of thermal energy within a system. But to simply equate entropy with “disorder” without further qualification is extremely misleading because it is far too easy to forget that entropy (and thermodynamics in general) applies only to molecular-level systems capable of exchanging thermal energy with the surroundings. Carrying these concepts over to macro systems may yield compelling analogies, but it is no longer science. it is far better to avoid the term “disorder” altogether in discussing entropy.
Entropy and Probability
The distribution of thermal energy in a system is characterized by the number of quantized microstates that are accessible (i.e., among which energy can be shared); the more of these there are, the greater the entropy of the system. This is the basis of an alternative (and more fundamental) definition of entropy
$\color{red} S = k \ln Ω \label{23.2.2}$
in which k is the Boltzmann constant (the gas constant per molecule, 1.3810–23 J K–1) and Ω (omega) is the number of microstates that correspond to a given macrostate of the system. The more such microstates, the greater is the probability of the system being in the corresponding macrostate. For any physically realizable macrostate, the quantity Ω is an unimaginably large number, typically around $10^{10^{25}}$ for one mole. By comparison, the number of atoms that make up the earth is about $10^{50}$. But even though it is beyond human comprehension to compare numbers that seem to verge on infinity, the thermal energy contained in actual physical systems manages to discover the largest of these quantities with no difficulty at all, quickly settling in to the most probable macrostate for a given set of conditions.
The reason S depends on the logarithm of Ω is easy to understand. Suppose we have two systems (containers of gas, say) with S1, Ω1 and S2, Ω2. If we now redefine this as a single system (without actually mixing the two gases), then the entropy of the new system will be
$S = S_1 + S_2$
but the number of microstates will be the product Ω1Ω2because for each state of system 1, system 2 can be in any of Ω2 states. Because
$\ln(Ω_1Ω_2) = \ln Ω_1 + \ln Ω_2$
Hence, the additivity of the entropy is preserved.
If someone could make a movie showing the motions of individual atoms of a gas or of a chemical reaction system in its equilibrium state, there is no way you could determine, on watching it, whether the film is playing in the forward or reverse direction. Physicists describe this by saying that such systems possess time-reversal symmetry; neither classical nor quantum mechanics offers any clue to the direction of time.
However, when a movie showing changes at the macroscopic level is being played backward, the weirdness is starkly apparent to anyone; if you see books flying off of a table top or tea being sucked back up into a tea bag (or a chemical reaction running in reverse), you will immediately know that something is wrong. At this level, time clearly has a direction, and it is often noted that because the entropy of the world as a whole always increases and never decreases, it is entropy that gives time its direction. It is for this reason that entropy is sometimes referred to as "time's arrow".
But there is a problem here: conventional thermodynamics is able to define entropy change only for reversible processes which, as we know, take infinitely long to perform. So we are faced with the apparent paradox that thermodynamics, which deals only with differences between states and not the journeys between them, is unable to describe the very process of change by which we are aware of the flow of time.
The direction of time is revealed to the chemist by the progress of a reaction toward its state of equilibrium; once equilibrium is reached, the net change that leads to it ceases, and from the standpoint of that particular system, the flow of time stops. If we extend the same idea to the much larger system of the world as a whole, this leads to the concept of the "heat death of the universe" that was mentioned briefly in the previous lesson.
Absolute Entropies
Energy values, as you know, are all relative, and must be defined on a scale that is completely arbitrary; there is no such thing as the absolute energy of a substance, so we can arbitrarily define the enthalpy or internal energy of an element in its most stable form at 298K and 1 atm pressure as zero. The same is not true of the entropy; since entropy is a measure of the “dilution” of thermal energy, it follows that the less thermal energy available to spread through a system (that is, the lower the temperature), the smaller will be its entropy. In other words, as the absolute temperature of a substance approaches zero, so does its entropy. This principle is the basis of the Third law of thermodynamics, which states that the entropy of a perfectly-ordered solid at 0 K is zero.
Third law of thermodynamics
The entropy of a perfectly-ordered solid at 0 K is zero.
The absolute entropy of a substance at any temperature above 0 K must be determined by calculating the increments of heat q required to bring the substance from 0 K to the temperature of interest, and then summing the ratios q/T. Two kinds of experimental measurements are needed:
1. The enthalpies associated with any phase changes the substance may undergo within the temperature range of interest. Melting of a solid and vaporization of a liquid correspond to sizeable increases in the number of microstates available to accept thermal energy, so as these processes occur, energy will flow into a system, filling these new microstates to the extent required to maintain a constant temperature (the freezing or boiling point); these inflows of thermal energy correspond to the heats of fusion and vaporization. The entropy increase associated with melting, for example, is just ΔHfusion/Tm.
2. The heat capacity C of a phase expresses the quantity of heat required to change the temperature by a small amount ΔT , or more precisely, by an infinitesimal amount dT . Thus the entropy increase brought about by warming a substance over a range of temperatures that does not encompass a phase transition is given by the sum of the quantities C dT/T for each increment of temperature dT . This is of course just the integral
$S_{0^o \rightarrow T^o} = \int _{o^o}^{T^o} \dfrac{C_p}{T} dt$
Because the heat capacity is itself slightly temperature dependent, the most precise determinations of absolute entropies require that the functional dependence of C on T be used in the above integral in place of a constant C.
$S_{0^o \rightarrow T^o} = \int _{o^o}^{T^o} \dfrac{C_p(T)}{T} dt$
When this is not known, one can take a series of heat capacity measurements over narrow temperature increments ΔT and measure the area under each section of the curve in Figure $3$.
The area under each section of the plot represents the entropy change associated with heating the substance through an interval ΔT. To this must be added the enthalpies of melting, vaporization, and of any solid-solid phase changes. Values of Cp for temperatures near zero are not measured directly, but can be estimated from quantum theory.
/ Tb are added to obtain the absolute entropy at temperature T. As shown in Figure $4$ above, the entropy of a substance increases with temperature, and it does so for two reasons:
• As the temperature rises, more microstates become accessible, allowing thermal energy to be more widely dispersed. This is reflected in the gradual increase of entropy with temperature.
• The molecules of solids, liquids, and gases have increasingly greater freedom to move around, facilitating the spreading and sharing of thermal energy. Phase changes are therefore accompanied by massive and discontinuous increase in the entropy.
Standard Entropies of substances
The standard entropy of a substance is its entropy at 1 atm pressure. The values found in tables are normally those for 298K, and are expressed in units of J K–1 mol–1. The table below shows some typical values for gaseous substances.
Table $2$: Standard entropies of some gases at 298 K, J K–1 mol–1
He 126 H2 131 CH4 186
Ne 146 N2 192 H2O(g) 187
Ar 155 CO 197 CO2 213
Kr 164 F2 203 C2H6 229
Xe 170 O2 205 n -C3H8 270
Cl2 223 n -C4H10 310
Note especially how the values given in this Table $2$:illustrate these important points:
• Although the standard internal energies and enthalpies of these substances would be zero, the entropies are not. This is because there is no absolute scale of energy, so we conventionally set the “energies of formation” of elements in their standard states to zero. Entropy, however, measures not energy itself, but its dispersal amongst the various quantum states available to accept it, and these exist even in pure elements.
• It is apparent that entropies generally increase with molecular weight. For the noble gases, this is of course a direct reflection of the principle that translational quantum states are more closely packed in heavier molecules, allowing of them to be occupied.
• The entropies of the diatomic and polyatomic molecules show the additional effects of rotational quantum levels.
Table $3$: Standard entropies of some solid elements at 298 K, J K–1 mol–1
C(diamond) C(graphite) Fe Pb Na S(rhombic) Si W
2.5 5.7 27.1 51.0 64.9 32.0 18.9 33.5
The entropies of the solid elements are strongly influenced by the manner in which the atoms are bound to one another. The contrast between diamond and graphite is particularly striking; graphite, which is built up of loosely-bound stacks of hexagonal sheets, appears to be more than twice as good at soaking up thermal energy as diamond, in which the carbon atoms are tightly locked into a three-dimensional lattice, thus affording them less opportunity to vibrate around their equilibrium positions. Looking at all the examples in the above table, you will note a general inverse correlation between the hardness of a solid and its entropy. Thus sodium, which can be cut with a knife, has almost twice the entropy of iron; the much greater entropy of lead reflects both its high atomic weight and the relative softness of this metal. These trends are consistent with the oft-expressed principle that the more “disordered” a substance, the greater its entropy.
Table $4$: Standard entropy of water at 298 K, J K–1 mol–1
solid liquid gas
41 70 186
Gases, which serve as efficient vehicles for spreading thermal energy over a large volume of space, have much higher entropies than condensed phases. Similarly, liquids have higher entropies than solids owing to the multiplicity of ways in which the molecules can interact (that is, store energy.)
How Entropy depends on Concentration
As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy per mole of solute (the molar entropy) will of course always increase as the solution becomes more dilute.
For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by
$\Delta S = R \ln \left( \dfrac{V_2}{V_1} \right) \label{23.2.4}$
Note: If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.
Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation to express the entropy change associated with a change in the pressure of a perfect gas:
$\Delta S = R \ln \left( \dfrac{P_1}{P_2} \right) \label{23.2.5}$
Expressing the entropy change directly in concentrations, we have the similar relation
$\Delta S = R \ln \left( \dfrac{c_1}{c_2} \right) \label{23.2.6}$
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
How thermal energy is stored in molecules
Thermal energy is the portion of a molecule's energy that is proportional to its temperature, and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation. Since there are three directions in space, all molecules possess three modes of translational motion.
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation; a linear molecule such as CO2 in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal vibrations. For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the quantized translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at a given temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels. It is readily apparent that virtually all the thermal energy resides in translational states.
Notice the greatly different spacing of the three kinds of energy levels. This is extremely important because it determines the number of energy quanta that a molecule can accept, and, as the following illustration shows, the number of different ways this energy can be distributed amongst the molecules.
The more closely spaced the quantized energy states of a molecule, the greater will be the number of ways in which a given quantity of thermal energy can be shared amongst a collection of these molecules.
Note
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states.
Quantum states, microstates, and energy spreading
At the atomic and molecular level, all energy is quantized; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations. Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10600— a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line: any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
• Addition of energy quanta (higher temperature),
• Increase in the number of molecules (resulting from dissociation, for example).
• the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Heat Death: Energy-spreading changes the world
Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery. The profundity of this conclusion was recognized around 1900, when it was first described at the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever happen.
Why do gases tend to expand, but never contract?
Everybody knows that a gas, if left to itself, will tend to expand and fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of microstates its thermal energy can occupy. Since all such states within the thermally accessible range of energies are equally probable, the expansion of the gas can be viewed as a consequence of the tendency of thermal energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
Imagine a gas initially confined to one half of a box (Figure $7$). The barrier is then removed so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space. In terms of the spreading of thermal energy, Figure 23.2.X may be helpful. The tendency of a gas to expand is due to the more closely-spaced thermal energy states in the larger volume .
Entropy of mixing and dilution
Mixing and dilution really amount to the same thing, especially for idea gases. Replace the pair of containers shown above with one containing two kinds of molecules in the separate sections (Figure $9$). When we remove the barrier, the "red" and "blue" molecules will each expand into the space of the other. (Recall Dalton's Law that "each gas is a vacuum to the other gas".) However, notice that although each gas underwent an expansion, the overall process amounts to what we call "mixing".
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. But bear in mind that whereas the enthalpy associated with the expansion of a perfect gas is by definition zero, ΔH's of mixing of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute. But what's really dramatic is that when just one molecule of a second gas is introduced into the container ( in Figure $8$), an unimaginably huge number of new configurations become possible, greatly increasing the number of microstates that are thermally accessible (as indicated by the pink shading above).
Why heat flows from hot to cold
Just as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler one always operates in the direction “warmer-to-cooler” because this allows thermal energy to populate a larger number of energy microstates as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”.
When the bodies are brought into thermal contact (b), thermal energy flows from the higher occupied levels in the warmer object into the unoccupied levels of the cooler one until equal numbers are occupied in both bodies, bringing them to the same temperature. As you might expect, the increase in the amount of energy spreading and sharing is proportional to the amount of heat transferred q, but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T, the degree of dilution of the thermal energy is given by
$\dfrac{q}{T}$
To understand why we have to divide by the temperature, consider the effect of very large and very small values of Tin the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are initially occupied, so the amount of energy spreading into vacant states can be very great. Conversely, if the temperature is initially large, more thermal energy is already spread around within it, and absorption of the additional energy will have a relatively small effect on the degree of thermal disorder within the body.
Chemical reactions: why the equilibrium constant depends on the temperature
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
1. The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H2→ 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H2. Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and this molecule would not exist.
2. In order for this dissociation to occur, however, a quantity of thermal energy (heat) qU must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in H, as indicated by the vertical displacement of the right half in each of the four panels below.
In Figure $11$ are schematic representations of the translational energy levels of the two components H and H2 of the hydrogen dissociation reaction. The shading shows how the relative populations of occupied microstates vary with the temperature, causing the equilibrium composition to change in favor of the dissociation product.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
• As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T1, the number of populated states of H2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
• At some temperature T2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H2 and “2H” in equal amounts; that is, the mole ratio of H2/H will be 1:2.
• As the temperature rises to T3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant, thus favoring dissociation.
The result is exactly what the LeChatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures.
The following table generalizes these relations for the four sign-combinations of ΔH and ΔS. (Note that use of the standard ΔH° and ΔS° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.)
Exothermic reaction, ΔS > 0
This combustion reaction, like most such reactions, is spontaneous at all temperatures. The positive entropy change is due mainly to the greater mass of CO2 molecules compared to those of O2.
< 0
• ΔH° = –46.2 kJ
• ΔS° = –389 J K–1
• ΔG° = –16.4 kJ at 298 K
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
> 0
• ΔH° = 55.3 kJ
• ΔS° = +176 J K–1
• ΔG° = +2.8 kJ at 298 K
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures.Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
< 0
• ΔH° = 33.2 kJ
• ΔS° = –249 J K1
• ΔG° = +51.3 kJ at 298 K
This reaction is not spontaneous at any temperature, meaning that its reverse is always spontaneous. But because the reverse reaction is kinetically inhibited, NO2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Phase changes
Everybody knows that the solid is the stable form of a substance at low temperatures, while the gaseous state prevails at high temperatures. Why should this be? The diagram in Figure $12$ shows that
1. the density of energy states is smallest in the solid and greatest (much, much greater) in the gas, and
2. the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.
Changes of phase involve exchange of energy with the surroundings (whose energy content relative to the system is indicated (with much exaggeration!) by the height of the yellow vertical bars in Figure $13$. When solid and liquid are in equilibrium (middle section of diagram below), there is sufficient thermal energy (indicated by pink shading) to populate the energy states of both phases. If heat is allowed to flow into the surroundings, it is withdrawn selectively from the more abundantly populated levels of the liquid phase, causing the quantity of this phase to decrease in favor of the solid. The temperature remains constant as the heat of fusion is returned to the system in exact compensation for the heat lost to the surroundings. Finally, after the last trace of liquid has disappeared, the only states remaining are those of the solid. Any further withdrawal of heat results in a temperature drop as the states of the solid become depopulated.
Colligative Properties of Solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare, connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of why these phenomena occur.
Basically, these all result from the effect of dilution of the solvent on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities. The temperatures at which this occurs are depicted by the shading.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases.
Effects of pressure on the entropy: Osmotic pressure
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised (Figure $16$). The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase.
Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
This phenomenon can explain osmotic pressure. Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this pressure $\Pi$ is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.02%3A_Entropy_Rules.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in green type.
• In any macroscopic change, the entropy of the world (that is, system + surroundings) always increases; it never decreases.
• Processes that do not exchange heat with the surroundings (such as the free expansion of a gas into a vacuum) involve entropy change of the system alone, and are always spontaneous.
• A heat engine is a device that converts heat into work. The fraction of heat that can be converted into work is limited by the fall in temperature between the input to the engine and the exhaust.
• According to the Second Law of Thermodynamics, complete conversion of heat into work by a spontaneous cyclic process is impossible.
The First Law of thermodynamics, expressed as ΔU = q + w, is essentially a statement of the law of conservation of energy. The significance of this law is that it tells us that any proposed process that would violate this condition can be dismissed as impossible, without even inquiring further into the details of the process.
Why is the First Law not enough?
For simple mechanical operations on macroscopic objects, the First Law, conservation of energy, is all we usually need to determine such things as how many joules of energy is required to lift a weight or to boil some water, how many grams of glucose you must metabolize in order to climb a hill, or how much fuel your car needs to drive a given distance.
But if you think about it, there are a number of "simple mechanical operations" that never occur, even though they would not violate energy conservation.
• Suppose you drop a book onto a table top. The kinetic energy contained in the falling book is dispersed as thermal energy, slightly warming the book and the table top. According to the First Law, there is no reason why placing pre-warmed book on a warmed table top should not be able to propel the book back into the air. Similarly, why can't the energy imparted to the nail (and to the wood) by a hammer not pop the nail back out?
• One might propose a scheme to propel a ship by means of a machine that takes in seawater, extracts part of its thermal energy which is used to rotate the propeller, and then tosses the resulting ice cubes overboard. As long as the work done to turn the propeller is no greater than the heat required to melt the ice, the First Law is satisfied.
• Because motion of the air molecules is completely random, there is no reason why all of the molecules in one half of a room cannot suddenly "decide" to move into the other half, asphyxiating the unfortunate occupants of that side. (To the extent that air behaves as a perfect gas, this doesn't involve the First Law at all.)
What do all these scenarios that conform to the First Law but are nevertheless never seen to occur have in common? In every case, energy becomes less spread out, less "diluted". In the first two examples, thermal energy (dispersed) gets concentrated into organized kinetic energy of a macroscopic object— a book, a propeller. In the third case, the thermal energy gets concentrated into a smaller volume as the gas contracts.
The second law of thermodynamics says in effect, that the extent to which any natural process can occur is limited by the dilution of thermal energy (increase in entropy) that accompanies it, and once the change has occurred, it can never be un-done without spreading even more energy around. This is one of the most profound laws of nature, and should be a part of every educated person's world view. It is unfortunate that this law is so widely misrepresented as simply ordaining the increase in "disorder". A more brief statement of the Second Law (for those who know the meaning of "entropy") is
Second Law of Thermodynamics: The entropy of the world only increases and never decreases.
The more formal and historical ways of stating the Second Law will be presented farther below after we introduce the topic of heat engines. It is also worth knowing this important consequence of the Second Law: Just because the energy is “there” does not mean it will be available to do anything useful.
Entropy and Spontaneous Change
We explained how processes that take place spontaneously always proceed in a direction that leads to the spreading and sharing of thermal energy.
• A book falls to the tabletop (rather than absorbing heat and jumping up from it) because its kinetic energy changes into thermal energy which is widely dispersed into the molecules of the book and the table.
• A gas expands and solutions mix because the thermal energy their molecules possess get spread over a larger volume of space.
• Hydrogen gas dissociates into H atoms which share thermal energy amongst more particles and a greater volume of space. (But only if the temperature is high enough to make the huge number of new microstates energetically accessible.)
Because all natural processes lead to the spreading and sharing of thermal energy, and because entropy is a measure of the extent to which energy is dispersed in the world, it follows that:
In any spontaneous macroscopic change, the entropy of the world increases.
All natural processes that allow the free exchange of thermal energy amongst chemically-significant numbers of particles are accompanied by a spreading or “dilution” of energy that leaves the world forever changed. In other words, all spontaneous change leads to an increase in the entropy of the world. At first sight, this might seem to be inconsistent with our observations of very common instances in which there is a clear decrease in entropy, such as the freezing of a liquid, the formation of a precipitate, or the growth of an organism.
System + surroundings = the world!
... but it’s the entropy of the system plus surroundings that counts! It is important to understand that the criterion for spontaneous change is the entropy change of the system and the surroundings— that is, of the “world”, which we denote by ΔStotal:
$ΔS_{total} = ΔS_{system} + ΔS_{surroundings} \label{23.1}$
The only way the entropy of the surroundings can be affected is by exchange of heat with the system:
$ΔS_{surroundings} = \dfrac{q_{surr}}{ T} \label{23.2}$
Thus the freezing of water is accompanied by a flow of heat (the heat of fusion) into the surroundings, causing ΔSsurr to increase. At temperatures below the freezing point, this increase more than offsets the decrease in the entropy of the water itself, so ΔSworld exceeds zero and the process is spontaneous. The problem example below works this out in detail for a specific example.
Note that it does not matter whether the change in the system occurs reversibly or irreversibly; as mentioned previously, it is always possible to define an alternative (irreversible) pathway in which the amount of heat exchanged with the surroundings is the same as qrev ; because ΔS is a state function, the entropy change of the surroundings will have the same value as for the unrealizable reversible pathway.
If there is no flow of heat into or out of the surroundings, the entropy change of the system and that of the world are identical. Examples of such processes, which are always spontaneous, are the free expansion of an ideal gas into a vacuum, and the mixing of two ideal gases. In practice, almost all processes involving mixing and diffusion can be regarded as driven exclusively by the entropy increase of the system.
Most processes involving chemical and phase changes involve the exchange of heat with the surroundings, so their tendency to occur cannot always be predicted by focusing attention on the system alone. Further, owing to the –q/T term in ΔSsurroundings, the spontaneity of all such processes will depend on the temperature, as we illustrated for the dissociation of H2 previously.
As a quantitative example, let us consider the freezing of water. We know that liquid water will spontaneously change into ice when the temperature drops below 0°C at 1 atm pressure. Since the entropy of the solid is less than that of the liquid, we know the entropy of the water (the system here) will decrease on freezing. The amount of decrease is found by dividing the heat of fusion of ice by the temperature for the reversible pathway, which occurs at the normal freezing point:
$ΔS_{system} = \dfrac{-6000 \; J/mol}{273 \;K} = -21.978 \; J/mol$
If the process is actually carried at 0°C, then the heat of fusion is transferred to the surroundings at the same temperature, and the entropy of the surroundings increases by
$ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 21.979\; J/mol$
so that ΔStotal = 0. Under these conditions the process can proceed in either direction (freezing or melting) without affecting the entropy of the world; this means that both ice and liquid water can be present simultaneously without any change occurring; the system is said to be in equilibrium.
Suppose now that the water is supercooled to –1°C before it freezes. The entropy change of the water still corresponds to the reversible value qrev/T = (–6000J)/(273K). The entropy change of the surroundings, however, is now given by
$ΔS_{surroundings} = \dfrac{6000 \; J/mol}{273 \;K} = 22.059\; J/mol$
The total entropy change is now
$ΔS_{total} = (–21.978 + 22.059) J;\ K^{–1} mol^{–1} = +0.081\; J \;K^{–1} mol^{–1}$
indicating that the process can now occur (“is spontaneous”) only in the one direction.
Why did we use 273 K when evaluating ΔSsystem and 272 K for calculating ΔSsurroundings? In the latter case it is possible to formulate a reversible pathway by which heat can be transferred to the surroundings at any temperature. ΔSsystem, however, is a state function of water, and will vary with temperature only slightly.
Note that in order to actually freeze water, it must be cooled to very slightly below its normal freezing point, a condition known as supercooling. Freezing of supercooled water is of course an irreversible process (once it starts, it cannot be stopped except by raising the temperature by a finite amount), and the positive value of $ΔS_{total}$ tells us that this process will occur spontaneously at temperatures below 273 K. Under these conditions, the process is driven by the entropy increase of the surroundings resulting from flow of the heat of fusion of water into the surroundings.
Does the entropy of the world ever decrease?
The principle that thermal energy (and the molecules carrying it) tends to spread out is based on simple statistics. It must be remembered, however, that the laws of probability have meaningful application only to systems made up of large numbers of independent actors. If you trap a hundred flies in a bottle, they will generally distribute themselves more or less uniformly throughout the container; if there are only four flies, however, it is quite likely that all of them will occasionally be located in one particular half of the bottle.
Why the sky is blue
Similarly, you can trust with complete certainty that the spontaneous movement of half the molecules of the air to one side of the room you now occupy will not occur, even though the molecules are moving randomly and independently. On the other hand, if we consider a box whose dimensions are only a few molecular diameters, then we would expect that the random and short-term displacement of the small number of particles it contains to one side of the box would occur quite frequently. This is, in fact, the cause of the blueness of the sky: random fluctuations in the air density over tiny volumes of space whose dimensions are comparable with the wavelength of light results in selective scattering of the shorter wavelengths, so that blue light is scattered out, leaving the red light for the enjoyment of sunset-watchers to the east.
Brownian motion
This refers to the irregular zig-zag-like movement of extremely small particles such as plant pollen when they are suspended in a drop of liquid. Any such particle is continually being buffeted by the thermal motions of the surrounding liquid molecules. If size of the particle is very large compared to that the the liquid molecules, the forces that result from collisions of these molecules with the particle will cancel out and the particle remains undisturbed. If the particle is very small, however (perhaps only a thousand times larger than a molecule of the liquid), then the chances that it will undergo sufficiently more hits from one direction than from another during a brief interval of time become significant.
In these two examples, the entropy of the system decreases without any compensating flow of heat into the surroundings, leading to a net (but only temporary) decrease in the entropy of the world. This does not represent a failure of the Second Law, however, because no one has ever devised a way to extract useful work from these processes.
Heat Engines
The Industrial Revolution of the 19th century was largely driven by the invention of the steam engine. The first major use of such engines was to pump water out of mines, whose flooding from natural seepage seriously limited the depths to which they could be driven, and thus the availability of the metal ores that were essential to the expansion of industrial activities. The steam engine is a type of heat engine, a device that converts heat, provided by burning a fuel, into mechanical work, typically delivered through the motion of a piston in opposition to an opposing force. An engine is therefore an energy conversion device in which, ideally, every joule of heat released by combustion of the fuel could be extracted as work at the output shaft; such an engine would operate at 100 percent efficiency.
However, engineers of the time were perplexed to find that the efficiencies of steam engines were rather low (usually around 20%), with most of the heat being exhausted uselessly to the environment. Everyone understood that an efficiency exceeding 100% would be impossible (that would violate conservation of energy, and thus the First Law), but it was not clear why efficiencies could not rise significantly beyond the small values observed even as mechanical designs improved
The answer was found by a young French engineer, Sadi Carnot, who in 1824 published an analysis of an idealized heat engine that is generally considered to be the foundation of the science of thermodynamics— notwithstanding the fact that Carnot still accepted the belief that heat is a fluid-like substance called “caloric”. We will not replicate his analysis here (this is normally done in more advanced courses in physical chemistry), but will simply state his conclusion in his own [translated] words:
"The production of motive power is then due in steam-engines not to an actual consumption of caloric, but to its transportation from a warm body to a cold body...the production of heat alone is not sufficient to give birth to the impelling power: it is necessary that there should also be cold; without it, the heat would be useless. The ultimate attainable efficiency of any heat engine will depend on the temperatures at which heat is supplied to and removed from it."
The left side of the figure represents a generalized heat engine into which a quantity of heat qH, extracted from a source or “reservoir” at temperature TH is partly converted into work w. The remainder of the heat qL is exhausted to a reservoir at a lower temperature TL. In practice, TH would be the temperature of the steam in a steam engine, or the temperature of the combustion mixture in an internal combustion or turbine engine. The low temperature reservoir is ordinarily that of the local environment. The efficiency ε (epsilon) of a heat engine is the fraction of the heat abstracted from the high temperature reservoir that can be converted into work:
$ε = \dfrac{w}{q_H} \label{3.3}$
Carnot’s crucial finding (for which he would certainly have deserved a Nobel prize if these had existed at the time) is that the efficiency is proportional to the "distance'' in temperature that the heat can “fall” as it passes through the engine:
$ε = 1 - \dfrac{T_L}{T_H} \label{3.4}$
This is illustrated graphically in the right half of the figure just above, in which the efficiency is simply the fraction of the “complete” fall (in temperature) to absolute zero (arrow b) that the heat undergoes in the engine (arrow a.) Clearly, the only way to attain 100% efficiency would be to set the temperature of the exhaust reservoir to 0°K, which would be impossible. For most terrestrial heat engines, TL is just the temperature of the environment, normally around 300 K, so the only practical way to improve the efficiency is to make TH as high as possible. This is the reason that high pressure (superheated) steam is favored in commercial thermal power plants. The highest temperatures (and the greatest operating efficiencies) are obtained in gas turbine engines. However, as operating temperatures rise, the costs of dealing with higher steam pressures and the ability of materials such as turbine blades to withstand high temperatures become significant factors, placing an upper limit of around 600K on TH, thus imposing a maximum of around 50 percent efficiency on thermal power generation.
For nuclear plants, in which safety considerations require lower steam pressures, the efficiency is lower. One consequence of this is that a larger fraction of the heat is exhausted to the environment, which may result in greater harm to aquatic organisms when the cooling water is returned to a stream or estuary.
Example $1$
Several proposals have been made for building a heat engine that makes use of the temperature differential between the surface waters of the ocean and cooler waters that, being more dense, reside at greater depth. If the exhaust temperature is 5°C, what is the maximum amount of work that could be extracted from 1000 L of surface water at 10°C? (The specific heat capacity of water is 4.184 J g–1K–1.)
Solution
The amount of heat (qH) that must be extracted to cool the water by 5 K is (4.184 J g–1K–1)(106 g)(5 K) = 2.09 × 107 J. The ideal thermodynamic efficiency is given by
$1 -\dfrac{278 \;K}{283\; K} = 0.018$
The amount of work that could be done would be
$(0.018)(2.09 \times 10^7 \;J) = 3.7 \times 10^6 \;J$
Comment: It may be only 1.8% efficient, but it’s free!
The drinking bird as a heat engine
Few toys illustrate as many principles of physical science as this popular device that has been around for many years. At first glance it might appear to be a perpetual motion machine, but it's really just a simple heat engine. Modern "dippy birds" (as they are sometimes called) utilize dichloromethane as the working fluid.
This liquid boils at 39° C, and therefore has a rather high vapor pressure at room temperature. The liquid (to which a dye is often added for dramatic effect) is stored in a reservoir at the bottom of the bird. The bird's beak is covered with felt which, when momentarily dipped in water, creates a cooling effect as the water evaporates. This causes some of the CH2Cl2 vapor to condense in the head, reducing the pressure inside the device, causing more liquid to boil off and re-condense in the head. The redistribution of fluid upsets the balance, causing the bird to dip its beak back into the water. Once the head fills with liquid, it drains back into the bottom, tipping the bird upright to repeat the cycle.
We will leave it to you to relate this to the heat engine diagram above by identifying the heat source and sink, and estimate the thermodynamic efficiency of the engine.
Heat Pumps
If a heat engine is run “in reverse” by performing work on it (that is, changing “work out” to “work in” in Fig 8), it becomes a device for transporting heat against a thermal gradient. Refrigerators and air conditioners are the most commonly-encountered heat pumps. A heat pump can also be used to heat the interior of a building. In this application, the low temperature reservoir can be a heat exchanger buried in the earth or immersed in a well. In this application heat pumps are more efficient than furnaces or electric heating, but the capital cost is rather high.
The Second Law: what it means
It was the above observation by Carnot that eventually led to the formulation of the Second Law of Thermodynamics near the end of the 19th Century. One statement of this law (by Kelvin and Planck) is as follows:
Definition: Second Law of Thermodynamics (Kelvin Definition)
It is impossible for a cyclic process connected to a reservoir at one temperature to produce a positive amount of work in the surroundings.
To help you understand this statement and how it applies to heat engines, consider the schematic heat engine in the figure in which a working fluid (combustion gases or steam) expands against the restraining force of a weight that is mechanically linked to the piston. From a thermodynamic perspective, the working fluid is the system and everything else is surroundings. Expansion of the fluid occurs when it absorbs heat from the surroundings; return of the system to its initial state requires that the surrounding do work on the system. Now re-read the above statement of the Second Law, paying special attention to the italicized phrases which are explained below:
• A cyclic process is one in which the system returns to its initial state. A simple steam engine undergoes an expansion step (the power stroke), followed by a compression (exhaust stroke) in which the piston, and thus the engine, returns to its initial state before the process repeats.
• At one temperature” means that the expansion and compression steps operate isothermally. This means that ΔU = 0; just enough heat is absorbed by the system to perform the work required to raise the weight, so for this step q = –w.
• A positive amount of work in the surroundings” means that the engine does more work on the surroundings than the surroundings do on the engine. Without this condition the engine would be useless.
Note carefully that the Second Law applies only to a cyclic process— isothermal expansion of a gas against a non-zero pressure always does work on the surroundings, but an engine must repeat this process continually; to do so it must be returned to its initial state at the end of every cycle. When operating isothermally, the work –w it does on the surroundings in the expansion step (power stroke) is nullified by the work +w the surroundings must do on the system in order to complete the cycle. The Second Law can also be stated in an alternative way:
Definition: Second Law of Thermodynamics (Planck Definition)
It is impossible to construct a machine operating in cycles that will convert heat into work without producing any other changes.
Thus the Second Law does allow an engine to convert heat into work, but only if “other changes” (transfer of a portion of the heat directly to the surroundings) are allowed. And since heat can only flow spontaneously from a source at a higher temperature to a sink at a lower temperature, the impossibility of isothermal conversion of heat into work is implied. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.03%3A_The_Second_Law_of_Thermodynamics.txt |
Learning Objectives
• Gibbs Energy is a state function defined as $G = H – TS$.
• The practical utility of the Gibbs function is that $ΔG$ for any process is negative if it leads to an increase in the entropy of the world. Thus spontaneous change at a given temperature and pressure can only occur when it would lead to a decrease in $G$.
• The sign of the standard free energy change $ΔG^o$ of a chemical reaction determines whether the reaction will tend to proceed in the forward or reverse direction.
• Similarly, the relative signs of $ΔH^o$ and $ΔH^o$ determine whether the spontaneity of a chemical reaction will be affected by the temperature, and if so, in what way.
• The existence of sharp melting and boiling points reflects the differing temperature dependancies of the free energies of the solid, liquid, and vapor phases of a pure substance, which are in turn reflect their differing entropies.
Previously, we saw that it is the sum of the entropy changes of the system and surroundings that determines whether a process will occur spontaneously. In chemical thermodynamics we prefer to focus our attention on the system rather than the surroundings, and would like to avoid having to calculate the entropy change of the surroundings explicitly.
In this unit we introduce a new thermodynamic function, the free energy, which turns out to be the single most useful criterion for predicting the direction of a chemical reaction and the composition of the system at equilibrium. However, the term "free energy", although still widely used, is rather misleading, so we will often refer to it as "Gibbs energy." The free energy enables us to do this for changes that occur at a constant temperature and pressure (the Gibbs energy) or constant temperature and volume (the Helmholtz energy.)
Free energy: the Gibbs function
The Gibbs energy (also known as the Gibbs function or Gibbs Potential) is defined as
$G = H – T S \label{23.4.1}$
in which $S$ refers to the entropy of the system. Since $H$, $T$ and $S$ are all state functions, so is $G$. Thus for any change in state (under constant temperature), we can write the extremely important relation
$ΔG = ΔH – T ΔS \label{23.4.2}$
How does this simple equation encompass the entropy change of the world $ΔS_{total}$, which we already know is the sole criterion for spontaneous change from the second law of thermodynamics? Starting with the definition
$ΔS_{total} = ΔS_{surr} + ΔS_{sys} \label{23.4.3}$
we would first like to get rid of $ΔS_{surr}$. How can a chemical reaction (a change in the system) affect the entropy of the surroundings? Because most reactions are either exothermic or endothermic, they are accompanied by a flow of heat qp across the system boundary. The enthalpy change of the reaction $ΔH$ is defined as the flow of heat into the system from the surroundings when the reaction is carried out at constant pressure, so the heat withdrawn from the surroundings will be $–q_p$ which will cause the entropy of the surroundings to change by $–q_p / T = –ΔH/T$. We can therefore rewrite Equation $\ref{23.4.3}$ as
$ΔS_{total} = \dfrac{- ΔH}{T} + ΔS_{sys} \label{23.4.4}$
Multiplying each side by $-T$, we obtain
$-TΔS_{total} = ΔH - TΔS_{sys} \label{23.4.5}$
which expresses the entropy change of the world in terms of thermodynamic properties of the system exclusively. If $-TΔS_{total}$ is denoted by $ΔG$, then we have Equation $\ref{23.4.2}$ which defines the Gibbs energy change for the process.
From the foregoing, you should convince yourself that $G$ will decrease in any process occurring at constant temperature and pressure which is accompanied by an overall increase in the entropy. The constant temperature is a consequence of the temperature and the enthalpy appearing in the preceding Equation $\ref{23.4.5}$. Since most chemical and phase changes of interest to chemists take place under such conditions, the Gibbs energy is the most useful of all the thermodynamic properties of a substance, and (as we shall see in the lesson that follows this one) it is closely linked to the equilibrium constant.
Some textbooks and teachers say that the free energy, and thus the spontaneity of a reaction, depends on both the enthalpy and entropy changes of a reaction, and they sometimes even refer to reactions as "energy driven" or "entropy driven" depending on whether $ΔH$ or the $TΔS$ term dominates. This is technically correct, but misleading because it disguises the important fact that $ΔS_{total}$, which this equation expresses in an indirect way, is the only criterion of spontaneous change.
Helmholtz Energy is also a "Free Energy"
We will deal only with the Gibbs energy in this course. The Helmholtz free energy is of interest mainly to chemical engineers (whose industrial-scale processes are often confined to tanks and reactors of fixed volume) and some geochemists whose interest is centered on the chemistry that occurs deep within the earth's surface.
Gibbs Energy and Chemical Change
Remember that $ΔG$ is meaningful only for changes in which the temperature and pressure remain constant. These are the conditions under which most reactions are carried out in the laboratory; the system is usually open to the atmosphere (constant pressure) and we begin and end the process at room temperature (after any heat we have added or which is liberated by the reaction has dissipated.) The importance of the Gibbs function can hardly be over-stated: it serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Thus if the free energy of the reactants is greater than that of the products, the entropy of the world will increase when the reaction takes place as written, and so the reaction will tend to take place spontaneously. Conversely, if the free energy of the products exceeds that of the reactants, then the reaction will not take place in the direction written, but it will tend to proceed in the reverse direction.
$ΔG$ serves as the single master variable that determines whether a given chemical change is thermodynamically possible. Moreover, it determines the direction and extent of chemical change.
In a spontaneous change, Gibbs energy always decreases and never increases. This of course reflects the fact that the entropy of the world behaves in the exact opposite way (owing to the negative sign in the $TΔS$ term).
$\ce{H_2O(l) \rightarrow H2O(s)} \label{23.5.6}$
water below its freezing point undergoes a decrease in its entropy, but the heat released into the surroundings more than compensates for this, so the entropy of the world increases, the free energy of the H2O diminishes, and the process proceeds spontaneously.
In a spontaneous change, Gibbs energy always decreases and never increases.
An important consequence of the one-way downward path of the free energy is that once it reaches its minimum possible value, all net change comes to a halt. This, of course, represents the state of chemical equilibrium. These relations are nicely summarized as follows:
• $ΔG < 0$: reaction can spontaneously proceed to the right: $A \rightarrow B \nonumber$
• $ΔG > 0$: reaction can spontaneously proceed to the left: $A \leftarrow B \nonumber$
• $ΔG = 0$: the reaction is at equilibrium and both $[A]$ and $[B]$ will not change: $A \rightleftharpoons B. \nonumber$
No need to find the value of ΔG for a Specific Reaction!
This might seem strange, given the key importance $ΔG$ in determining whether or not a reaction will take place in a given direction. It turns out, however, that it is almost never necessary to explicitly evaluate $ΔG$. As we will show in the lesson that follows this one, it is far more convenient to work with the equilibrium constant of a reaction, within which $ΔG$ is "hidden". This is just as well, because for most reactions (those that take place in solutions or gas mixtures) the value of $ΔG$ depends on the proportions of the various reaction components in the mixture; it is not a simple sum of the "products minus reactants" type, as is the case with $ΔH$.
Recalling the condition for spontaneous change
$ΔG = ΔH – TΔS < 0$
it is apparent that the temperature dependence of ΔG depends almost entirely on the entropy change associated with the process. (We say "almost" because the values of $ΔH$ and $ΔS$ are themselves slightly temperature dependent; both gradually increase with temperature). In particular, notice that in the above equation the sign of the entropy change determines whether the reaction becomes more or less spontaneous as the temperature is raised. For any given reaction, the sign of $ΔH$ can also be positive or negative. This means that there are four possibilities for the influence that temperature can have on the spontaneity of a process:
Case 1: ΔH < 0 and ΔS > 0
Both enthalpic $\Delta H$ and entropic $-T\Delta S$ terms will be negative, so $ΔG$ will be negative regardless of the temperature. An exothermic reaction whose entropy increases will be spontaneous at all temperatures.
Case 2: ΔH < 0 and ΔS < 0
If the reaction is sufficiently exothermic it can force $ΔG$ negative only at temperatures below which $|TΔS| < |ΔH|$. This means that there is a temperature $T = ΔH / ΔS$ at which the reaction is at equilibrium; the reaction will only proceed spontaneously below this temperature. The freezing of a liquid or the condensation of a gas are the most common examples of this condition.
Case 3: ΔH > 0 and ΔS > 0
This is the reverse of the previous case; the entropy increase must overcome the handicap of an endothermic process so that $TΔS > ΔH$. Since the effect of the temperature is to "magnify" the influence of a positive $ΔS$, the process will be spontaneous at temperatures above $T = ΔH / ΔS$. (Think of melting and boiling.)
Case 4: ΔH > 0 and ΔS < 0
With both $ΔH$ and $ΔS$ working against it, this kind of process will not proceed spontaneously at any temperature. Substance A always has a greater number of accessible energy states, and is therefore always the preferred form.
The plots above are the important ones; do not try to memorize them, but make sure you understand and can explain or reproduce them for a given set of ΔH and ΔS.
• Their most important differentiating features are the position of the ΔH line (above or below the is TΔS line), and the slope of the latter, which of course depends on the sign of ΔS.
• The reaction A → B will occur spontaneously only when ΔG is negative (blue arrows pointing down.)
• TΔS plots are not quite straight lines as shown here. Similarly, the lines representing ΔH are even more curved.
The other two plots on each diagram are only for the chemistry-committed.
• Each pair of energy-level diagrams depicts the relative spacing of the microscopic energy levels in the reactants and products as reflected by the value of ΔS°. (The greater the entropy, the more closely-spaced are the quantized microstates.)
• The red shading indicates the range of energy levels that are accessible to the system at each temperature. The spontaneous direction of the reaction will always be in the direction in which the red shading overlaps the greater number of energy levels, resulting in the maximum dispersal of thermal energy.
• Note that the vertical offsets correspond to ΔH° for the reaction.
• Never forget that it is the ability of thermal energy to spread into as many of these states as possible that determines the tendency of the process to take place. None of this is to scale, of course!
The Standard Gibbs Energy
You have already been introduced to the terms such as $ΔU^o$ and $ΔH^o$ in which the $^o$ sign indicates that all components (reactants and products) are in their standard states. This concept of standard states is especially important in the case of the free energy, so let's take a few moments to review it. More exact definitions of the conventional standard states can be found in most physical chemistry textbooks. In specialized fields such as biochemistry and oceanography, alternative definitions may apply. For example, the "standard pH" of zero (corresponding to $[H^{+}] = 1\,M$) is impractical in biochemistry, so pH = 7 is commonly employed. For most practical purposes, the following definitions are good enough:
• gases: 1 atmosphere partial pressure
• pure liquids: the liquid under a total (hydrostatic) pressure of 1 atm.
• solutes: an effective concentration of 1 mol L–1 (1 mol dm–3). ("Effective" concentrations approach real concentrations as the latter approach zero; for practical purposes, these can be considered identical at real concentrations smaller than about 10–4 molar.)
• solids: the pure solid under 1 atm pressure
Reminder on Standard States
• There is actually no "standard temperature", but because most thermodynamics tables list values for 298.15 K (25° C), this temperature is usually implied.
• These same definitions apply to standard enthalpies and internal energies.
• Do not confuse these thermodynamic standard states with the "standard temperature and pressure" (STP) widely employed in gas law calculations.
To make use of Gibbs energies to predict chemical changes, we need to know the free energies of the individual components of the reaction. For this purpose we can combine the standard enthalpy of formation and the standard entropy of a substance to get its standard free energy of formation
$ΔG_f^o = ΔH_f^o – TΔS_f^o \label{23.4.7}$
Recall that the symbol ° refers to the standard state of a substance measured under the conditions of 1 atm pressure or an effective concentration of 1 mol L–1 and a temperature of 298 K. Then determine the standard Gibbs energy of the reaction according to
$ΔG^o = \sum ΔG_f^o \;(\text{products})– \sum ΔG_f^o \;(\text{reactants}) \label{24.4.8}$
As with standard heats of formation, the standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under the standard conditions of 1 atm pressure and 298 K. Standard Gibbs free energies of formation are normally found directly from tables. Once the values for all the reactants and products are known, the standard Gibbs energy change for the reaction is found by Equation $\ref{23.4.7}$. Most tables of thermodynamic values list $ΔG_f^o$ values for common substances (e.g., Table T2), which can, of course, always be found from values of ΔHf° and ΔSf°.
Example $1$
Find the standard Gibbs energy change for the reaction
$\ce{CaCO3(s) \rightarrow CaO (s) + CO2(g)} \nonumber$
The $ΔG_f^o°$ values for the three components of this reaction system are $\ce{CaCO3(s)}$: –1128 kJ mol–1, CaO(s): –603.5 kJ mol–1, CO2(g): –137.2 kJ mol–1.
Solution
Substituting into Equation $\ref{23.4.7}$, we have
$ΔG^o = (–603.5 –137.2) – (–1128) kJ\, mol^{–1} = +130.9\, kJ\, mol^{–1} \nonumber$
This indicates that the process is not spontaneous under standard conditions (i.e., solid calcium carbone will not form solid calcium oxide and CO2 at 1 atm partial pressure at 25° C).
Comment: This reaction is carried out on a huge scale to manufacture cement, so it is obvious that the process can be spontaneous under different conditions.
The practical importance of the Gibbs energy is that it allows us to make predictions based on the properties (Δ values) of the reactants and products themselves, eliminating the need to experiment. But bear in mind that while thermodynamics always correctly predicts whether a given process can take place (is spontaneous in the thermodynamic sense), it is unable to tell us if it will take place at an observable rate.
When thermodynamics says "no", it means exactly that. When it says "yes", it means "maybe".
Example $2$
The reaction
$\ce{ 1/2 O2(g) + H2(g) → H2O(l)} \nonumber$
is used in fuel cells to produce an electrical current. The reaction can also be carried out by direct combustion.
Thermodynamic data: molar entropies in J mol–1 K–1: O2(g) 205.0; H2(g)130.6; H2O(l) 70.0; H2O(l) ΔH°f = –285.9 kJ mol–1.
Use this information to find
1. The amount of heat released when the reaction takes place by direct combustion;
2. The amount of electrical work the same reaction can perform when carried out in a fuel cell at 298 K under reversible conditions;
3. The amount of heat released under the same conditions.
Solution
First, we need to find $ΔH^o$ and $ΔS^o$ for the process. Recalling that the standard enthalpy of formation of the elements is zero,
\begin{align*} ΔH^o &= ΔH^p_f(\text{products}) – ΔH^°_f(\text{reactants}) \[4pt] &= –285.9\, kJ\, mol^{–1} – 0 \[4pt] &= –285.9 \,kJ \,mol^{–1} \end{align*}.
Similarly,
\begin{align*} ΔS^o &= S^o_f(\text{products}) – S^o_f(\text{reactants}) \[4pt] &= (70.0) – (½ \times 205.0 + 130.6) \[4pt] &= –163\, J\, K^{–1}mol^{–1} \end{align*}
1. When the hydrogen and oxygen are combined directly, the heat released will be $ΔH^o = –285.9\, kJ\, mol^{–1}$.
2. The maximum electrical work the fuel cell can perform is given by \begin{align*}ΔG^o &= ΔH^o – TΔS^o \[4pt] &= –285.9 \,kJ\, mol^{–1} – (298\, K)(–163\, JK^{–1}mol^{–1}) \[4pt] &= –237.2 \,kJ\, mol^{–1}.\end{align*}.
3. The heat released in the fuel cell reaction is the difference between the enthalpy change (the total energy available) and the reversible work that was expended: \begin{align*} ΔH^o – ΔG^o &= TΔS^o \[4pt] &= (298\, K)(–163\, JK^{–1}mol^{–1}) \[4pt] &= –48,800\, J\, mol^{–1} \[4pt] &=–48.8 \,kJ\, mol^{–1}.\end{align*}.
The foregoing example illustrates an important advantage of fuel cells. Although direct combustion of a mole of hydrogen gas yields more energy than is produced by the same net reaction within the fuel cell, the latter, in the form of electrical energy, can be utilized at nearly 100-percent energy efficiency by a motor or some other electrical device. If the thermal energy released by direct combustion were supplied to a heat engine, second-law considerations would require that at least half of this energy be "wasted" to the surroundings.
ΔG vs. ΔG°: what's the difference?
Δrefer to single, specific chemical changes in which all components (reactants and products) are in their standard states.
The $ΔG_f^o$ of a substance, like $ΔH_f^o$, refers to the reaction in which that substance is formed from the elements as they exist in their most stable forms at 1 atm pressure and (usually) 298 K. Both of these terms are by definition zero for the elements in their standard states. There are only a few common cases in which this might create some ambiguity:
Table $1$: Standard Gibbs Energies of Select Substances
Stable Form $ΔG_f^o$ (kJ mol–1) Unstable Form $ΔG_f^o$ (kJ mol–1)
$\ce{O2(g)}$ 0 $\ce{O3(g)}$ 163.2
$\ce{C(graphite)}$ 0 $\ce{C(diamond)}$ 2.9
$\ce{S(rhombic)}$ 0 $\ce{S(monoclinic)}$ 0.1
$\ce{P(white)}$ 0 $\ce{P4(g)}$ 24.4
Ions in aqueous solution are a special case; their standard free energies are relative to the hydrated hydrogen ion $\ce{H^{+}(aq)}$ which is assigned $ΔG_f^o = 0$.
$ΔG$ is very different from ΔG°. The distinction is nicely illustrated in Figure $5$ in which ΔG is plotted on a vertical axis for two hypothetical reactions having opposite signs of Δ. The horizontal axis schematically expresses the relative concentrations of reactants and products at any point of the process. Note that the origin corresponds to the composition at which half of the reactants have been converted into products.
Take careful note of the following:
• for the Δ > 0 reaction. Notice that there are an infinite number of these values, depending on the progress of the reaction. In contrast there is only a single value of Δ, corresponding to the composition at which ΔG = 0 ().
• At this point, some products have been formed, but the composition is still dominated by reactants.
• If we begin at a composition to the left of , ΔG will be negative and the composition will move to the right. Similarly if we begin with a composition to the right of , ΔG will be positive and the composition will move to the left.
• The plot on the right is for the ΔG° < 0 reaction, for which ΔG° is shown at . At its equilibrium point , there are more products than reactants. If we start at a composition to the right of , the composition will tend to move to the left. If the initial composition is to the left of , the reaction will tend to proceed to the right.
• What would happen if Δ were 0? The equilibrium point of such a reaction would be at the origin, corresponding to half the reactants being converted to products.
The important principle you should understand from this is that a negative Δ does not mean that the reactants will be completely transformed into products. By the same token, a positive Δ does not mean that no products are formed at all.
It should now be clear from the discussion above that a given reaction carried out under standard conditions is characterized by a single value of Δ.
The reason for the Gibbs energy minimum at equilibrium relates to the increase in entropy when products and reactants coexist in the same phase. As seen in the plot, even a minute amount of "contamination" of products by reactants reduces the free energy below that of the pure products. In contrast, composition of a chemical reaction system undergoes continual change until the equilibrium state is reached. So the a single reaction can have an infinite number of ΔG values, reflecting the infinite possible compositions between the extremes of pure reactants (zero extent of reaction) and pure products (unity extent of reaction).
In the example of a reaction A → B, depicted in the above diagram, the standard free energy of the products is smaller than that of the reactants , so the reaction will take place spontaneously. This does not mean that each mole of pure A will be converted into one mole of pure B. For reactions in which products and reactants occupy a single phase (gas or solution), the meaning of "spontaneous" is that the equilibrium composition will correspond to an extent of reaction greater than 0.5 but smaller than unity.Note, however, that for Δ values in excess of about ±50 kJ mol–1, the equilibrium composition will be negligibly different from zero or unity extent-of-reaction. The physical meaning of ΔG is that it tells us how far the free energy of the system has changed from G° of the pure reactants . As the reaction proceeds to the right, the composition changes, and ΔG begins to fall. When the composition reaches , ΔG reaches its minimum value and further reaction would cause it to rise. But because free energy can only decrease but never increase, this does not happen. The composition of the system remains permanently at its equilibrium value.
A G vs. extent-of-reaction diagram for a non-spontaneous reaction can be interpreted in a similar way; the equilibrium composition will correspond to an extent of reaction greater than zero but less than 0.5. In this case, the minimum at reflects the increase in entropy when the reactants are "contaminated" by a small quantity of products.
If all this detail about ΔG seems a bit overwhelming, do not worry: it all gets hidden in the equilibrium constant and reaction quotient that we discuss in the next lesson!
Interpretation of Standard Gibbs energy changes
Although it is $ΔG$ rather than $ΔG^o$ that serves as a criterion for spontaneous change at constant temperature and pressure, $ΔG^o$ values are so readily available that they are often used to get a rough idea of whether a given chemical change is possible. This is practical to do in some cases, but not in others:
Example
It generally works for reactions such as
$\ce{4 NH_3(g) + 5 O_2(g) → 4 NO(g) + 6 H_2O(g)} \nonumber$
with $ΔG^o = –1,010\, kJ$.
(industrially important for the manufacture of nitric acid) because $ΔG^o$ is so negative that the reaction will be spontaneous and virtually complete under just about any reasonable set of conditions.
Example
The following reaction expresses the fact that the water molecule is thermodynamically stable:
$\ce{2 H_2(g) + 1/2 O_2(g)→ H_2O(l)} \nonumber$
with $ΔG^o = –237.2 \,kJ$.
Note that this refers to liquid water (the standard state of H2O at 25°). If you think about it, a negative standard Gibbs energy of formation (of which this is an example) can in fact be considered a definition of molecular stability.
Example
Similarly, dissociation of dihydrogen into its atoms is highly unlikely under standard conditions:
$\ce{H_2O(g) → 2 H(g) + O(g)} \nonumber$
with $ΔG^o = +406.6\, kJ$.
Again, an analogous situation would apply to any stable molecule.
Example
Now consider the dissociation of dinitrogen tetroxide
$\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber$
with $ΔG^o = +2.8 kJ$.
in which the positive value of Δ tells us that N2O4 at 1 atm pressure will not change into two moles of NO2 at the same pressure, but owing to the small absolute value of Δ, we can expect the spontaneity of the process to be quite sensitive to both the temperature (as shown in the table below) and to the pressure in exactly the way the Le Chatelier principle predicts.
Example
For reactions involving dissolved ions, one has to be quite careful. Thus for the dissociation of the weak hydrofluoric acid
$\ce{HF(aq) → H^+(aq) + F^–(aq)} \nonumber$
with $ΔG^o = –317 \,kJ$.
it is clear that a 1 mol/L solution of HF will not dissociate into 1M ions, but this fact is not very useful because if the HF is added to water, the initial concentration of the fluoride ion will be zero (and that of H+ very close to zero), and the Le Chatelier principle again predicts that some dissociation will be spontaneous.
Example
It is common knowledge that dissociation of water into hydrogen- and hydroxyl ions occurs only very sparingly:
$\ce{H_2O(l) → H^+(aq) + OH^–(aq) } \nonumber$
with $ΔG^o = 79.9 \,kJ$.
which correctly predicts that the water will not form 1M (effective concentration) of the ions, but this is hardly news if you already know that the product of these ion concentrations can never exceed 10–14 at 298K.
Example
Finally, consider this most familiar of all phase change processes, the vaporization of liquid water:
$\ce{H_2O(l) → H_2O(g)} \nonumber$
with $ΔG^o = 8.58 \,kJ$.
Conversion of liquid water to its vapor at 1 atm partial pressure does not take place at 25° C, at which temperature the equilibrium partial pressure of the vapor (the "vapor pressure") is only 0.031 atm (23.8 torr.) Gaseous H2O at a pressure of 1 atm can only exist at 100° C. Of course, water left in an open container at room temperature will spontaneously evaporate if the partial pressure of water vapor in the air is less than 0.031 atm, corresponding to a relative humidity of under 100%
Finding the Equilibrium Temperature
A reaction is in its equilibrium state when
$ΔG = ΔH – TΔS = 0 \label{23.4.1a}$
The temperature at which this occurs is given by
$T = \dfrac{ΔH}{TΔS} \label{23.4.1b}$
If we approximate $ΔH$ by $ΔH^o$ and $ΔS$ by $ΔS^o$, so Equation \ref{23.4.1a} would be
$ΔG \approx ΔH^o – TΔS^o = 0 \label{23.4.1aa}$
We can then estimate the normal boiling point of a liquid. From the following thermodynamic data for water:
Caution!
Because ΔH° values are normally expressed in kilojoules while ΔS° is given in joules, a very common student error is to overlook the need to express both in the same units.
We find that liquid water is in equilibrium with water vapor at a partial pressure of 1 atm when the temperature is
$T = \dfrac{44,100\, J}{118.7\, J\, K^{–1}} = 371.5\, K$
But "the normal boiling point of water is 373 K", you say? Very true. The reason we are off here is that both ΔH ° and ΔS ° have their own temperature dependencies; we are using the "standard" 25° values without correcting them to 100° C. Nevertheless, if you think about it, the fact that we can estimate the boiling point of a liquid from a table of thermodynamic data should be rather impressive! Of course, the farther one gets from 298 K, the more unreliable will be the result. Thus for the dissociation of dihydrogen into its atoms,
All one can say here is that H2 will break down at something over 3000 K or so. (You may already know that all molecules will dissociate into their atoms at high temperatures.) We tend to think of high temperatures as somehow "forcing" molecules to dissociate into their atoms, but this is wrong. In order to get the H–H bond to vibrate so violently through purely thermal excitation that the atoms would fly apart, a temperature more like 30,000 K would be required. The proper interpretation is at the temperature corresponding to ΔH/TΔS, the molecule spontaneously absorbs energy from the surroundings sufficient to overcome the H-H bond strength.
Predicting the Effects of Temperature
The $T\Delta S$ term interacts with the $ΔH$ term in $\Delta G$ to determine whether the reaction can take place at a given temperature. This can be more clearly understood by examining plots of $TΔS^o$ and $ΔH^o$ as functions of the temperature for some actual reactions. Of course these parameters refer to standard states that generally do not correspond to the temperatures, pressures, or concentrations that might be of interest in an actual case. Nevertheless, these quantities are easily found and they can usefully predict the way that temperature affects these systems.
Case 1: Exothermic reaction, ΔS° > 0
$\ce{C(graphite) + O_2(g) → CO_2(g)} \nonumber$
• $ΔH^o = –393\, kJ$
• $ΔG^o = –394 \,kJ$ at $298\, K$
This combustion reaction, like most such reactions, is spontaneous at all temperatures. The positive entropy change is due mainly to the greater mass of $\ce{CO2}$ molecules compared to those of $\ce{O2}$.
Case 2: Exothermic reaction, ΔS° < 0
$\ce{3 H_2 + N_2 → 2 NH_3(g) } \nonumber$
• $ΔH^o = –46.2\, kJ$
• $ΔG^o = –16.4\, kJ$ at $298\, K$
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
Case 3: Endothermic reaction, ΔS° > 0
$\ce{N_2O_4(g) → 2 NO_2(g)} \nonumber$
• $ΔH^o = 55.3\, kJ$
• $ΔG^o = +2.8\, kJ$ at $298\, K$
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures. Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
Case 4: Endothermic Reaction, ΔS° < 0
$\ce{ 1/2 N_2 (g) + O_2 (g)→ NO_2(g)} \nonumber$
• $ΔH^° = 33.2\, kJ$
• $ΔS^o = –249\, J\, K^{–1}$
• $ΔG^o = +51.3\, kJ$ at $298\, K$
This reaction is not spontaneous at any temperature, meaning that its reverse is always spontaneous. But because the reverse reaction is kinetically inhibited, NO2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Concluding remarks on Gibbs Energy
The appellation “free energy” for $G$ has led to so much confusion that many scientists now refer to it simply as the Gibbs energy. The “free” part of the older name reflects the steam-engine origins of thermodynamics with its interest in converting heat into work: $ΔG$ is the maximum amount of energy, which can be “freed” from the system to perform useful work. By "useful", we mean work other than that which is associated with the expansion of the system. This is most commonly in the form of electrical work (moving electric charge through a potential difference), but other forms of work (osmotic work, increase in surface area) are also possible.
A much more serious difficulty with the Gibbs function, particularly in the context of chemistry, is that although $G$ has the units of energy (joules, or in its intensive form, J mol–1), it lacks one of the most important attributes of energy in that it is not conserved. Thus, although the free energy always falls when a gas expands or a chemical reaction takes place spontaneously, there need be no compensating increase in energy anywhere else. Referring to $G$ as an energy also reinforces the false but widespread notion that a fall in energy must accompany any change. But if we accept that energy is conserved, it is apparent that the only necessary condition for change (whether the dropping of a weight, expansion of a gas, or a chemical reaction) is the redistribution of energy. The quantity $–ΔG$ associated with a process represents the quantity of energy that is “shared and spread”, which as we have already explained is the meaning of the increase in the entropy. The quotient $–ΔG/T$ is in fact identical with $ΔS_{total}$, the entropy change of the world, whose increase is the primary criterion for any kind of change.
$G$ differs from the thermodynamic quantities H and S in another significant way: it has no physical reality as a property of matter, whereas $H$ and $S$ can be related to the quantity and distribution of energy in a collection of molecules. The free energy is simply a useful construct that serves as a criterion for change and makes calculations easier.
What Gibbs Energy is not...
• Gibbs Energy is not free energy
• Gibbs Energy is not energy
• Gibbs Energy is not even "real" | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.04%3A_Free_Energy_and_the_Gibbs_Function.txt |
The free energy of a pure liquid or solid at 1 atm pressure is just its molar free energy of formation ΔG multiplied by the number of moles present. For gases and substances in solution, we have to take into account the concentration (which, in the case of gases, is normally expressed in terms of the pressure). We know that the lower the concentration, the greater the entropy, and thus the smaller the free energy. The following excerpt from this lesson serves as the starting point for the rest of the present lesson.
Entropy Depends on Concentration
As a substance becomes more dispersed in space, the thermal energy it carries is also spread over a larger volume, leading to an increase in its entropy. Because entropy, like energy, is an extensive property, a dilute solution of a given substance may well possess a smaller entropy than the same volume of a more concentrated solution, but the entropy per mole of solute (the molar entropy) will of course always increase as the solution becomes more dilute.
For gaseous substances, the volume and pressure are respectively direct and inverse measures of concentration. For an ideal gas that expands at a constant temperature (meaning that it absorbs heat from the surroundings to compensate for the work it does during the expansion), the increase in entropy is given by
$\Delta S = R\ln \left( \dfrac{V_2}{V_1} \right) \label{2-4}$
(If the gas is allowed to cool during the expansion, the relation becomes more complicated and will best be discussed in a more advanced course.)
Because the pressure of a gas is inversely proportional to its volume, we can easily alter the above relation to express the entropy change associated with a change in the pressure of a perfect gas:
$\Delta S = R\ln \left( \dfrac{P_1}{P_2} \right) \label{2-5}$
Expressing the entropy change directly in concentrations $c$, we have the similar relation
$\Delta S = R\ln \left( \dfrac{C_1}{C_2} \right) \label{2-6}$
Although these equations strictly apply only to perfect gases and cannot be used at all for liquids and solids, it turns out that in a dilute solution, the solute can often be treated as a gas dispersed in the volume of the solution, so the last equation can actually give a fairly accurate value for the entropy of dilution of a solution. We will see later that this has important consequences in determining the equilibrium concentrations in a homogeneous reaction mixture.
The Gibbs energy of a Gas: standard states
The pressure of a perfect gas does not affect its enthalpy, but it does affect the entropy (box at left), and thus, through the –TΔS term, the free energy. When the pressure of such a gas changes from $P_1$ to $P_2$, the Gibbs energy change is
$\Delta G = \Delta H - T \Delta S = 0 - RT \ln \left( \dfrac{P_1}{P_2} \right) \label{4.8}$
How can we evaluate the free energy of a specific sample of a gas at some arbitrary pressure? First, recall that the standard molar free energy G° that you would look up in a table refers to a pressure of 1 atm. The free energy per mole of our sample is just the sum of this value and any change in free energy that would occur if the pressure were changed from 1 atm to the pressure of interest
$G = G^o + RT \ln \left( \dfrac{P_1}{1\; atm} \right) \label{4.9}$
which we normally write in abbreviated form
$G = G^o + RT \ln P \label{4-10}$
Escaping Tendency
The term escaping tendency is not commonly used in traditional thermodynamics because it is essentially synonymous with the free energy, but it is worth knowing because it helps us appreciate the physical significance of free energy in certain contexts. The higher the pressure of a gas, the greater will be the tendency of its molecules to leave the confines of the container; we will call this the escaping tendency. The above equation tells us that the pressure of a gas is a directly observable measure of its free energy (G, not G°!). Combining these two ideas, we can say that the free energy of a gas is also a measure of its escaping tendency.
From Gases to Solutions: Mixing and Dilution
All substances, given the opportunity to form a homogeneous mixture with other substances, will tend to become more dilute. This can be rationalized simply from elementary statistics; there are more equally probable ways of arranging one hundred black marbles and one hundred white marbles, than two hundred marbles of a single color. For massive objects like marbles this has nothing to do with entropy, of course. However, when we are dealing with huge numbers of molecules capable of storing, exchanging and spreading thermal energy, mixing and expansion are definitely entropy-driven processes.
It can be argued, in fact, that mixing and expansion are really very similar; after all, when we mix two gases, each is expanding into the space formerly occupied exclusively by the other.
Suppose, for example, that we have a gas initially confined to one half of a box, and we then remove the barrier so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space; the actual increase, according to Equations $\ref{2-4}$ and $\ref{2-5}$ above, is $R \ln 2$. And from Equation $\ref{4-10}$, the change in $G$ will be $–RT \ln 2$.
Now let us repeat the experiment, but starting this time with "red" molecules in one half of the container and "blue" ones in the right half. Because we now have two gases expanding into twice their initial volumes, the changes in S and G will be twice as great:
$ΔS = 2 R \ln 2$
$ΔG = –2 RT \ln 2$
However, notice that although each gas underwent an expansion, the overall process in this case is equivalent to mixing.
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. An important qualification here is that the solution must be an ideal one, meaning that the strength of interactions between all type of molecules (solutes A and B, and the solvent) must be the same. Remember that the enthalpy associated with the expansion of a perfect gas is by definition zero. In contrast, the $ΔH_{mixing}$ of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute.
Given this proviso, we can define the Gibbs energy of dilution or mixing by substituting this equation into the definition of ΔG:
$\Delta G_{dilution} = \Delta H_{dilution} - RT\ln \left( \dfrac{C_1}{C_2} \right) \label{4-11}$
If the substance in question forms an ideal solution with the other components, then $ΔH_{diution}$ is zero, and we can write
$\Delta G_{dilution} = RT\ln \left( \dfrac{C_2}{C_1} \right) \label{4-12}$
These relations tell us that the dilution of a substance from an initial concentration $C_1$ to a more dilute concentration $C_2$ is accompanied by a decrease in the free energy, and thus will occur spontaneously. By the same token, the spontaneous “un-dilution” of a solution will not occur (we do not expect the tea to diffuse back into the tea bag!) However, un-dilution can be forced to occur if some means can be found to supply to the system an amount of energy (in the form of work) equal to $\Delta G_{dilution}$.
An important practical example of this is the metabolic work performed by the kidneys in concentrating substances from the blood for excretion in the urine.
To find the Gibbs energy of a solute at some arbitrary concentration, we proceed in very much the same way as we did for a gas: we take the sum of the standard free energy plus any change in the free energy that would accompany a change in concentration from the standard state to the actual state of the solution. From Equation $\ref{4-12}$ it is easy to derive an expression analogous to Equation $\ref{4-10}$:
$G = G^o + RT \ln C \label{4-13}$
which gives the free energy of a solute at some arbitrary concentration $C$ in terms of its value $G^o$ in its standard state.
Although Equation $\ref{4-13}$ has the same simple form as Equation $\ref{4-10}$, its practical application is fraught with difficulties, the major one being that it does not usually give values of $G$ that are consistent with experiment, especially for solutes that are ionic or are slightly soluble. This is because most solutions (especially those containing dissolved ions) are far from ideal; intermolecular interactions between solute molecules and between solute and solvent bring back the enthalpy term that we left out in deriving Equation $\ref{4-12}$. In addition, the structural organization of the solution becomes concentration dependent, so that the entropy depends on concentration in a more complicated way than is implied by the concentration analog of Equation $\ref{4-12}$.
Chemical Reactions and Mixing
We characterize the tendency for a chemical reaction A → B to occur at constant temperature and pressure by the value of its standard Gibbs energy change ΔG°. If this quantity is negative, we know that the reaction will take place spontaneously. However, have you ever wondered why it is that substance A is not completely transformed into B if the latter is thermodynamically more stable? The answer is that if the reaction takes place in a single phase (gas or liquid), something else is going on: A and B are mixing together, and this process creates its own Gibbs energy change $ΔG_{mixing}$. For a simple binary mixture of A and B (without any reaction), the changes in $S$ and $G$ can be represented by these simple plots:
We will not try to prove it here, but it turns out that no matter how much lower the Gibbs energy of the products compared to that of the reactants, the free energy of the system can always be reduced even more if some of the reactants remain in the solution to contribute a ΔGmixing term. This is the reason that a plot of G as a function of the composition of such a system has a minimum at some point short of complete conversion.
Diffusion refers to the transport of a substance across a concentration gradient. The direction is always toward the region of lower concentration. You should now see that from a thermodynamic standpoint, these processes are identical in that they both represent the spontaneous "escape" of molecules from a region of higher concentration (lower entropy, higher Gibbs energy) to a region of lower concentration.
Activity and Standard State of the Solute
Instead of complicating G° by trying to correct for all of these effects, chemists have chosen to retain its simple form by making a single small change in the form of $\ref{4-13}$:
$G = G^o + RT \ln a \label{4-14}$
This equation is guaranteed to work, because a, the activity of the solute, is its thermodynamically effective concentration. The relation between the activity and the concentration is given by
$a = \gamma c \label{4-15}$
where $\gamma$ is the activity coefficient. As the solution becomes more dilute, the activity coefficient approaches unity:
$\lim _{c \rightarrow 0} \gamma =1 \label{4.16}$
The price we pay for this simplicity is that the relation between the concentration and the activity at higher concentrations can be quite complicated, and must be determined experimentally for every different solution.
The question of what standard state we choose for the solute (that is, at what concentration is G° defined, and in what units is it expressed?) is one that you will wish you had never asked. We might be tempted to use a concentration of 1 molar, but a solution this concentrated would be subject to all kinds of intermolecular interaction effects, and would not make a very practical standard state. These effects could be eliminated by going to the opposite extreme of an “infinitely dilute” solution, but by Equation $\ref{4-12}$ this would imply a free energy of minus infinity for the solute, which would be awkward. Chemists have therefore agreed to define the standard state of a solute as one in which the concentration is 1 molar, but all solute-solute interactions are magically switched off, so that $\gamma$ is effectively unity. Since this is impossible, no solution corresponding to this standard state can actually exist, but this turns out to be only a small drawback, and seems to be the best compromise between convenience, utility, and reality. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.05%3A_Thermodynamics_of_Mixing_and_Dilution.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in italic type.
• As a homogeneous chemical reaction proceeds, the Gibbs energies of the reactants become more negative and those of the products more positive as the composition of the system changes.
• The total Gibbs energy of the system (reactants + products) always becomes more negative as the reaction proceeds. Eventually it reaches a minimum value at a system composition that defines the equilibrium composition of the system, after which time no further net change will occur.
• The equilibrium constant for the reaction is determined the standard Gibbs energy change:
ΔG° = -RT ln Kp
• The sign of the temperature dependence of the equilibrium constant is governed by the sign of ΔH°. This is the basis of the Le Chatelier Principle.
• The Gibbs energies of solid and liquid components are constants that do not change with composition. Thus in heterogeneous reactions such as phase changes, the total Gibbs energy does not pass through a minimum and when the system is not at equilibrium only all-products or all-reactants will be stable.
• Two reactions are coupled when the product of one reaction is consumed in the other. If ΔG° for the first reaction is positive, the overall process can still be spontaneous if ΔG° for the second reaction is sufficiently negative— in which case the second reaction is said to "drive" the first reaction.
Under conditions of constant temperature and pressure, chemical change will tend to occur in whatever direction leads to a decrease in the value of the Gibbs energy. In this lesson we will see how G varies with the composition of the system as reactants change into products. When G falls as far as it can, all net change comes to a stop. The equilibrium composition of the mixture is determined by ΔG° which also defines the equilibrium constant K.
The Road to Equilibrium is Down the Gibbs Energy Hill
This means, of course, that if the total Gibbs energy $G$ of a mixture of reactants and products goes through a minimum value as the composition changes, then all net change will cease— the reaction system will be in a state of chemical equilibrium. You will recall that the relative concentrations of reactants and products in the equilibrium state is expressed by the equilibrium constant. In this lesson we will examine the relation between the Gibbs energy change for a reaction and the equilibrium constant.
To keep things as simple as possible, we will consider a homogeneous chemical reaction of the form
$A + B \rightleftharpoons C + D$
in which all components are gases at the temperature of interest. If the sum of the standard Gibbs energies of the products is less than that of the reactants, ΔG° for the reaction will be negative and the reaction will proceed to the right. But how far? If the reactants are completely transformed into products, the equilibrium constant would be infinity. The equilibrium constants we actually observe all have finite values, implying that even if the products have a lower Gibbs energy than the reactants, some of the latter will always remain when the process comes to equilibrium.
A homogeneous reaction is one in which everything takes place in a single gas or liquid phase.
To understand how equilibrium constants relate to ΔG° values, assume that all of the reactants are gases, so that the Gibbs energy of gas A, for example, is given at all times by
$G_A = G_A^° + RT \ln P_A \label{5-1}$
The Gibbs energy change for the reaction is sum of the Gibbs energies of the products, minus the sum of Gibbs energies of the reactants:
$\Delta G = \underbrace{G_C + G_D}_{\text{products}} \underbrace{– G_A – G_B}_{\text{reactants}} \label{5-2}$
Using Equation $\ref{5-1}$ to expand each term on the right of Equation \ref{5-2}, we have
$\Delta G = (G^°_C + RT \ln P_C) + (G^°_D + RT \ln P_D) – (G^°_B + RT \ln P_B) – (G^°_A + RT \ln P+A) \label{5-3}$
We can now express the $G^°$ terms collectively as $\Delta G^°$, and combine the logarithmic pressure terms into a single fraction
$\Delta G = \Delta G° + RT \ln \left( \dfrac{P_CP_D}{P_AP_B} \right) \label{5-4}$
which is more conveniently expressed in terms of the reaction quotient $Q$.
$\Delta{G} = \Delta G^° + RT \ln Q \label{5-5}$
The Gibbs energy $G$ is a quantity that becomes more negative during the course of any natural process. Thus as a chemical reaction takes place, $G$ only falls and will never become more positive. Eventually a point is reached where any further transformation of reactants into products would cause $G$ to increase. At this point $G$ is at a minimum (see below), and no further net change can take place; the reaction is then at equilibrium.
Although Equations \ref{5-1}-\ref{5-5} are strictly correct only for perfect gases, we will see later that equations of similar form can be applied to many liquid solutions by substituting concentrations for pressures.
Example $1$: Dissociation of Dinitrogen Tetroxide
Consider the gas-phase dissociation reaction
$\ce{N_2O_4 \rightarrow 2 NO_2 } \nonumber$
which is a simple example of the Gibbs energy relationships in a homogeneous reaction.
The Gibbs energy of 1 mole of N2O4 (1) is smaller than that of 2 moles of NO2 (2) by 5.3 kJ; thus $\Delta G^o = +5.3\, \text{kJ}$ for the complete transformation of reactants into products. The straight diagonal line shows the Gibbs energy of all possible compositions if the two gases were prevented from mixing. The red curved line show the Gibbs energy of the actual reaction mixture. This passes through a minimum at (3) where 0.814 mol of $N_2O_4$ are in equilibrium with 0.372 mol of $NO_2$. The difference (4) corresponds to the Gibbs energy of mixing of reactants and products which always results in an equilibrium mixture whose Gibbs energy is lower than that of either pure reactants or pure products. Thus some amount of reaction will occur even if ΔG° for the process is positive.
What’s the difference between ΔG and ΔG°?
It’s very important to be aware of this distinction; that little ° symbol makes a world of difference! First, the standard Gibbs energy change ΔG° has a single value for a particular reaction at a given temperature and pressure; this is the difference
$\sum G^°_{f} (\text{products}) – \sum G^°_{f}(\text{reactants})$
that are tabulated in thermodynamic tables. It corresponds to the Gibbs energy change for a process that never really happens: the complete transformation of pure N2O4 into pure NO2 at a constant pressure of 1 atm.
The other quantity $\Delta G$, defined by Equation $\ref{5-5}$, represents the total Gibbs energies of all substances in the reaction mixture at any particular system composition. In contrast to $\Delta G^°$ which is a constant for a given reaction, $\Delta G$ varies continuously as the composition changes, finally reaching zero at equilibrium. $\Delta G$ is the “distance” (in Gibbs energy) from the equilibrium state of a given reaction. Thus for the limiting cases of pure $\ce{N_2O_4}$ or $\ce{NO_2}$ (as far from the equilibrium state as the system can be!),
$Q = \dfrac{[NO_2]^2}{[N_2O_4]} = \pm\infty$
which makes the logarithm in Equation $\ref{5-5}$, and thus the value of $\Delta G$, approach the same asymptotic limits (1) or (2). As the reaction proceeds in the appropriate direction $\Delta G$ approaches zero; once there (3), the system is at its equilibrium composition and no further net change will occur.
Example $2$: Isomerization of Butane
The standard molar Gibbs energy change for this very simple reaction is –2.26 kJ, but mixing of the unreacted butane with the product brings the Gibbs energy of the equilibrium mixture down to about –3.1 kJ mol–1 at the equilibrium composition corresponding to 77 percent conversion.
Notice particularly that
• The sum of the Gibbs energies of the two gases (n-butane and iso-butane) separately varies linearly with the composition of the mixture (red line ).
• The green curve adds the Gibbs energy of mixing to the above sum; its minimum defines the equilibrium composition.
• As the composition approaches the equilibrium value , $ΔG$ (which denotes how much farther the Gibbs energy of the system can fall) approaches zero.
The detailed calculations that lead to the values shown above can be found here.
Why reactions lead to mixtures of reactants and products
We are now in a position to answer the question posed earlier: if ΔG° for a reaction is negative, meaning that the Gibbs energies of the products are more negative than those of the reactants, why will some of the latter remain after equilibrium is reached? The answer is that no matter how low the Gibbs energy of the products, the Gibbs energy of the system can be reduced even more by allowing some of the products to be "contaminated" (i.e., diluted) by some reactants. Owing to the entropy associated with mixing of reactants and products, no homogeneous reaction will be 100% complete. An interesting corollary of this is that any reaction for which a balanced chemical equation can be written can in principle take place to some extent, however minute that might be.
Gibbs energies of mixing of products with reactants tend to be rather small, so for reactions having ΔG° values that are highly negative or positive (±20 kJ mol–1, say), the equilibrium mixture will, for all practical purposes, be either [almost] "pure" reactants or products.
The Equilibrium Constant
Now let us return to Equation $\ref{5-5}$ which we reproduce here:
$\Delta{G} = \Delta{G^°} + RT \ln Q$
As the reaction approaches equilibrium, $\Delta G$ becomes less negative and finally reaches zero. At equilibrium $\Delta{G} = 0$ and $Q = K$, so we can write (must know this!)
$\Delta{G^°} = –RT \ln K_p \label{5-6}$
in which $K_p$, the equilibrium constant expressed in pressure units, is the special value of $Q$ that corresponds to the equilibrium composition.
This equation is one of the most important in chemistry because it relates the equilibrium composition of a chemical reaction system to measurable physical properties of the reactants and products. If you know the entropies and the enthalpies of formation of a set of substances, you can predict the equilibrium constant of any reaction involving these substances without the need to know anything about the mechanism of the reaction.
Instead of writing Equation $\ref{5-6}$ in terms of Kp, we can use any of the other forms of the equilibrium constant such as Kc (concentrations), Kx (mole fractions), Kn(numbers of moles), etc. Remember, however, that for ionic solutions especially, only the Ka, in which activities are used, will be strictly valid.
It is often useful to solve Equation $\ref{5-6}$ for the equilibrium constant, yielding
$K = \exp {\left ( {-\Delta G \over RT} \right )} \label{5-7}$
This relation is most conveniently plotted against the logarithm of $K$ as shown in Figure $3$, where it can be represented as a straight line that passes through the point (0,0).
Example $3$
Calculate the equilibrium constant for the reaction from the following thermodynamic data:
$\ce{H^{+}(aq) + OH^{–}(aq) <=> H_2O(l)} \nonumber$
$H^+(aq)$
$OH^–(aq)$
$H_2O(l)$
ΔHf°, kJ mol–1
0
–230.0
–285.8
S°, J K–1 mol–1
0*
–10.9
70.0
* Note that the standard entropy of the hydrogen ion is zero by definition. This reflects the fact that it is impossible to carry out thermodynamic studies on a single charged species. All ionic entropies are relative to that of $\ce{H^{+}(aq)}$, which explains why some values (as for aqueous hydroxide ion) are negative.
Solution
From the above data, we can evaluate the following quantities:
\begin{align*} \Delta{H}^o &= \sum \Delta H^o_{f}(\text{products}) - \sum \Delta H^o_{f}(\text{reactants}) \[4pt] &= (–285.8) - (-230) \[4pt] &= –55.8\, kJ \; mol^{-1} \end{align*}
\begin{align*}\Delta{S}^o &= \sum \Delta S^o (\text{products}) - \sum \Delta S° (\text{reactants}) \[4pt] &= (70.0) – (–10.9) \[4pt] &= +80.8\, J \; K^{-1}\; mol^{-1} \end{align*}
The value of $\Delta{G}°$ at 298 K is
\begin{align*} \Delta H^o – T\Delta S^o &= (–55800) – (298)(80.8) \[4pt] &= –79900\, J\, mol^{–1} \end{align*}
From Equation $\ref{5-7}$ we have
\begin{align*} K &= \exp\left(\dfrac{–79900}{8.314 \times 298}\right) \[4pt] &= e^{32.2} = 1.01 \times 10^{–14} \end{align*}
Equilibrium and Temperature
We have already discussed how changing the temperature will increase or decrease the tendency for a process to take place, depending on the sign of ΔS°. This relation can be developed formally by differentiating the relation
$\Delta G^° = \Delta H^° – T\Delta S^° \label{5-8}$
with respect to the temperature:
$\dfrac{d(-\Delta G^°)}{dT} = -\Delta S^° \label{5-9}$
Hence, the sign of the entropy change determines whether the reaction becomes more or less allowed as the temperature increases.
We often want to know how a change in the temperature will affect the value of an equilibrium constant whose value is known at some fixed temperature. Suppose that the equilibrium constant has the value $K_1$ at temperature $T_1$ and we wish to estimate $K_2$ at temperature $T_2$. Expanding Equation $\ref{5-7}$ in terms of $\Delta H^°$ and $\Delta S^°$, we obtain
$–RT_1 \ln K_1 = \Delta H^ ° – T_1 \Delta S^°$
and
$–RT_2 \ln K_2 = \Delta H ^° – T_2 \Delta S^°$
Dividing both sides by RT and subtracting, we obtain
$\ln K_1 - \ln K_2 = - \left( \dfrac{\Delta H^°}{RT_1} -\dfrac{\Delta H^°}{RT_2} \right) \label{5-10}$
Which is most conveniently expressed as the ratio
$\ln \dfrac{K_1}{K_2} = - \dfrac{\Delta H^°}{R} \left( \dfrac{1}{T_1} -\dfrac{1}{T_2} \right) \label{5-11}$
This is its theoretical foundation of Le Chatelier's Principle with respect to the effect of the temperature on equilibrium:
• if the reaction is exothermic $\Delta H^° < 0$, then increasing temperature will make the second exponential term smaller and $K$ will decrease. The equilibrium will then “shift to the left”.
• If $\Delta H^° > 0$, then increasing T will make the exponent less negative and $K$ will increase and the equilibrium will “shift to the right”.
This is an extremely important relationship, but not just because of its use in calculating the temperature dependence of an equilibrium constant. Even more important is its application in the “reverse” direction to experimentally determine ΔH° from two values of the equilibrium constant measured at different temperatures. Direct calorimetric determinations of heats of reaction are not easy to make; relatively few chemists have the equipment and experience required for this rather exacting task. Measurement of an equilibrium constant is generally much easier, and often well within the capabilities of anyone who has had an introductory Chemistry course. Once the value of ΔH° is determined it can be combined with the Gibbs energy change (from a single observation of K, through Equation $\ref{5-7}$) to allow ΔS° to be calculated through Equation $\ref{5-9}$.
Equilibrium Without Mixing: it's all or nothing
You should now understand that for homogeneous reactions (those that take place entirely in the gas phase or in solution) the equilibrium composition will never be 100% products, no matter how much lower their Gibbs energy relative to the reactants. As was summarized in the N2O4-dissociation example discussed previously. This is due to "dilution" of the products by the reactants. In heterogeneous reactions (those which involve more than one phase) this dilution, and the effects that flow from it, may not be possible.
A particularly simple but important type of a heterogeneous process is phase change. Consider, for example, an equilibrium mixture of ice and liquid water. The concentration of H2O in each phase is dependent only on the density of the phase; there is no way that ice can be “diluted” with water, or vice versa. This means that at all temperatures other than the freezing point, the lowest Gibbs energy state will be that corresponding to pure ice or pure liquid. Only at the freezing point, where the Gibbs energies of water and ice are identical, can both phases coexist, and they may do so in any proportion.
Gibbs energy of the ice-water system
Only at 0°C can ice and liquid water coexist in any proportion. Note that in contrast to the homogeneous N2O4 example, there is no Gibbs energy minimum at intermediate compositions.
Coupled Reactions
Two reactions are said to be coupled when the product of one of them is the reactant in the other:
$A \rightarrow B \nonumber$
and
$B \rightarrow C \nonumber$
If the standard Gibbs energy of the first reaction is positive but that of the second reaction is sufficiently negative, then for the overall process will be negative and we say that the first reaction is “driven” by the second one. This, of course, is just another way of describing an effect that you already know as the Le Chatelier principle: the removal of substance B by the second reaction causes the equilibrium of the first to “shift to the right”. Similarly, the equilibrium constant of the overall reaction is the product of the equilibrium constants of the two steps.
1 Cu2S(s) → 2 Cu(s) + S(s) ΔG° = + 86.2 kJ ΔH° = + 76.3 kJ
2 S(s) + O2(g)→ SO2(g) ΔG° = –300.1 kJ ΔH° = + 296.8 kJ
3 Cu2S(s)→ 2 Cu(s) + SO2(g) ΔG° = –213.9 kJ ΔH° = – 217.3 kJ
In the above example, reaction 1 is the first step in obtaining metallic copper from one of its principal ores. This reaction is endothermic and it has a positive Gibbs energy change, so it will not proceed spontaneously at any temperature. If Cu2S is heated in the air, however, the sulfur is removed as rapidly as it is formed by oxidation in the highly spontaneous reaction 2, which supplies the Gibbs energy required to drive 1. The combined process, known as roasting, is of considerable industrial importance and is one of a large class of processes employed for winning metals from their ores. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.06%3A_Free_energy_and_Equilibrium.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in bold
• What do all of the colligative properties of solutions have in common in terms of entropy?
• At the melting and boiling points, the two phases in equilibrium have equal numbers of energetically accessible microstates. How does the addition of a nonvolatile solute upset this balance in the two cases?
• Describe the effect of increasing the hydrostatic pressure on a liquid has on its vapor pressure, and suggest a reason for this.
• Under what conditions does osmotic flow occur? What is the fundamental driving force? What is the definition of osmotic pressure?
• Explain the critical role of temperature in the extraction of metals from their oxide ores.
• Describe the bioenergetic cycle of free energy in terms of the roles of glucose, electron-acceptors, ATP, and photosynthesis.
• Define aerobic and non-aerobic oxidation, and explain why they differ in efficiency.
• Sketch a simple proton-free energy diagram showing why an acid reacts with a base.
• Sketch a simple electron-free energy diagram showing why a metal reacts with an oxidizing agent that is below it in the electromotive series.
• What thermodynamic factors are involved in the spontaneous contraction of an elastic material such as a rubber band?
Thermodynamics may appear at first to be a rather esoteric subject, but when you think about it, almost every chemical (and biological) process is governed by changes in entropy and free energy. Examples such as those given below should help you connect these concepts with the real world.
Colligative properties of solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare, connected to.) The key role of the solvent concentration is obscured by the expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered in the unit on solutions. Our purpose here is to offer a more complete explanation of why these phenomena occur.
The conventional explanation is that dilution of a liquid by a non-volatile solute reduces the vapor pressure or "escaping tendency" of the liquid in that phase, leading to a net transport of material into that phase. Equilibrium can then be re-established by adjusting the temperature of applying a hydrostatic pressure to the solution. The stable phase at any temperature will be the one having the lowest free energy, and from which molecules have the smallest escaping tendency.
A more fundamental approach is to recall that dilution of a liquid creates uncountable numbers of new microstates, increasing the density of quantum states in the solution compared to that in the pure liquid. To the extent that these new states are thermally accessible, they will become populated at the expense of some of the microstates of the other phase. Equilibrium between two phases can be restored by adjusting the temperature or pressure so that equal numbers of microstates are occupied in each phase.
Boiling point elevation and freezing point depression
These effects are readily understood in terms of the schematic diagrams shown below. The red shading indicates the temperatures required to make equal numbers of microstates thermally accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases. The shifts of the boiling- and and freezing points in opposite directions reflects the higher energies of vapor microstates and lower energies of solid microstates in relation to those of the liquid. Note that these diagrams are purely schematic and nowhere near to scale!
As mentioned previously, the more conventional explanation of bp elevation and fp depression is given in terms of vapor pressures. The latter are measures of the free energies of molecules in a phase, and the relationships are best understood by showing plots of –TΔS° for each phase in terms of the temperature:
Melting and boiling points of a pure substance.
The stable phase at any temperature will be the one having the lowest free energy, and from which molecules have the smallest escaping tendency.
Melting- and boiling points of a solvent containing a non-volatile solute.
The solute "dilutes" the solvent, reducing its free energy (purple line) and shifting the transition temperatures in opposite directions.
The pressure on a liquid affects its volatility
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised. The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25° C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of the quantized energy states in the liquid phase.
Why pressure affects the volatility of a liquid. Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
Osmotic flow and osmotic pressure
Molecules in a fluid are always subject to random thermal motions. If a given kind of molecule is present in greater concentration in one region of a fluid, the effect of these random motions will be to produce a net migration into the region of lower concentration; this migration is known as thermal diffusion.
Osmotic flow (commonly known simply as osmosis) occurs when molecules of a solvent diffuse through a membrane that is permeable only to those molecules, and into a solution in which the solvent is "diluted" by the presence of solute molecules. In the simplest example, there is pure solvent on the left side of the membrane; on the other side the same solvent contains a solute whose molecules are too large to pass through the membrane. Because the solvent concentration on the right side will always be smaller than that on the left, osmotic flow will continue indefinitely if the right side can accommodate the increased volume.
If, however, the liquid on the right side of the membrane is placed under hydrostatic pressure, the driving force for osmotic flow will diminish. If the pressure is raised sufficiently high, osmotic flow will come to a halt; the system is then said to be in osmotic equilibrium. The pressure required to bring about osmotic equilibrium (and thus to stop osmotic flow) is known as the osmotic pressure, commonly denoted by Π (Greek capital pi).
Osmotic equilibrium, like any kind of equilibrium, occurs when the free energies (that is, the escaping tendencies) of the diffusible molecules are the same on the two sides of the membrane. The free energy on the right side, initially lower due to the lower solvent concentration, is raised by the hydrostatic pressure, making it identical with that of the pure liquid on the left.
From the standpoint of microscopic energy states, the effect of the applied pressure is to slightly increase the spacing of solvent energy states on right side of the membrane so as to equalize the number of accessible states on the two sides, as shown here in a very schematic way.
Extraction of Metals from their Oxides
Since ancient times, the recovery of metals from their ores has been one of the most important applications of chemistry to civilization and culture. The oldest, and still the most common smelting process for oxide ores involves heating them in the presence of carbon. Originally, charcoal was used, but industrial-scale smelting uses coke, a crude form of carbon prepared by pyrolysis (heating) of coal. The basic reactions are:
$MO + C \rightleftharpoons M + CO \label{2.1}$
$MO + ½ O_2 \rightleftharpoons M + ½ CO_2 \label{2.2}$
$MO + CO \rightleftharpoons M + CO_2 \label{2.3}$
Each of these can be regarded as a pair of coupled reactions in which the metal M and the carbon are effectively competing for the oxygen atom. Using reaction $\ref{2.1}$ as an example, it can be broken down into the following two parts:
$MO \rightleftharpoons M + ½ O_2 \;\;\; ΔG^o > 0 \label{2.4}$
$C + ½ O_2 \rightleftharpoons CO \;\;\; ΔG^o < 0 \label{2.5}$
At ordinary environmental temperatures, reaction $\ref{2.4}$ is always spontaneous in the reverse direction (that is why ores form in the first place!), so ΔG° of Reaction $\ref{2.4}$ will be positive. ΔG° for reaction $\ref{2.5}$ is always negative, but at low temperatures it will not be sufficiently negative to drive $\ref{2.4}$.
The smelting process depends on the different ways in which the free energies of reactions like $\ref{2.4}$ and $\ref{2.4}$ vary with the temperature. This temperature dependence is almost entirely dominated by the TΔS° term in the Gibbs function, and thus by the entropy change. The latter depends mainly on Δng, the change in the number of moles of gas in the reaction. Removal of oxygen from the ore is always accompanied by a large increase in the system volume so ΔS for this step is always positive and the reaction becomes more spontaneous at higher temperatures. The temperature dependences of the reactions that take up oxygen vary, however (Table 23.6.X).
Table 23.6.X: The temperature dependences of the reactions
Reaction
Δng
dG°)/dT
C + ½ O2 → CO
0.5
<0>
C + O2 → CO2
0
0
CO + ½ O2 →CO2
–0.5
>0
A plot of the temperature dependences of the free energies of these reactions, superimposed on similar plots for the oxygen removal reactions $\ref{2.4}$ is called an Ellingham diagram. For a given oxide MO to be smeltable, the temperature must be high enough that reaction $\ref{2.4}$ falls below that of at least one of the oxygen-consuming reactions. The slopes of the lines on this diagram are determined by the sign of the entropy change.
Examination of the Ellingham diagram shown above illustrates why the metals known to the ancients were mainly those such as copper and lead, which can be obtained by smelting at the relatively low temperatures that were obtainable by the methods available at the time in which a charcoal fire supplied both the heat and the carbon. Thus the bronze age preceded the iron age; the latter had to await the development of technology capable of producing higher temperatures, such as the blast furnace. Smelting of aluminum oxide by carbon requires temperatures too high to be practical; commercial production of aluminum is accomplished by electrolysis of the molten ore.
Bioenergetics
Many of the reactions that take place in living organisms require a source of free energy to drive them. The immediate source of this energy in heterotrophic organisms, which include animals, fungi, and most bacteria, is the sugar glucose. Oxidation of glucose to carbon dioxide and water is accompanied by a large negative free energy change
$\ce{C6H12O6 + 6 O2 \rightarrow 6 CO2 + 6 H2O} \quad ΔG^o = – 2,880 \,kJ\, mol^{–1} \label{6-1}$
Of course it would not do to simply “burn” the glucose in the normal way; the energy change would be wasted as heat, and rather too quickly for the well-being of the organism! Effective utilization of this free energy requires a means of capturing it from the glucose and then releasing it in small amounts when and where it is needed. This is accomplished by breaking down the glucose in a series of a dozen or more steps in which the energy liberated in each stage is captured by an “energy carrier” molecule, of which the most important is adenosine diphosphate, known as ADP. At each step in the breakdown of glucose, an ADP molecule reacts with inorganic phosphate (denoted by Pi) and changes into ATP:
$ADP + P_i \rightarrow ATP$
$ΔG^° = +30 \;kJ\; mol^{–1} \label{(6-2}$
The 30 kJ mol–1 of free energy stored in each ATP molecule is released when the molecule travels to a site where it is needed and loses one of its phosphate groups, yielding inorganic phosphate and ADP, which eventually finds its way back the site of glucose metabolism for recycling back into ATP. The complete breakdown of one molecule of glucose is coupled with the production of 38 molecules of ATP according to the overall reaction
$\ce{C6H12O6 + 6 O2 + 38 Pi + 38 ADP 38 ATP + 6CO2 + 44 H2O} \label{6-3}$
For each mole of glucose metabolized, 38 × (30 kJ) = 1140 kJ of free energy is captured as ATP, representing an energy efficiency of 1140/2880 = 0.4. That is, 40% of the free energy obtainable from the oxidation of glucose is made available to drive other metabolic processes. The rest is liberated as heat.
Where does the glucose come from? Animals obtain their glucose from their food, especially cellulose and starches that, like glucose, have the empirical formula {CH2O}. Animals obtain this food by eating plants or other animals. Ultimately, all food comes from plants, most of which are able to make their own glucose from CO2and H2O through the process of photosynthesis. This is just the reverse of Eq 40 in which the free energy is supplied by the quanta of light absorbed by chlorophyll and other photosynthetic pigments.
This describes aerobic respiration, which evolved after the development of photosynthetic life on Earth began to raise the concentration of atmospheric oxygen. Oxygen is a poison to most life processes at the cellular level, and it is believed that aerobic respiration developed as a means to protect organisms from this peril. Those that did not adapt have literally “gone underground” and constitute the more primitive anaerobic bacteria.
The function of oxygen in respiration is to serve as an acceptor of the electrons that glucose loses when it undergoes oxidation. Other electron acceptors can fulfill the same function when oxygen is not available, but none yields nearly as much free energy. For example, if oxygen cannot be supplied to mammalian muscle cells as rapidly as it is needed, they switch over to an anaerobic process yielding lactic acid instead of CO2:
$\ce{C6H12O6 + 2 ADP → 2 CH3CH(OH)COOH} \quad ΔG^o = –218\, kJ\, mol^{–1} \label{6-4}$
In this process, only (2 × 30 kJ) = 60 kJ of free energy is captured, so the efficiency is only 28% on the basis of this reaction, and it is even lower in relation to glucose. In “aerobic” exercising, one tries to maintain sufficient lung capacity and cardiac output to supply oxygen to muscle cells at a rate that promotes the aerobic pathway.
Respiration and many other metabolic processes involve electron-transfer reactions.
Acid-base reactions: the fall of the proton
According to the widely useful Brønsted-Lowry concept, an acid is a proton donor and a base is a proton acceptor. In 1953, Gurney showed how this idea could be made even more useful by placing acid-base conjugate pairs on a proton-free energy scale.
In this view, acids are proton sources and bases are proton sinks. Protons fall spontaneously from acids to fill sinks in which the proton free energy levels are lower. The pH is a measure of the average proton free energy in the solution; when this quantity is the same as the proton free energy level of a conjugate pair, the two species are present in equal concentrations (this corresponds, of course to the equality of pH and pKa in the conventional theory.)
The proton-free energy concept is commonly employed in aquatic environmental chemistry in which multiple acid-base systems must be dealt with on a semi-quantitative bases. It is, however, admirably adapted to any presentation of acid-base chemistry, even at the first-year college level, and it seems a shame that it never seems to have made its way into the ordinary curriculum.
Oxidation-reduction: the fall of the electron
Oxidation-reduction reactions proceed in a direction that allows the electron to “fall” (in free energy) from a “source” to a “sink”. Later on when you study electrochemistry you will see how this free energy can manifest itself as an electrical voltage and be extracted from the system as electrical work.
Electron-free energy levels and the electromotive series
You may already have seen an electromotive force table that shows the relative tendencies of different reducing agents to donate electrons. The same information can be presented in another way that relates electron-donating (reducing) power to the "fall" in free energy that electrons undergo when they are transferred to an oxidizing agent.
Any available sink on the right side will tend to drain electrons from a source above it. For example, immersion of metallic zinc in a solution of CuSO4 allow electrons to "fall" from the high free energy they possess in Zn down to the lower free energy level in the newly-introduced Cu2+ ions. This will result in the reduction of Cu2+ to metallic copper and the oxidation of the zinc (red arrows.)
Similarly, addition of chlorine to water will introduce a new electron sink (Cl2) that lies below the free energy of the accessible electrons in H2O, draining them away and producing O2 and Cl (blue arrows.) Note especially the positions of the H2/ H+ and H2O/O2,H+ couples on this chart, as they define the range of E°s for substances that will not decompose water. The zero for G° is arbitrarily set at the electron activity at which the H2/H+ couple is at equilibrium; this corresponds to E° = 0 volts on the ordinary electromotive scale.
Coupled redox reactions are central to bioenergetics
$\{CH_2O\} → CO_2 + H_2O \label{6-5}$
In the above chart, the zero for G° is arbitrarily set at the electron free energy required to maintain the H2/H+ couple is at equilibrium; this corresponds to E° = 0 volts on the ordinary electromotive scale.
Organisms that live in environments where oxygen is lacking, such as marshes, muddy soils, and the intestinal tracts of animals, must utilize other electron acceptors to extract free energy from carbohydrate. A wide variety of inorganic ions such as sulfate and nitrate, as well as other carbon compounds can serve as electron acceptors, yielding the gaseous products like H2S, NH3 and CH4 which are commonly noticed in such locations. From the location of these acceptors on the scale, it is apparent that the amount of energy they can extract from a given quantity of carbohydrate is much less than for O2. One reason that aerobic organisms have dominated the earth is believed to be the much greater energy-efficiency of oxygen as an electron acceptor.
What did organisms use for food before there was a widespread supply of carbohydrate in the world? Any of the electron sources near the upper left of the table can in theory serve this function, although at reduced energy efficiency. As a matter of fact, there are still a number of these autotrophic bacteria around whose “food” is CH4 , CH3OH, FeCO3, and even H2!
Thermodynamics of Rubber Bands
Rubber is composed of random-length chains of polymerized isoprene molecules. The poly(isoprene) chains are held together partly by weak intermolecular forces, but are joined at irregular intervals by covalent disulfide bonds so as to form a network. The intermolecular forces between the chain fragments tend to curl them up, but application of a tensile force can cause them to elongate. The disulfide cross-links prevent the chains from slipping apart from one another, thus maintaining the physical integrity of the material. Without this cross-linking, the polymer chains would behave more like a pile of spaghetti.
The ability of rubber bands and other elastic substances to undergo a change in physical dimensions in response to a change in the applied stretching force is subject to the same laws of thermodynamics as any other physical process. You can investigate this for yourself.
Example $1$
Hold a rubber band (the thicker the better) against your upper lip, and notice how the temperature changes when the band is stretched, and then again when it is allowed to contract.
1. Use the results of this observation to determine the signs of ΔH, ΔG and ΔS for the process $rubber_{stretched} \rightarrow rubber_{unstretched}$
2. How will the tendency of the stretched rubber to contract be changed if the temperature is raised?
Solution
1. Contraction is obviously a spontaneous process, so ΔG is negative. You will have observed that heat is given off when the band is stretched, meaning that contraction is endothermic, so ΔH > 0. Thus according to ΔG = ΔH – TΔS, ΔS for the contraction process must be positive.
2. Because ΔS > 0, contraction of the rubber becomes more spontaneous as the temperature is raised.
Why is Rubber Elastic?
When an ordinary material is placed under tension, the strain energy is taken up by bond distortions and is stored as electrostatic potential energy which rises very rapidly so as to greatly inhibit further elongation. In rubber-like polymers, this does not happen; the strain energy is instead stored as thermal (kinetic) energy.
Free polymer chains naturally tend to curl up in complex and ever-changing ways as thermal energy allows random bond rotations to take place. In a rubber-like material in its relaxed state, the portions of the polymer chains between cross-links are continually jumping between different randomly-coiled configurations.
When the rubber is stretched, the polymer segments straighten out as the applied force overcomes the weak dispersion force interactions that caused the strands to curl. Each chain segment is pulled into an almost-straight conformation, thus greatly reducing the quantity of thermal energy it can store. The excess thermal energy spreads into the material and is lost in the form of heat. When the rubber relaxes, the polymer strands curl up again and soak up thermal energy.
The spontaneous contraction of rubber is largely an entropy-driven process. The number of energetically-equivalent ways of distributing thermal energy amongst the nearly-linear polymer chains of the stretched state of rubber is insignificant compared to those available when the chains are curled up in random ways, so the un-stretched form of rubber is statistically the most likely one by overwhelming odds. As noted in part (b) of the above problem example, the gain in entropy when the rubber contracts drives ΔG more negative at higher temperatures. This means that a rubber band, held at constant tension in stretched state, will contract when it is heated.
This fact can be put to use in an interesting way. Replace the spokes of a bicycle wheel with rubber bands, and shine a heat lamp on one side of the wheel. The contraction of the heated bands will shift the wheel off-center, causing it to rotate. This rotation will continue indefinitely as long as the heat source is present. The device has become a heat engine whose working "fluid" is rubber!
This recalls the classic perpetual motion machine design in which a wheel is caused to rotate by continually-shifting unbalanced weights (the one depicted here uses hinged vials of mercury). That, as we saw, would violate the Second Law by producing work in a cyclic process without degrading heat to a lower temperature. The rubber-band heat engine avoids this pitfall by absorbing heat from an external source on one side of the wheel, and releasing it at a lower temperature on the unheated side. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.07%3A_Some_Applications_of_Entropy_and_Free_Energy.txt |
Learning Objectives
You are expected to be able to define and explain the significance of terms identified in bold.
energy within a system.
• The entropy of a substance increases with its molecular weight and complexity and with temperature. The entropy also increases as the pressure or concentration becomes smaller. Entropies of gases are much larger than those of condensed phases.
The Physical Meaning of Entropy
Entropy is a measure of the degree of spreading and sharing of thermal energy within a system. This “spreading and sharing” can be spreading of the thermal energy into a larger volume of space or its sharing amongst previously inaccessible microstates of the system. The following table shows how this concept applies to a number of common processes.
Table \(1\): Observations and Explanations in terms of Entropy changes
system and process source of entropy increase of system
A deck of cards is shuffled, or 100 coins, initially heads up, are randomly tossed. This has nothing to do with entropy because macro objects are unable to exchange thermal energy with the surroundings within the time scale of the process
Two identical blocks of copper, one at 20°C and the other at 40°C, are placed in contact. The cooler block contains more unoccupied microstates, so heat flows from the warmer block until equal numbers of microstates are populated in the two blocks.
A gas expands isothermally to twice its initial volume. A constant amount of thermal energy spreads over a larger volume of space
1 mole of water is heated by 1C°. The increased thermal energy makes additional microstates accessible. (The increase is by a factor of about 1020,000,000,000,000, 000,000,000.)
Equal volumes of two gases are allowed to mix. The effect is the same as allowing each gas to expand to twice its volume; the thermal energy in each is now spread over a larger volume.
One mole of dihydrogen, H2, is placed in a container and heated to 3000K. Some of the H2 dissociates to H because at this temperature there are more thermally accessible microstates in the 2 moles of H.
The above reaction mixture is cooled to 300K. The composition shifts back to virtually all H2because this molecule contains more thermally accessible microstates at low temperatures.
Entropy is an extensive quantity; that is, it is proportional to the quantity of matter in a system; thus 100 g of metallic copper has twice the entropy of 50 g at the same temperature. This makes sense because the larger piece of copper contains twice as many quantized energy levels able to contain the thermal energy.
Thermal energy is the portion of a molecule's energy that is proportional to its temperature, and thus relates to motion at the molecular scale. What kinds of molecular motions are possible? For monatomic molecules, there is only one: actual movement from one location to another, which we call translation. Since there are three directions in space, all molecules possess three modes of translational motion.
For polyatomic molecules, two additional kinds of motions are possible. One of these is rotation; a linear molecule such as CO2 in which the atoms are all laid out along the x-axis can rotate along the y- and z-axes, while molecules having less symmetry can rotate about all three axes. Thus linear molecules possess two modes of rotational motion, while non-linear ones have three rotational modes. Finally, molecules consisting of two or more atoms can undergo internal vibrations. For freely moving molecules in a gas, the number of vibrational modes or patterns depends on both the number of atoms and the shape of the molecule, and it increases rapidly as the molecule becomes more complicated.
The relative populations of the quantized translational, rotational and vibrational energy states of a typical diatomic molecule are depicted by the thickness of the lines in this schematic (not-to-scale!) diagram. The colored shading indicates the total thermal energy available at an arbitrary temperature. The numbers at the top show order-of-magnitude spacings between adjacent levels.
Notice that the spacing between the quantized translational levels is so minute that they can be considered nearly continuous. This means that at all temperatures, the thermal energy of a collection of molecules resides almost exclusively in translational microstates. At ordinary temperatures (around 25° C), most of the molecules are in their zero-level vibrational and rotational states (corresponding to the bottom-most bars in the diagram.) The prevalence of vibrational states is so overwhelming that we can effectively equate the thermal energy of molecules with translational motions alone.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics. A very important point to bear in mind is that the number of discrete microstates that can be populated by an arbitrary quantity of energy depends on the spacing of the states. As a very simple example, suppose that we have two molecules (depicted by the orange dots) in a system total available thermal energy is indicated by the yellow shading.
In the system with the more closely-spaced energy levels, there are three possible microstates, while in the one with the more widely-spaced levels, only two possibilities are available.
The spacing of molecular energy states becomes closer as the mass and number of bonds in the molecule increases, so we can generally say that the more complex the molecule, the greater the density of its energy states.
At the atomic and molecular level, all energy is quantized; each particle possesses discrete states of kinetic energy and is able to accept thermal energy only in packets whose values correspond to the energies of one or more of these states. Polyatomic molecules can store energy in rotational and vibrational motions, and all molecules (even monatomic ones) will possess translational kinetic energy (thermal energy) at all temperatures above absolute zero. The energy difference between adjacent translational states is so minute that translational kinetic energy can be regarded as continuous (non-quantized) for most practical purposes.
The number of ways in which thermal energy can be distributed amongst the allowed states within a collection of molecules is easily calculated from simple statistics, but we will confine ourselves to an example here. Suppose that we have a system consisting of three molecules and three quanta of energy to share among them. We can give all the kinetic energy to any one molecule, leaving the others with none, we can give two units to one molecule and one unit to another, or we can share out the energy equally and give one unit to each molecule. All told, there are ten possible ways of distributing three units of energy among three identical molecules as shown here:
Each of these ten possibilities represents a distinct microstate that will describe the system at any instant in time. Those microstates that possess identical distributions of energy among the accessible quantum levels (and differ only in which particular molecules occupy the levels) are known as configurations. Because all microstates are equally probable, the probability of any one configuration is proportional to the number of microstates that can produce it. Thus in the system shown above, the configuration labeled ii will be observed 60% of the time, while iii will occur only 10% of the time.
As the number of molecules and the number of quanta increases, the number of accessible microstates grows explosively; if 1000 quanta of energy are shared by 1000 molecules, the number of available microstates will be around 10600— a number that greatly exceeds the number of atoms in the observable universe! The number of possible configurations (as defined above) also increases, but in such a way as to greatly reduce the probability of all but the most probable configurations. Thus for a sample of a gas large enough to be observable under normal conditions, only a single configuration (energy distribution amongst the quantum states) need be considered; even the second-most-probable configuration can be neglected.
The bottom line: any collection of molecules large enough in numbers to have chemical significance will have its therrmal energy distributed over an unimaginably large number of microstates. The number of microstates increases exponentially as more energy states ("configurations" as defined above) become accessible owing to
• Addition of energy quanta (higher temperature),
• Increase in the number of molecules (resulting from dissociation, for example).
• the volume of the system increases (which decreases the spacing between energy states, allowing more of them to be populated at a given temperature.)
Using Entropy to Understand Spontaneous Processes
Heat Death
Energy is conserved; if you lift a book off the table, and let it fall, the total amount of energy in the world remains unchanged. All you have done is transferred it from the form in which it was stored within the glucose in your body to your muscles, and then to the book (that is, you did work on the book by moving it up against the earth’s gravitational field). After the book has fallen, this same quantity of energy exists as thermal energy (heat) in the book and table top.
What has changed, however, is the availability of this energy. Once the energy has spread into the huge number of thermal microstates in the warmed objects, the probability of its spontaneously (that is, by chance) becoming un-dispersed is essentially zero. Thus although the energy is still “there”, it is forever beyond utilization or recovery. The profundity of this conclusion was recognized around 1900, when it was first described at the “heat death” of the world. This refers to the fact that every spontaneous process (essentially every change that occurs) is accompanied by the “dilution” of energy. The obvious implication is that all of the molecular-level kinetic energy will be spread out completely, and nothing more will ever happen.
Gases Expansions
Everybody knows that a gas, if left to itself, will tend to expand and fill the volume within which it is confined completely and uniformly. What “drives” this expansion? At the simplest level it is clear that with more space available, random motions of the individual molecules will inevitably disperse them throughout the space. But as we mentioned above, the allowed energy states that molecules can occupy are spaced more closely in a larger volume than in a smaller one. The larger the volume available to the gas, the greater the number of microstates its thermal energy can occupy. Since all such states within the thermally accessible range of energies are equally probable, the expansion of the gas can be viewed as a consequence of the tendency of thermal energy to be spread and shared as widely as possible. Once this has happened, the probability that this sharing of energy will reverse itself (that is, that the gas will spontaneously contract) is so minute as to be unthinkable.
Imagine a gas initially confined to one half of a box (Figure \(4\)). The barrier is then removed so that it can expand into the full volume of the container. We know that the entropy of the gas will increase as the thermal energy of its molecules spreads into the enlarged space. In terms of the spreading of thermal energy, Figure \(5\) may be helpful. The tendency of a gas to expand is due to the more closely-spaced thermal energy states in the larger volume .
Mixing and dilution
Mixing and dilution really amount to the same thing, especially for idea gases. Replace the pair of containers shown above with one containing two kinds of molecules in the separate sections (Figure \(6\)). When we remove the barrier, the "red" and "blue" molecules will each expand into the space of the other. (Recall Dalton's Law that "each gas is a vacuum to the other gas".) However, notice that although each gas underwent an expansion, the overall process amounts to what we call "mixing".
What is true for gaseous molecules can, in principle, apply also to solute molecules dissolved in a solvent. But bear in mind that whereas the enthalpy associated with the expansion of a perfect gas is by definition zero, ΔH's of mixing of two liquids or of dissolving a solute in a solvent have finite values which may limit the miscibility of liquids or the solubility of a solute.
It's unfortunate the the simplified diagrams we are using to illustrate the greater numbers of energetically accessible microstates in an expanded gas or a mixture of gases fail to convey the immensity of this increase. Only by working through the statistical mathematics of these processes (beyond the scope of first-year Chemistry!) can one gain an appreciation of the magnitude of the probabilities of these spontaneous processes.
It turns out that when just one molecule of a second gas is inroduced into the container of another gas, an unimaginably huge number of new configurations becom available. This happens because the added molecule (indicated by the blue arrow in the diagram) can in principle replace any one of the old (red) ones, each case giving rise to a new microstate.
Heat Flow
ust as gases spontaneously change their volumes from “smaller-to-larger”, the flow of heat from a warmer body to a cooler one always operates in the direction “warmer-to-cooler” because this allows thermal energy to populate a larger number of energy microstates as new ones are made available by bringing the cooler body into contact with the warmer one; in effect, the thermal energy becomes more “diluted”. In this simplified schematic diagram, the "cold" and "hot" bodies differ in the numbers of translational microstates that are occupied, as indicated by the shading. When they are brought into thermal contact, a hugely greater number of microstates are created, as is indicated by their closer spacing in the rightmost section of the diagram, which represents the combined bodies in thermal equilibrium. The thermal energy in the initial two bodies fills these new microstates to a level (and thus, temperature) that is somewhere between those of the two original bodies.
Note that this explanation applies equally well to the case of two solids brought into thermal contact, or two the mixing of two fluids having different temperatures.
As you might expect, the increase in the amount of energy spreading and sharing is proportional to the amount of heat transferred q, but there is one other factor involved, and that is the temperature at which the transfer occurs. When a quantity of heat q passes into a system at temperature T, the degree of dilution of the thermal energy is given by
\[\dfrac{q}{T}\]
To understand why we have to divide by the temperature, consider the effect of very large and very small values of \(T\) in the denominator. If the body receiving the heat is initially at a very low temperature, relatively few thermal energy states are initially occupied, so the amount of energy spreading into vacant states can be very great. Conversely, if the temperature is initially large, more thermal energy is already spread around within it, and absorption of the additional energy will have a relatively small effect on the degree of thermal disorder within the body.
Chemical Reactions and Equilibrium
When a chemical reaction takes place, two kinds of changes relating to thermal energy are involved:
1. The ways that thermal energy can be stored within the reactants will generally be different from those for the products. For example, in the reaction H2→ 2 H, the reactant dihydrogen possesses vibrational and rotational energy states, while the atomic hydrogen in the product has translational states only— but the total number of translational states in two moles of H is twice as great as in one mole of H2. Because of their extremely close spacing, translational states are the only ones that really count at ordinary temperatures, so we can say that thermal energy can become twice as diluted (“spread out”) in the product than in the reactant. If this were the only factor to consider, then dissociation of dihydrogen would always be spontaneous and this molecule would not exist.
2. In order for this dissociation to occur, however, a quantity of thermal energy (heat) \(qU\) must be taken up from the surroundings in order to break the H–H bond. In other words, the ground state (the energy at which the manifold of energy states begins) is higher in \(H\), as indicated by the vertical displacement of the right half in each of the four panels below.
In Figure \(8\) are schematic representations of the translational energy levels of the two components H and H2 of the hydrogen dissociation reaction. The shading shows how the relative populations of occupied microstates vary with the temperature, causing the equilibrium composition to change in favor of the dissociation product.
The ability of energy to spread into the product molecules is constrained by the availability of sufficient thermal energy to produce these molecules. This is where the temperature comes in. At absolute zero the situation is very simple; no thermal energy is available to bring about dissociation, so the only component present will be dihydrogen.
• As the temperature increases, the number of populated energy states rises, as indicated by the shading in the diagram. At temperature T1, the number of populated states of H2 is greater than that of 2H, so some of the latter will be present in the equilibrium mixture, but only as the minority component.
• At some temperature T2 the numbers of populated states in the two components of the reaction system will be identical, so the equilibrium mixture will contain H2 and “2H” in equal amounts; that is, the mole ratio of H2/H will be 1:2.
• As the temperature rises to T3 and above, we see that the number of energy states that are thermally accessible in the product begins to exceed that for the reactant, thus favoring dissociation.
The result is exactly what the Le Chatelier Principle predicts: the equilibrium state for an endothermic reaction is shifted to the right at higher temperatures.
The following table generalizes these relations for the four sign-combinations of ΔH and ΔS. (Note that use of the standard ΔH° and ΔS° values in the example reactions is not strictly correct here, and can yield misleading results when used generally.)
> 0
This combustion reaction, like most such reactions, is spontaneous at all temperatures. The positive entropy change is due mainly to the greater mass of CO2 molecules compared to those of O2.
< 0
• ΔH° = –46.2 kJ
• ΔS° = –389 J K–1
• ΔG° = –16.4 kJ at 298 K
The decrease in moles of gas in the Haber ammonia synthesis drives the entropy change negative, making the reaction spontaneous only at low temperatures. Thus higher T, which speeds up the reaction, also reduces its extent.
> 0
• ΔH° = 55.3 kJ
• ΔS° = +176 J K–1
• ΔG° = +2.8 kJ at 298 K
Dissociation reactions are typically endothermic with positive entropy change, and are therefore spontaneous at high temperatures.Ultimately, all molecules decompose to their atoms at sufficiently high temperatures.
< 0
• ΔH° = 33.2 kJ
• ΔS° = –249 J K1
• ΔG° = +51.3 kJ at 298 K
This reaction is not spontaneous at any temperature, meaning that its reverse is always spontaneous. But because the reverse reaction is kinetically inhibited, NO2 can exist indefinitely at ordinary temperatures even though it is thermodynamically unstable.
Phase Changes
Everybody knows that the solid is the stable form of a substance at low temperatures, while the gaseous state prevails at high temperatures. Why should this be? The diagram in Figure \(9\) shows that
1. the density of energy states is smallest in the solid and greatest (much, much greater) in the gas, and
2. the ground states of the liquid and gas are offset from that of the previous state by the heats of fusion and vaporization, respectively.
Changes of phase involve exchange of energy with the surroundings (whose energy content relative to the system is indicated (with much exaggeration!) by the height of the yellow vertical bars in Figure \(13\). When solid and liquid are in equilibrium (middle section of diagram below), there is sufficient thermal energy (indicated by pink shading) to populate the energy states of both phases. If heat is allowed to flow into the surroundings, it is withdrawn selectively from the more abundantly populated levels of the liquid phase, causing the quantity of this phase to decrease in favor of the solid. The temperature remains constant as the heat of fusion is returned to the system in exact compensation for the heat lost to the surroundings. Finally, after the last trace of liquid has disappeared, the only states remaining are those of the solid. Any further withdrawal of heat results in a temperature drop as the states of the solid become depopulated.
Colligative Properties of Solutions
Vapor pressure lowering, boiling point elevation, freezing point depression and osmosis are well-known phenomena that occur when a non-volatile solute such as sugar or a salt is dissolved in a volatile solvent such as water. All these effects result from “dilution” of the solvent by the added solute, and because of this commonality they are referred to as colligative properties (Lat. co ligare, connected to.) The key role of the solvent concentration is obscured by the greatly-simplified expressions used to calculate the magnitude of these effects, in which only the solute concentration appears. The details of how to carry out these calculations and the many important applications of colligative properties are covered elsewhere. Our purpose here is to offer a more complete explanation of why these phenomena occur.
Basically, these all result from the effect of dilution of the solvent on its entropy, and thus in the increase in the density of energy states of the system in the solution compared to that in the pure liquid. Equilibrium between two phases (liquid-gas for boiling and solid-liquid for freezing) occurs when the energy states in each phase can be populated at equal densities. The temperatures at which this occurs are depicted by the shading.
Dilution of the solvent adds new energy states to the liquid, but does not affect the vapor phase. This raises the temperature required to make equal numbers of microstates accessible in the two phases.
Dilution of the solvent adds new energy states to the liquid, but does not affect the solid phase. This reduces the temperature required to make equal numbers of states accessible in the two phases.
Osmotic Pressure: Effects of Pressure on the Entropy
When a liquid is subjected to hydrostatic pressure— for example, by an inert, non-dissolving gas that occupies the vapor space above the surface, the vapor pressure of the liquid is raised (Figure \(13\)). The pressure acts to compress the liquid very slightly, effectively narrowing the potential energy well in which the individual molecules reside and thus increasing their tendency to escape from the liquid phase. (Because liquids are not very compressible, the effect is quite small; a 100-atm applied pressure will raise the vapor pressure of water at 25°C by only about 2 torr.) In terms of the entropy, we can say that the applied pressure reduces the dimensions of the "box" within which the principal translational motions of the molecules are confined within the liquid, thus reducing the density of energy states in the liquid phase.
Applying hydrostatic pressure to a liquid increases the spacing of its microstates, so that the number of energetically accessible states in the gas, although unchanged, is relatively greater— thus increasing the tendency of molecules to escape into the vapor phase. In terms of free energy, the higher pressure raises the free energy of the liquid, but does not affect that of the gas phase.
This phenomenon can explain osmotic pressure. Osmotic pressure, students must be reminded, is not what drives osmosis, but is rather the hydrostatic pressure that must be applied to the more concentrated solution (more dilute solvent) in order to stop osmotic flow of solvent into the solution. The effect of this pressure \(\Pi\) is to slightly increase the spacing of solvent energy states on the high-pressure (dilute-solvent) side of the membrane to match that of the pure solvent, restoring osmotic equilibrium. | textbooks/chem/General_Chemistry/Chem1_(Lower)/15%3A_Thermodynamics_of_Chemical_Equilibria/15.08%3A_Quantum_states_Microstates_and_Energy_spreading_in_Reactions.txt |
Electrochemistry is the study of electricity and how it relates to chemical reactions. In electrochemistry, electricity can be generated by movements of electrons from one element to another in a reaction known as redox reaction, or oxidation-reduction reaction.
• 16.1: Chemistry and Electricity
The connection between chemistry and electricity is a very old one, going back to ALESSANDRO VOLTA'S discovery, in 1793, that electricity could be produced by placing two dissimilar metals on opposite sides of a moistened paper.
• 16.2: Galvanic cells and Electrodes
We can measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials. This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, each immersed each in a solution containing a dissolved salt of the corresponding metal.
• 16.3: Cell Potentials and Thermodynamics
It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. For Example, zinc is more active because it can displace (precipitate) copper from solution. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
• 16.4: The Nernst Equation
The standard cell potentials we discussed in a previous section refer to cells in which all dissolved substances are at unit activity, which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure (known as the fugacity) of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know.
• 16.5: Applications of the Nernst Equation
We ordinarily think of the oxidation potential being controlled by the concentrations of the oxidized and reduced forms of a redox couple, as given by the Nernst equation. Under certain circumstances it becomes more useful to think of E as an independent variable that can be used to control the value of Q in the Nernst equation. This usually occurs when two redox systems are present, one being much more concentrated or kinetically active than the other.
• 16.6: Batteries and Fuel Cells
One of the oldest and most important applications of electrochemistry is to the storage and conversion of energy. You already know that a galvanic cell converts chemical energy to work; similarly, an electrolytic cell converts electrical work into chemical free energy. Devices that carry out these conversions are called batteries. In ordinary batteries the chemical components are contained within the device itself. If the reactants are supplied from an external source, the device is a fuel cell.
• 16.7: Timeline of Battery Development
Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive.
• 16.8: Electrochemical Corrosion
Corrosion can be defined as the deterioration of materials by chemical processes. Of these, the most important by far is electrochemical corrosion of metals, in which the oxidation process M → M+ + e– is facilitated by the presence of a suitable electron acceptor, sometimes referred to in corrosion science as a depolarizer. In a sense, corrosion can be viewed as the spontaneous return of metals to their ores.
• 16.9: Corrosion Gallery
A gallery of corrosion in different situations.
• 16.10: Electrolytic Cells and Electrolysis
Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds.
16: Electrochemistry
Learning Objectives
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Electroneutrality principle - Bulk matter cannot have a chemically-significant unbalance of positive and negative ions.
• Dissolution of a metal in water can proceed to a measurable extent only if some means is provided for removing the excess negative charge that remains. This can be by electron-acceptor ions in solution, or by drawing electrons out of the metal through an external circuit.
• Interfacial potentials - these exist at all phase boundaries. In the case of a metal in contact with an electrolyte solution, the interfacial region consists of an electric double layer.
• The potential difference between a metal and the solution is almost entirely located across the very thin double layer, leading to extremely large potential gradients in this region.
The connection between chemistry and electricity is a very old one, going back to ALESSANDRO VOLTA'S discovery, in 1793, that electricity could be produced by placing two dissimilar metals on opposite sides of a moistened paper.
Electroneutrality
Nature seems to strongly discourage any process that would lead to an excess of positive or negative charge in matter. Suppose, for example, that we immerse a piece of zinc metal in pure water. A small number of zinc atoms go into solution as $Zn^{2+}$ ions, leaving their electrons behind in the metal:
$Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.1}$
As this process goes on, the electrons which remain in the zinc cause a negative charge to build up within the metal which makes it increasingly difficult for additional positive ions to leave the metallic phase. A similar buildup of positive charge in the liquid phase adds to this inhibition. Very soon, therefore, the process comes to a halt, resulting in a solution in which the concentration of $Zn^{2+}$ is still too low (around 10–10 M) to be detected by ordinary chemical means.
There would be no build-up of this opposing charge in the two phases if the excess electrons could be removed from the metal or the positive ions consumed as the electrode reaction proceeds. For example, we could drain off the electrons left behind in the zinc through an external circuit that forms part of a complete electrochemical cell; this we will describe later. Another way to remove these same electrons is to bring a good electron acceptor (that is, an oxidizing agent) into contact with the electrode. A suitable acceptor would be hydrogen ions; this is why acids attack many metals. For the very active metals such as sodium, water itself is a sufficiently good electron acceptor.
The degree of charge unbalance that is allowed produces differences in electric potential of no more than a few volts, and corresponds to unbalances in the concentrations of oppositely charged particles that are not chemically significant. There is nothing mysterious about this prohibition, known as the electroneutrality principle; it is a simple consequence of the thermodynamic work required to separate opposite charges, or to bring like charges into closer contact. The additional work raises the free energy change of the process, making it less spontaneous.
The only way we can get the oxidation of the metal to continue is to couple it with some other process that restores electroneutrality to the two phases. A simple way to accomplish this would be to immerse the zinc in a solution of copper sulfate instead of pure water. The zinc metal quickly becomes covered with a black coating of finely-divided metallic copper. The reaction is a simple reduction-oxidation reaction involving a transfer of two electrons from the zinc to the copper:
$Zn(s) \rightarrow Zn^{2+} + 2e^– \label{1.1.2}$
and
$Cu^{2+} + 2e^– \rightarrow Cu(s) \label{1.1.3}$
The dissolution of the zinc is no longer inhibited by a buildup of negative charge in the metal, because the excess electrons are removed from the zinc by copper ions that come into contact with it. At the same time, the solution remains electrically neutral, since for each $Zn$ ion introduced to the solution, one $Cu$ ion is removed. The net reaction
$Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s) \label{1.1.4}$
quickly goes to completion.
Potential differences at interfaces
The transition region between two phases consists of a region of charge unbalance known as the electric double layer. As its name implies, this consists of an inner monomolecular layer of adsorbed water molecules and ions, and an outer diffuse region that compensates for any local charge unbalance that gradually merges into the completely random arrangement of the bulk solution. In the case of a metal immersed in pure water, the electron fluid within the metal causes the polar water molecules to adsorb to the surface and orient themselves so as to create two thin planes of positive and negative charge. If the water contains dissolved ions, some of the larger (and more polarizable) anions will loosely bond (chemisorb) to the metal, creating a negative inner layer which is compensated by an excess of cations in the outer layer.
Electrochemistry is the study of reactions in which charged particles (ions or electrons) cross the interface between two phases of matter, typically a metallic phase (the electrode) and a conductive solution, or electrolyte. A process of this kind can always be represented as a chemical reaction and is known generally as an electrode process. Electrode processes (also called electrode reactions) take place within the double layer and produce a slight unbalance in the electric charges of the electrode and the solution. Much of the importance of electrochemistry lies in the ways that these potential differences can be related to the thermodynamics and kinetics of electrode reactions. In particular, manipulation of the interfacial potential difference affords an important way of exerting external control on an electrode reaction.
The interfacial potential differences which develop in electrode-solution systems are limited to only a few volts at most. This may not seem like very much until you consider that this potential difference spans a very small distance. In the case of an electrode immersed in a solution, this distance corresponds to the thin layer of water molecules and ions that attach themselves to the electrode surface, normally only a few atomic diameters. Thus a very small voltage can produce a very large potential gradient. For example, a potential difference of one volt across a typical 10–8 cm interfacial boundary amounts to a potential gradient of 100 million volts per centimeter— a very significant value indeed!
Interfacial potentials are not confined to metallic electrodes immersed in solutions; they can in fact exist between any two phases in contact, even in the absence of chemical reactions. In many forms of matter, they are the result of adsorption or ordered alignment of molecules caused by non-uniform forces in the interfacial region. Thus colloidal particles in aqueous suspensions selectively adsorb a given kind of ion, positive for some colloids, and negative for others. The resulting net electric charge prevents the particles from coming together and coalescing, which they would otherwise tend to do under the influence of ordinary van der Waals attractions.
Interfacial potential differences are not directly observable. The usual way of measuring a potential difference between two points is to bring the two leads of a voltmeter into contact with them. It's simple enough to touch one lead of the meter to a metallic electrode, but there is no way you can connect the other lead to the solution side of the interfacial region without introducing a second electrode with its own interfacial potential, so you would be measuring the sum of two potential differences. Thus single electrode potentials, as they are commonly known, are not directly observable.
What we can observe, and make much use of, are potential differences between pairs of electrodes in electrochemical cells. This is the topic of the next page in this series.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Electroneutrality principle - Bulk matter cannot have a chemically-significant unbalance of positive and negative ions.
• Dissolution of a metal in water can proceed to a measurable extent only if some means is provided for removing the excess negative charge that remains. This can be by electron-acceptor ions in solution, or by drawing electrons out of the metal through an external circuit.
• Interfacial potentials - these exist at all phase boundaries. In the case of a metal in contact with an electrolyte solution, the interfacial region consists of an electric double layer.
• The potential difference between a metal and the solution is almost entirely located across the very thin double layer, leading to extremely large potential gradients in this region. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.01%3A_Chemistry_and_Electricity.txt |
It is physically impossible to measure the potential difference between a piece of metal and the solution in which it is immersed. We can, however, measure the difference between the potentials of two electrodes that dip into the same solution, or more usefully, are in two different solutions. In the latter case, each electrode-solution pair constitutes an oxidation-reduction half cell, and we are measuring the sum of the two half-cell potentials.
This arrangement is called a galvanic cell. A typical cell might consist of two pieces of metal, one zinc and the other copper, each immersed each in a solution containing a dissolved salt of the corresponding metal. The two solutions are separated by a porous barrier that prevents them from rapidly mixing but allows ions to diffuse through.
If we connect the zinc and copper by means of a metallic conductor, the excess electrons that remain when Zn2+ ions emerge from the zinc in the left cell would be able to flow through the external circuit and into the right electrode, where they could be delivered to the Cu2+ ions which become "discharged", that is, converted into Cu atoms at the surface of the copper electrode. The net reaction is the oxidation of zinc by copper(II) ions:
$Zn_{(s)} + Cu^{2+} \rightarrow Zn^{2+} + Cu_{(s)}$
but this time, the oxidation and reduction steps (half reactions) take place in separate locations:
left electrode:
Zn(s) → Zn2+ + 2e oxidation
right electrode:
Cu2+ + 2e→ Cu(s) reduction
Electrochemical cells allow measurement and control of a redox reaction
The reaction can be started and stopped by connecting or disconnecting the two electrodes. If we place a variable resistance in the circuit, we can even control the rate of the net cell reaction by simply turning a knob. By connecting a battery or other source of current to the two electrodes, we can force the reaction to proceed in its non-spontaneous, or reverse direction. By placing an ammeter in the external circuit, we can measure the amount of electric charge that passes through the electrodes, and thus the number of moles of reactants that get transformed into products in the cell reaction.
Electric charge q is measured in coulombs. The amount of charge carried by one mole of electrons is known as the Faraday, which we denote by F. Careful experiments have determined that 1 F = 96467 C. For most purposes, you can simply use 96,500 Coulombs as the value of the faraday. When we measure electric current, we are measuring the rate at which electric charge is transported through the circuit. A current of one ampere corresponds to the flow of one coulomb per second.
Charge Transport within the Cell
For the cell to operate, not only must there be an external electrical circuit between the two electrodes, but the two electrolytes (the solutions) must be in contact. The need for this can be understood by considering what would happen if the two solutions were physically separated. Positive charge (in the form of Zn2+) is added to the electrolyte in the left compartment, and removed (as Cu2+) from the right side, causing the solution in contact with the zinc to acquire a net positive charge, while a net negative charge would build up in the solution on the copper side of the cell. These violations of electroneutrality would make it more difficult (require more work) to introduce additional Zn2+ ions into the positively-charged electrolyte or for electrons to flow into right compartment where they are needed to reduce the Cu2+ ions, thus effectively stopping the reaction after only a chemically insignificant amount has taken place.
In order to sustain the cell reaction, the charge carried by the electrons through the external circuit must be accompanied by a compensating transport of ions between the two cells. This means that we must provide a path for ions to move directly from one cell to the other. This ionic transport involves not only the electroactive species Cu2+ and Zn2+, but also the counterions, which in this example are nitrate, NO3-. Thus an excess of Cu2+ in the left compartment could be alleviated by the drift of these ions into the right side, or equally well by diffusion of nitrate ions to the left. More detailed studies reveal that both processes occur, and that the relative amounts of charge carried through the solution by positive and negative ions depends on their relative mobilities, which express the velocity with which the ions are able to make their way through the solution. Since negative ions tend to be larger than positive ions, the latter tend to have higher mobilities and carry the larger fraction of charge.
In the simplest cells, the barrier between the two solutions can be a porous membrane, but for precise measurements, a more complicated arrangement, known as a salt bridge, is used. The salt bridge consists of an intermediate compartment filled with a concentrated solution of KCl and fitted with porous barriers at each end. The purpose of the salt bridge is to minimize the natural potential difference, known as the junction potential, that develops (as mentioned in the previous section) when any two phases (such as the two solutions) are in contact. This potential difference would combine with the two half-cell potentials so as introduce a degree of uncertainty into any measurement of the cell potential. With the salt bridge, we have two liquid junction potentials instead of one, but they tend to cancel each other out.
Cell description conventions
In order to make it easier to describe a given electrochemical cell, a special symbolic notation has been adopted. In this notation the cell we described above would be
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
There are several other conventions relating to cell notation and nomenclature that you are expected to know:
• The anode is where oxidation occurs, and the cathode is the site of reduction. In an actual cell, the identity of the electrodes depends on the direction in which the net cell reaction is occurring.
• If electrons flow from the left electrode to the right electrode (as depicted in the above cell notation) when the cell operates in its spontaneous direction, the potential of the right electrode will be higher than that of the left, and the cell potential will be positive.
• "Conventional current flow" is from positive to negative, which is opposite to the direction of the electron flow. This means that if the electrons are flowing from the left electrode to the right, a galvanometer placed in the external circuit would indicate a current flow from right to left.
Electrodes and Electrode Reactions
An electrode reaction refers to the net oxidation or reduction process that takes place at an electrode. This reaction may take place in a single electron-transfer step, or as a succession of two or more steps. The substances that receive and lose electrons are called the electroactive species.
This process takes place within the very thin interfacial region at the electrode surface, and involves quantum-mechanical tunneling of electrons between the electrode and the electroactive species. The work required to displace the H2O molecules in the hydration spheres of the ions constitutes part of the activation energy of the process.
In the example of the Zn/Cu cell we have been using, the electrode reaction involves a metal and its hydrated cation; we call such electrodes metal-metal ion electrodes. There are a number of other kinds of electrodes which are widely encountered in electrochemistry and analytical chemistry.
Ion-ion Electrodes
Many electrode reactions involve only ionic species, such as $Fe^{2+}$ and $Fe^{3+}$. If neither of the electroactive species is a metal, some other metal must serve as a conduit for the supply or removal of electrons from the system. In order to avoid complications that would arise from electrode reactions involving this metal, a relatively inert substance such as platinum is commonly used. Such a half cell would be represented as
Pt(s) | Fe3+(aq), Fe2+(aq) || ...
and the half-cell reaction would be
$Fe^{2+}(aq) \rightarrow Fe^{3+} (aq) + e^-$
The reaction occurs at the surface of the electrode (Fig 4 above). The electroactive ion diffuses to the electrode surface and adsorbs (attaches) to it by van der Waals and Coulombic forces. In doing so, the waters of hydration that are normally attached to any ionic species must be displaced. This process is always endothermic, sometimes to such an extent that only a small fraction of the ions be able to contact the surface closely enough to undergo electron transfer, and the reaction will be slow. The actual electron-transfer occurs by quantum-mechanical tunnelling.
Gas Electrodes
Some electrode reactions involve a gaseous species such as $H_2$, $O_2$, or $Cl_2$. Such reactions must also be carried out on the surface of an electrochemically inert conductor such as platinum. A typical reaction of considerable commercial importance is
$Cl^-(aq) \rightarrow ½ Cl_2(g) + e^-$
Similar reactions involving the oxidation of $Br_2$ or $I_2$ also take place at platinum surfaces.
Insoluble–salt Electrodes
A typical electrode of this kind consists of a silver wire covered with a thin coating of silver chloride, which is insoluble in water. The electrode reaction consists in the oxidation and reduction of the silver:
$AgCl(s) + e^– → Ag(s) + Cl^–(aq)$
The half cell would be represented as
$... || Cl^– (aq) | AgCl (s) | Ag (s)$
Although the usefulness of such an electrode may not be immediately apparent, this kind of electrode finds very wide application in electrochemical measurements, as we shall see later.
Reference Electrodes
In most electrochemical experiments our interest is concentrated on only one of the electrode reactions. Since all measurements must be on a complete cell involving two electrode systems, it is common practice to employ a reference electrode as the other half of the cell. The major requirements of a reference electrode are that it be easy to prepare and maintain, and that its potential be stable. The last requirement essentially means that the concentration of any ionic species involved in the electrode reaction must be held at a fixed value. The most common way of accomplishing this is to use an electrode reaction involving a saturated solution of an insoluble salt of the ion. One such system, the silver-silver chloride electrode has already been mentioned:
$Ag | AgCl(s) | Cl^–(aq) || ...$
$Ag(s) + Cl^–(aq) →AgCl(s) + e^–$
This electrode usually takes the form of a piece of silver wire coated with AgCl. The coating is done by making the silver the anode in an electrolytic cell containing HCl; the Ag+ ions combine with Cl ions as fast as they are formed at the silver surface.
The other common reference electrode is the calomel electrode; calomel is the common name for mercury(I) chloride. Such a half cell would be represented as
$Hg | Hg^{2+}(aq) | KCl || ...$
and the half-cell reaction would be
$Hg(l) + Cl^– → ½ HgCl2(s) + e^–$
The potentials of both of these electrodes have been very accurately determined against the hydrogen electrode. The latter is seldom used in routine electrochemical measurements because it is more difficult to prepare; the platinum surface has to be specially treated by preliminary electrolysis. Also, there is need for a supply of hydrogen gas which makes it somewhat cumbersome and hazardous.
Summary and additional notes
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• A galvanic cell (sometimes more appropriately called a voltaic cell) consists of two half-cells joined by a salt bridge or some other path that allows ions to pass between the two sides in order to maintain electroneutrality.
• The conventional way of representing an electrochemical cell of any kind is to write the oxidation half reaction on the left and the reduction on the right. Thus for the reaction
Zn(s) + Cu2+ → Zn2+ + Cu(s)
we write
Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s)
in which the single vertical bars represent phase boundaries. The double bar denotes a liquid-liquid boundary which in laboratory cells consists of a salt bridge or in ion-permeable barrier. If the net cell reaction were written in reverse, the cell notation would become
Cu(s) | Cu2+(aq) || Zn 2+(aq) | Zn (s)
Remember: the Reduction process is always shown on the Right.
at the electrode surface. The energy required to displace water molecules from the hydration shell of an ion as it approaches the electrode surface constitutes an activation energy which can slow down the process. Even larger activation energies (and slower reactions) occur when a molecule such as O2 is formed or consumed. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.02%3A_Galvanic_cells_and_Electrodes.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.
• When we refer to the "standard potential of a half-cell" or "couple" M2+/M, we mean the potential difference ErightEleft of the cell
• If the potential difference of this cell is positive (ErightEleft >0), electrons will flow through an external circuit from the Pt/H2 electrode to the M electrode and the cell reaction will spontaneously proceed in the direction written. The more positive the cell potential, the greater the tendency of this reaction to occur and the stronger the oxidizing agent M2+.
• Through the relation = – Δ/nF it is apparent that a standard half-cell reduction potential is simply the decrease in the free energy per mole of electrons transferred to H+ ions under the conditions that define the SHE. Strong reducing agents (good electron donors) have more negative s, while strong oxidizing agents (good acceptors) have more positive s.
• For a more general cell X(s) | X+ || M2+ | M(s) , is similarly the fall in free energy per electron-mole when M2+ is reduced by X. This reaction can proceed spontaneously only if the cell potential is positive (Δ negative.)
• An electron free energy diagram that displays various redox couples on a vertical scale of free energies relative to H+ serves as a convenient means of visualizing the possible reactions when two or more redox-active pairs are present in a solution. The position of a redox couple in relation to those of the H2/ H+ and H2O/O2,H+ couples is especially significant because it indicates whether a given species will be thermodynamically stable in water.
• Latimer diagrams provide a convenient means of correlating the various oxidation states of a particular element.
It has long been known that some metals are more "active" than others in the sense that a more active metal can "displace" a less active one from a solution of its salt. The classic example is the one we have already mentioned on the preceding page:
$\ce{ Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)} \nonumber$
Here zinc is more active because it can displace (precipitate) copper from solution. If you immerse a piece of metallic zinc in a solution of copper sulfate, the surface of the zinc quickly becomes covered with a black coating of finely-divided copper, and the blue color of the hydrated copper(II) ion diminishes. Similar comparisons of other metals made it possible to arrange them in the order of their increasing electron-donating (reducing) power. This sequence became known as the electromotive or activity series of the metals.
The most active (most strongly reducing) metals appear on top, and least active metals appear on the bottom. A more active metal (such as Zn) will donate electrons to the cation of a less active metal (Cu2+, for example.) Notice the special role of hydrogen here; although H2 does not have the physical properties of a metal, it is capable of being "displaced" (a rather archaic term seldom used in modern chemistry) from H2O or H+-containing (acidic) solutions. Note that the "active" metals are all "attacked by acids"; what this really means is that they are capable of donating electrons to H+.
displace H2 from water, steam, or acids Li 2 Li(s) + 2 H2O(l) → 2 LiOH(aq) + H2(g)
K 2 K(s) + 2 H2O(l) → 2 KOH(aq) + H2(g)
Ca Ca(s) + 2 H2O(l) → Ca(OH)2(s) + H2(g)
Na 2 Na(s) + 2 H2O() → 2 NaOH(aq) + H2(g)
displace H2 from steam or acids Mg Mg(s) + 2 H2O(g) → Mg(OH)2(s) + H2(g)
Al 2 Al(s) + 6 H2O(g) → 2 Al(OH)3(s) + 3 H2(g)
Mn Mn(s) + 2 H2O(g) → Mn(OH)2(s) + H2(g)
Zn Zn(s) + 2 H2O(g) → Zn(OH)2(s) + H2(g)
Fe Fe(s) + 2 H2O(g) → Fe(OH)2(s) + H2(g)
displace H2 from acids only Ni Ni(s) + 2 H+(aq) → Ni2+(aq) + H2(g)
Sn Sn(s) + 2 H+(aq) → Sn2+(aq) + H2(g)
Pb Pb(s) + 2 H+(aq) → Pb2+(aq) + H2(g)
H2
can't displace H2 Cu
Ag
Pt
Au
The activity series has long been used to predict the direction of oxidation-reduction reactions. Consider, for example, the oxidation of Cu by metallic zinc that we have mentioned previously. The fact that zinc is near the top of the activity series means that this metal has a strong tendency to lose electrons. By the same token, the tendency of Zn to accept electrons is relatively small. Copper, on the other hand, is a poorer electron donor, and thus its oxidized form, Cu, is a fairly good electron acceptor. We would therefore expect the reaction
$Zn(s) + Cu^{2+} \rightarrow Zn^{2+} + Cu(s)$
to proceed in the direction indicated, rather than in the reverse direction. An old-fashioned way of expressing this is to say that "zinc will displace copper from solution". The above table is of limited practical use because it does not take into account the concentrations of the dissolved species. In order to treat these reactions quantitatively, it is convenient to consider the oxidation and reduction steps separately.
Standard half-cell potentials
When a net reaction proceeds in an electrochemical cell, oxidation occurs at one electrode (the anode) and reduction takes place at the other electrode (the cathode.) We can think of the cell as consisting of two half-cells joined together by an external circuit through which electrons flow and an internal pathway that allows ions to migrate between them so as to preserve electroneutrality.
Reduction potentials
Each half-cell has associated with it an electrode-solution potential difference whose magnitude depends on the nature of the particular electrode reaction and on the concentrations of the dissolved electroactive species. The sign of this potential difference depends on the direction (oxidation or reduction) in which the electrode reaction proceeds. To express them in a uniform way, we adopt the convention that half-cell potentials are always defined for the reduction direction. Thus the half-cell potential for the Zn/Zn2+ electrode (or couple as it is sometimes called) always refers to the reduction reaction
$Zn^{2+} + 2e^– \rightarrow Zn (s)$
In the cell Zn(s) | Zn2+(aq) || Cu2+(aq) | Cu(s) the zinc appears on the left side, indicating that it is being oxidized, not reduced. For this reason, the potential difference contributed by the left half-cell has the opposite sign to its conventional half-cell potential. More generally, we can define the cell potential (or cell EMF) as
$E_{cell} = \Delta V = E_{right} – E_{left} \label{1}$
in which "right" and "left" refer to the cell notation convention ("reduction on the right") and not, of course, to the physical orientation of a real cell in the laboratory. If we expand the above expression we see that the cell potential
$E_{cell} = V_{Cu} – V_{solution} + V_{solution} – V_{Zn}$
is just the difference between the two half-cell potentials $E_{right}$ and $E_{left}$.
Reference half-cells
The fact that individual half-cell potentials are not directly measurable does not prevent us from defining and working with them. Although we cannot determine the absolute value of a half-cell potential, we can still measure its value in relation to the potentials of other half cells. In particular, if we adopt a reference half-cell whose potential is arbitrarily defined as zero, and measure the potentials of various other electrode systems against this reference cell, we are in effect measuring the half-cell potentials on a scale that is relative to the potential of the reference cell.
The reference cell that has universally been adopted for this purpose is the hydrogen half-cell
$Pt | ½ H_{2(g)} | H^+_{(aq)} || \;...$
in which hydrogen gas is allowed to bubble over a platinum electrode having a specially treated surface which catalyzes the reaction
$½ H_{2(g)} → H^+ + e^–$
When this electrode is operated under standard conditions of 1 atm H2 pressure, 25°C, and pH = 0, it becomes the standard hydrogen electrode, sometimes abbreviated SHE.
To measure the relative potential of some other electrode couple M2+/M, we can set up a cell
$Pt | H_{2(g)} | H^+ || M^{2+}_{ (aq)} | M_{(s)}$
whose net reaction is
H2(g) + M2+(aq) → 2H+ + M(s)
the potential difference between the platinum and M electrodes will be
Ecell = VM – Vsolution + Vsolution – V Pt
but since the difference Vsolution – V Pt is by definition zero for the hydrogen half-cell, the cell potential we measure corresponds to
Ecell = VM – Vsolution
which is just the potential (relative to that of the SHE) of the half-cell whose reaction is
$M^{2+} + 2e^– → M_{(s)}$
Measurement of a standard reduction potential.
The M2+/M half-cell is on the left, and the standard hydrogen cell is on the right. The two half-cells are joined through the salt bridge in the middle. The more "active" the metal M (the greater its tendency to donate electrons to H+), the more negative will be Ecell = ΔV = ErightEleft
Standard [reduction] potentials are commonly denoted by the symbol . values for hundreds of electrodes have been determined (mostly in the period 1925-45, during which time they were referred to as "oxidation potentials") and are usually tabulated in order of increasing tendency to accept electrons (increasing oxidizing power.)
Table 2: some standard reduction potentials
oxidant
(electron acceptor)
reductant
(elecron donor)
E°, volts
Na+
Na(s)
–2.71
Zn2+
Zn(s)
–.76
Fe2+
Fe(s)
–.44
Cd2+
Cd(s)
–.40
Pb2+
Pb(s)
–.126
H2(g)
0.000
AgCl(s)
Ag(s) + Cl(aq)
+.222
Hg2Cl2(s)
2Cl(aq) + 2Hg()
+.268
Cu2+
Cu(s)
+.337
I2(s)
2 I
+.535
Fe3+
Fe2+
+.771
Ag+
Ag(s)
+.799
O2(g) + 4H+
2 H2O()
+1.23
Cl2(g)
2 Cl
+1.36
Note particularly that
• Unlike the activity series mentioned above, this table includes non-metallic substances, and it is quantitative.
• The more positive the half-cell EMF, the greater the tendency of the reductant to donate electrons, and the smaller the tendency of the oxidant to accept electrons.
• A species in the leftmost column can act as an oxidizing agent to any species above it in the reductant column.
• Oxidants such as Cl2 that are below H2O will tend to decompose water.
Given the values for two half reactions, you can easily predict the potential difference of the corresponding cell: simply add the reduction potential of the reduction half-cell to the negative of the reduction potential (that is, to the oxidation potential) of the oxidation reaction.
Example $1$
Find the standard potential of the cell
Cu(s) | Cu2+ || Cl | AgCl(s) | Ag(s)
and predict the direction of electron flow when the two electrodes are connected.
Solution
The net reaction corresponding to this cell will be:
Cu(s) + 2AgCl(s) → 2 Ag(s) + 2 Cl(aq) + Cu2+(aq)
Where Cu(s)/Cu2+ is being Oxidized and Ag(s)/Ag+ is being Reduced
Ecell = Ereduction + Eoxidation Or Written another way Ecell = Eright - Eleft =(.222 + (– .337)) v = -0.115 v
Since Ecell is negative, the reaction will run in the opposite direction. The correct net reaction will be:
2 Ag(s) + 2 Cl(aq) + Cu2+(aq) → 2AgCl(s) + Cu(s)
Where Ag(s)/Ag+ is being Oxidized and Cu(s)/Cu2+ is being Reduced
Ecell = Ereduction + Eoxidation Or Written another way Ecell = Eright - Eleft =(.337 + (– .222)) v = +0.115 v
Since this potential is positive, the reaction will proceed to the right; electrons will be withdrawn from the silver electrode and flow through the external circuit into the copper electrode. Note carefully that in combining these half-cell potentials, we did not multiply for the Cu2+/Cu couple by two. The reason for this will be explained later.
Cell potentials and Free (Gibbs) energy
From the above, it should be apparent that the potential difference between the electrodes of a cell is a measure of the tendency for the cell reaction to take place: the more positive the cell potential, the greater the tendency for the reaction to proceed to the right. But we already know that the standard free energy change expresses the tendency for any kind of process to occur under the conditions of constant temperature and pressure. Thus ΔG° and measure the same thing, and are related in a simple way:
$\Delta G° = –nFE° \label{2}$
... or in more detail (see below for explanations of the units given for voltage)
A few remarks are in order about this very fundamental and important relation:
• The negative sign on the right indicates that a positive cell potential (according to the sign convention discussed previously) implies a negative free energy change, and thus that the cell reaction will spontaneously proceed to the right.
• Electrical work is done when an electric charge q moves through a potential difference ΔV. The right side of Eq. 2 refers to the movement of n moles of charge across the cell potential , and thus has the dimensions of work.
• The value of Δ expresses the maximum useful work that a system can do on the surroundings. "Useful" work is that which can be extracted from the cell by electrical means to operate a lamp or some other external device. This excludes any P-V work that is simply a consequence of volume change (which could conceivably be put to some use!) and which would be performed in any case, even if the reactants were combined directly. This quantity of work –ΔG° can only be extracted from the system under the limiting conditions of a thermodynamically reversible change, which for an electrochemical cell implies zero current. The more rapidly the cell operates, the less electrical work it can supply.
• If F is expressed in coulombs per mole of electrons, the electrical work is in joules per mole. To relate these units to electrical units, recall that the coulomb is one amp-sec, and that power, which is the rate at which work is done, is measured in watts, which is the product of amps and volts
1 J = 1 watt-sec = 1 (amp-sec) × volts
Thus the volt has the dimensions of joules/coulomb– the energy produced per quantity of charge passing through the cell. Because voltage is the quotient of two extensive quantities, it is itself intensive. When we multiply the anodic and cathodic half-reactions by the stoichiometric factors required to ensure that each involves the same quantity of charge, the free energy change and the number of coulombs both increase by the same factor, leaving the potential (voltage) unchanged. This explains why we do not have to multiply the s of the anode and cathode reactions by stoichiometric factors when we are finding the potential of a complete cell.
If Eq. 2 is solved for , we have
$E° =\dfrac{\Delta G°}{nF} \label{3}$
This states explicitly that the cell potential is a measure of the free energy change per mole of electrons transferred, which is a brief re-statement of the principle explained immediately above. To see this more clearly, consider the cell
Cu(s) | Cu2+ || Cl | AgCl(s) | Ag(s)
for which we list the standard reduction potentials and ΔG°s of the half-reactions:
reaction
-nFE°= ΔG°
cathode: 2 × [AgCl(s) + e → Ag(s) + Cl]
anode: Cu(s) → Cu2+ + 2 e
+.222 v
–(+.337) V
–42800 J
+65000 J
net: 2 Ag(s) + 2 Cl(aq) + Cu2+(aq) → AgCl(s) + Cu(s)
cell: Cu(s) | Cu2+(aq) || AgCl(s) | Cl(aq) | Ag(s)
–.115 v
+22200 J
Here we multiply the cathodic reaction by two in order to balance the charge. Because the anodic reaction is written as an oxidation, we reverse the sign of its and obtain Ecell = ErightEleft = –.115 volt for the cell potential. The negative cell potential tells us that this reaction will not proceed spontaneously.
When the electrons don't cancel out
Note, however, that if we are combining two half reactions to obtain a third half reaction, the values are not additive, since this third half-reaction is not accompanied by another half reaction that causes the charges to cancel. Free energies are always additive, so we combine them, and use Δ = –nFE° to find the cell potential.
Example $2$
Calculate for the electrode Fe3+/Fe(s) from the standard potential of the couples Fe3+/Fe2+ and Fe2+/Fe(s)
Solution
Tabulate the values and calculate the Δs as follows:
(i) Fe3+ + e → Fe2+ 1 = .771 v , Δ1 = –.771 F
(ii) Fe2+ + 2 e → Fe(s) 2= –.440 v , Δ2 = +.880 F
(iii) Fe3+ + 3 e → Fe(s) 3 = ? , Δ3 = +.109 F
The Gibbs energy for half-reaction (iii) is 0.109nF, so 3 = –.109/3 = –.036 v
The fall of the electron
A table of standard half-cell potentials summarizes a large amount of chemistry, for it expresses the relative powers of various substances to donate and accept electrons by listing reduction half-reactions in order of increasing E° values, and thus of increasing spontaneity. The greater the value of E°, the greater the tendency of the substance on the left to acquire electrons, and thus the stronger this substance is as an oxidizing agent.
If you have studied elementary chemical thermodynamics, you will have learned about the role that a quantity called the Gibbs free energy, usually referred to as simply the "free energy", plays in determining the direction of any chemical change. The rule is that all spontaneous change (that is, all reactions that proceed to the "right") is associated with a fall in the free energy, and the greater the degree of that fall (Δ), the greater will be the tendency for the reaction to take place.
If you are not familiar with the concept of free energy, just think of it as something like potential energy, which similarly decreases when spontaneous mechanical events occur, such as the dropping of a weight.
Since oxidation-reduction processes involve the transfer of an electron from a donor to an acceptor, it makes sense to focus on the electron and to consider that it falls from a higher-free energy environment (the reductant, or "source") to a lower-free energy one (the oxidant, or "sink".)
As can be seen from the diagram below, this model makes it far easier to predict what will happen when two or more oxidants and reducants are combined; the electron "falls" as far as it can, filling up oxidizing agents (sinks) from the bottom up, very much in the same way as electrons fill atomic orbitals as we build up larger atoms.
Electron-free energy diagram of redox couples
This chart is essentially an abbreviated form of a table of standard potentials in which the various couples are displayed on a vertical scale corresponding to
E° = –Δ/nF. Any available sink on the right side will tend to drain electrons from a source above it. For example, immersion of metallic zinc in a solution of CuSO4 will result in the reduction of Cu2+ to metallic copper (red arrows.) Similarly, addition of chlorine to water will tend to oxidize the water, producing O2 and Cl (blue arrows.) Note especially the positions of the H2/ H+ and H2O/O2,H+ couples on this chart, as they define the range of E°s for substances that will not decompose water (green region.)
A more detailed table with a more complete explanation can be seen on the "Fall of the electron" tutorial page; it is strongly recommended that you take the time to acquire a thorough understanding of this concept.
At this point, it might be worth calling your attention to the similar way of depicting acid-base reactions as representing the "fall of the proton" as shown below and described much more thoroughly here.
Proton-free energy diagram of acid-base systems
Acids are proton sources (donors), bases are proton sinks. Protons "fall" (in free energy) whenever a base is present that presents proton-empty free energy levels. The red arrows show what happens when acetic acid is titrated with a strong base; the results are acetate ion and water. Note here again the crucial role of water, both as a proton acceptor (forming hydronium ion) and as a proton donor (forming hydroxide ion.) Note also that the pH of a solution is a direct measure of the average free energy of protons in the solution (relative to H3O+.)
An important difference between proton transfer and electron transfer reactions is that the latter can vary greatly in speed, from almost instantaneous to so slow as to be unobservable. Acid-base reactions are among the fastest known.
Considerable insight into the chemistry of a single element can be had by comparing the standard electrode potentials (and thus the relative free energies) of the various oxidation states of the element. The most convenient means of doing this is the Latimer diagram. As examples, diagrams for iron and chlorine are shown below.
The formulas of the species that represent each oxidation state of the element are written from left to right in order of decreasing oxidation number, and the standard potential for the reduction of each species to the next on the right is written in between the formulas. Potentials for reactions involving hydrogen ions will be pH dependent, so separate diagrams are usually provided for acidic and alkaline solutions (effective hydrogen ion concentrations of 1M and 10–14 M, respectively).
The more positive the reduction potential, the greater will be the tendency of the species on the left to be reduced to the one on the right. To see how Latimer diagrams are used, look first at the one for iron in acid solution. The line connecting Fe3+ and Fe2+ represents the reaction
Fe3+ + e → Fe2+
whose positive E° (.440 v) indicates that metallic iron will dissolve in acidic solution to form Fe2+. Because the oxidation of this species to the +3 state has a negative potential (-0.771v; moving to the left on the diagram reverses the sign), the +2 state will be the stable oxidation state of iron under these conditions.
Disproportionation
An important condition to recognize in a Latimer diagram is when the potential on the left of a species is less positive than that on the right. This indicates that the species can oxidize and reduce itself, a process known as disproportionation. As an example, consider Cl2 in alkaline solution. The potential for its reduction to Cl is sufficiently positive (+1.35 v) to supply the free energy necessary for the oxidation of one atom of chlorine to hypochlorite. Thus elemental chlorine is thermodynamically unstable with respect to disproportionation in alkaline solution, and the same it true of the oxidation product, ClO (hypochlorite ion).
Behavior of chlorine in water
Cl2 can oxidize water (green arrows, top) and also undergo disproportionation (purple arrows, bottom). In the latter process, one Cl2 molecule donates electrons to another. Bear in mind that many oxidation-reduction reactions, unlike most acid-base reactions, tend to be very slow, so the fact that a species is thermodynamically unstable does not always mean that it will quickly decompose. Thus the two reactions shown in the figure are normally very slow.
Thermodnamics of Galvanic cells
The free energy change for a process represents the maximum amount of non-PV work that can be extracted from it. In the case of an electrochemical cell, this work is due to the flow of electrons through the potential difference between the two electrodes. Note, however, that as the rate of electron flow (i.e., the current) increases, the potential difference must decrease; if we short-circuit the cell by connecting the two electrodes with a conductor having negligible resistance, the potential difference is zero and no work will be done. The full amount of work can be realized only if the cell operates at an infinitessimal rate; that is, reversibly.
You should recall that this is exactly analogous to the expansion of an ideal gas. The full amount of work w = PdV is extracted only under the special condition that the external pressure P opposing expansion is only infinitessimally smaller than the pressure of the gas itself. If the gas is allowed to expand into a vacuum (P = 0), no work will be done.
The total amount of energy a reaction can supply under standard conditions at constant pressure and temperature is given by ΔH°. If the reaction takes place by combining the reactants directly (no cell) or in a short-circuited cell, no work is done and the heat released is ΔH. If the reaction takes place in a cell that performs electrical work, then the heat released is diminished by the amount of electrical work done. In the limit of reversible operation, the heat released becomes
ΔH = Δ + T ΔS
Note
Pt | H2(g) | H+ || M2+ (aq) | M(s)
whose left half consists of a standard hydrogen electrode (SHE) and whose net reaction is
H2(g) + M2+(aq) → 2H+ + M(s) | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.03%3A_Cell_potentials_and_Thermodynamics.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• The Nernst equation relates the effective concentrations (activities) of the components of a cell reaction to the standard cell potential. For a simple reduction of the form Mn+ + ne → M, it tells us that a half-cell potential will change by 59/n mV per 10-fold change in the activity of the ion.
• Ionic concentrations can usually be used in place of activities when the total concentration of ions in the solution does not exceed about about 0.001M.
• In those reactions in which H+ or OH ions take part, the cell potential will also depend on the pH. Plots of E vs. pH showing the stability regions of related species are very useful means of summarizing the redox chemistry of an element.
The standard cell potentials we discussed in a previous section refer to cells in which all dissolved substances are at unit activity, which essentially means an "effective concentration" of 1 M. Similarly, any gases that take part in an electrode reaction are at an effective pressure (known as the fugacity) of 1 atm. If these concentrations or pressures have other values, the cell potential will change in a manner that can be predicted from the principles you already know.
Cell Potentials Depend on Concentrations
Suppose, for example, that we reduce the concentration of $Zn^{2+}$ in the $Zn/Cu$ cell from its standard effective value of 1 M to an to a much smaller value:
$Zn(s) | Zn^{2+}(aq, 0.001\,M) || Cu^{2+}(aq) | Cu(s)$
This will reduce the value of $Q$ for the cell reaction
$Zn(s) + Cu^{2+} → Zn^{2+} + Cu(s)$
thus making it more spontaneous, or "driving it to the right" as the Le Chatelier principle would predict, and making its free energy change $\Delta G$ more negative than $\Delta G°$, so that $E$ would be more positive than $E^°$. The relation between the actual cell potential $E$ and the standard potential $E^°$ is developed in the following way. We begin with the equation derived previously which relates the standard free energy change (for the complete conversion of products into reactants) to the standard potential
$\Delta G° = –nFE°$
By analogy we can write the more general equation
$\Delta G = –nFE$
which expresses the change in free energy for any extent of reaction— that is, for any value of the reaction quotient $Q$. We now substitute these into the expression that relates $\Delta G$ and $\Delta G°$ which you will recall from the chapter on chemical equilibrium:
$\Delta G = \Delta G° + RT \ln Q$
which gives
$–nFE = –nFE° + RT \ln Q$
which can be rearranged to
$\color{red} {\underbrace{E=E° -\dfrac{RT}{nF} \ln Q}_{\text{applicable at all temperatures}}} \label{Nernst Long}$
This is the Nernst equation that relates the cell potential to the standard potential and to the activities of the electroactive species. Notice that the cell potential will be the same as $E°$ only if $Q$ is unity. The Nernst equation is more commonly written in base-10 log form and for 25 °C:
$\color{red} {\underbrace{E=E° -\dfrac{0.059}{n} \log_{10} Q}_{\text{Applicable at only 298K}}} \label{Nernst Short}$
Significance of the Nernst Equation
The Nernst equation tells us that a half-cell potential will change by 59 millivolts per 10-fold change in the concentration of a substance involved in a one-electron oxidation or reduction; for two-electron processes, the variation will be 28 millivolts per decade concentration change. Thus for the dissolution of metallic copper
$Cu_{(s)} \rightarrow Cu^{2+} + 2e^–$
the potential
$E = (– 0.337) – 0.0295 \log_{10} [Cu^{2+}]$
becomes more positive (the reaction has a greater tendency to take place) as the cupric ion concentration decreases. This, of course, is exactly what the Le Chatelier Principle predicts; the more dilute the product, the greater the extent of the reaction.
Electrodes with poise
The equation just above for the Cu/Cu2+ half-cell raises an interesting question: suppose you immerse a piece of copper in a solution of pure water. With Q = [Cu2+] = 0, the potential difference between the electrode and the solution should be infinite! Are you in danger of being electrocuted? You need not worry; without any electron transfer, there is no charge to zap you with. Of course it won't be very long before some Cu2+ ions appear in the solution, and if there are only a few such ions per liter, the potential reduces to only about 20 volts. More to the point, however, the system is so far from equilibrium (for example, there are not enough ions to populate the electric double layer) that the Nernst equation doesn't really give meaningful results. Such an electrode is said to be unpoised. What ionic concentration is needed to poise an electrode? I don't really know, but I would be suspicious of anything much below 10–6 M.
The Nernst Equation works only in Dilute Ionic Solutions
Ions of opposite charge tend to associate into loosely-bound ion pairs in more concentrated solutions, thus reducing the number of ions that are free to donate or accept electrons at an electrode. For this reason, the Nernst equation cannot accurately predict half-cell potentials for solutions in which the total ionic concentration exceeds about 10–3 M.
How the cell potential really depends on concentration! The Nernst equation accurately predicts cell potentials only when the equilibrium quotient term Q is expressed in activities. Ionic activities depart increasingly from concentrations when the latter exceed 10–4 to 10–5 M, depending on the sizes and charges of the ions.
If the Nernst equation is applied to more concentrated solutions, the terms in the reaction quotient Q must be expressed in "effective concentrations" or activities of the electroactive ionic species. The activity coefficient $\gamma$) relates the concentration of an ion to its activity a in a given solution through the relation
$a = \gamma c$
Since electrode potentials measure activities directly, activity coefficients can be determined by carrying out appropriate EMF measurements on cells in which the concentration of the ion of interest is known. The resulting Es can then be used to convert concentrations into activities for use in other calculations involving equilibrium constants.
Cell potentials and pH: stability diagrams
As most of us recall from our struggles with balancing redox equations in beginning chemistry courses, many electron-transfer reactions involve hydrogen ions and hydroxide ions. The standard potentials for these reactions therefore refer to the pH, either 0 or 14, at which the appropriate ion has unit activity. Because multiple numbers of H+ or OH ions are often involved, the potentials given by the Nernst equation can vary greatly with the pH. It is frequently useful to look at the situation in another way by considering what combinations of potential and pH allow the stable existence of a particular species. This information is most usefully expressed by means of a E-vs.-pH diagram, also known as a Pourbaix diagram.
As was noted in connection with the shaded region in the figure below, water is subject to decomposition by strong oxidizing agents such as Cl2 and by reducing agents stronger than H2. The reduction reaction can be written either as
$2H^+ + 2e^– \rightarrow H_{2}(g)$
or, in neutral or alkaline solutions as
$2H_2O + 2 e^– \rightarrow H_{2}(g) + 2 OH^–$
These two reactions are equivalent and follow the same Nernst equation
$E_{H^+/H_2} = E_{H^+/H_2}^o + \dfrac{RT}{nF} \ln \left( \dfrac{[H^+]^2} {P_{H_2}} \right)$
which, at 25°C and unit H2 partial pressure reduces to
$E = E° - \dfrac{0.059}{2} × 2 pH = –0.059\; pH$
Similarly, the oxidation of water
$2H_2O \rightarrow O_{2}(g) + 4 H^+ + 2 e^–$
is governed by the Nernst equation
$E_{O_2/H_2O} = E_{O_2/H_2O}^o + \dfrac{RT}{nF} \ln \left( P_{O_2}[H^+]^4 \right)$
which similarly becomes E = 1.23 – 0.059 pH, so the E-vs-pH plots for both processes have identical slopes and yield the stability diagram for water in Figure $2$. This Pourbaix diagram has special relevance to electrochemical corrosion of metals. Thus metals above hydrogen in the activity series will tend to undergo oxidation (corrosion) by reducing H+ ions or water.
Chlorine in water
Because chlorine is widely used as a disinfectant for drinking water, swimming pools, and sewage treatment, it is worth looking at its stability diagram. Note that the effective bactericidal agent is not Cl2 itself, but its oxidation product hypochlorous acid HOCl which predominates at pH values below its pKa of 7.3. Note also that
• Cl2 is unstable in water except at very low pH; it decomposes into HOCl and Cl.
• Hypochlorous acid and its anion are stronger oxidants than O2 and thus subject to decomposition in water. The only stable chlorine species in water is Cl.
• Decomposition of HOCl occurs very slowly in the dark, but is catalyzed by sunlight. For this reason the chlorine in outside swimming pools must be frequently renewed.
• Decomposition of Cl2 and HOCl by reaction with organic material in municipal water supply systems sometimes makes it necessary to inject additional chlorine at outlying locations.
Each solid line represents a combination of E and pH at which the two species on either side of it can coexist; at all other points, only a single species is stable. Note that equilibria between species separated by diagonal lines are dependent on both E and pH, while those separated by horizontal or vertical lines are affected by pH only or E only, respectively.
Iron
Stability diagrams are able to condense a great amount of information into a compact representation, and are widely employed in geochemistry and corrosion engineering. The Pourbaix diagram for iron is one of the more commonly seen examples.
Pourbaix diagram for iron. Three oxidation states of iron (0, +2 and +3) are represented on this diagram. The stability regions for the oxidized iron states are shown only within the stability region of H2O. Equilibria between species separated by vertical lines are dependent on pH only.
The +3 oxidation state is the only stable one in environments in which the oxidation level is controlled by atmospheric O2. This is the reason the Earth’s crust contains iron oxides, which developed only after the appearance of green plants which are the source of O2. Iron is attacked by H+ to form H2 and Fe(II); the latter then reacts with O2 to form the various colored Fe(III) oxides that constitute “rust”. Numerous other species such as oxides and hydrous oxides are not shown. A really “complete” diagram for iron would need to have at least two additional dimensions showing the partial pressures of O2 and CO2.
Concentration Cells
From your study of thermodynamics you may recall that the process
solute (concentrated) → solute (dilute)
is accompanied by a fall in free energy, and therefore is capable of doing work on the surroundings; all that is required is some practical way of capturing this work. One way of doing this is by means of a concentration cell such as
Cu(s) | CuNO3(.1 M) || CuNO3(.01 M) | Cu(s)
cathode: Cu2+(.1 M) + 2e → Cu(s)
anode: Cu(s) → Cu2+(.01 M) + 2e
net: Cu2+(.1 M) → Cu2+(.01 M)
which represents the transport of cupric ion from a region of higher concentration to one of lower concentration.
The driving force for this process is the free energy change ΔG associated with the concentration gradient (C2 – C1), sometimes known as the free energy of dilution:
$ΔG_{dilution} = RT \ln(C_2 – C_1)$
Note, however, that Cu2+ ions need not physically move between the two compartments; electron flow through the external circuit creates a "virtual" flow as copper ions are created in the low-concentration side and discharged at the opposite electrode. Nitrate ions must also pass between the cells to maintain electroneutrality. The Nernst equation for this cell is
$E = E^° - \left(\dfrac{0.059}{n}\right) \log_{10} Q = 0 - 0.29 \log_{10} 0.1 = +0.285 \,V$
Note that $E^°$ for a concentration cell is always zero, since this would be the potential of a cell in which the electroactive species are at unit activity in both compartments.
$E^°$ for a concentration cell is always zero, | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.04%3A_The_Nernst_Equation.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Atmospheric oxygen is a strong oxidizing agent, but in the absence of a suitable catalyst this reaction is ordinarily too slow to be observable.
• The pE is a useful index of electron availability, especially in natural waters and biological systems in which multiple redox systems are usually present.
• Solubility products can often be measured by constructing a cell in which one of the electrodes involves the sparingly soluble salt, and the net cell equation corresponds to the solubility reaction.
• Potentiometric titrations are widely used to measure concentrations of species that are readily oxidized or reduced.
• Measurement of pH is accomplished with an electrode consisting of a thin membrane of glass in which Na+ ions are exchanged with H+.
• Membrane (Donnan) potentials arise when passage of an ion across a semipermeable membrane is selectively facilitated or inhibited.
• Nerve conduction occurs not by a flow of charge through the nerve, but by a wave of depolarization resulting from the concerted action of active channels that govern the passage of ions (mainly K+ and Na+) across the membrane enclosing the nerve.
We ordinarily think of the oxidation potential being controlled by the concentrations of the oxidized and reduced forms of a redox couple, as given by the Nernst equation. Under certain circumstances it becomes more useful to think of E as an independent variable that can be used to control the value of Q in the Nernst equation. This usually occurs when two redox systems are present, one being much more concentrated or kinetically active than the other. By far the most important example of this is the way atmospheric oxygen governs the composition of the many redox systems connected with biological activity.
Oxygen and the Aquatic Environment
The presence of oxygen in the atmosphere has a profound effect on the redox properties of the aquatic environment— that is, on natural waters exposed directly or indirectly to the atmosphere, and by extension, on organisms that live in an aerobic environment.This is due, of course, to its being an exceptionally strong oxidizing agent and thus a low-lying sink for electrons from most of the elements and all organic compounds. Those parts of the environment that are protected from atmospheric oxygen are equally important because it is only here that electrons are sufficiently available to produce the "reducing" conditions that are essential for processes varying from photosynthesis to nitrogen fixation.
Example $1$
Estimate the redox potential of a natural water that is in equilibrium with the atmosphere at pH 7 and 298 K. What fraction of a dilute solution Fe2+ will be in its oxidized form Fe3+ in such a water?
Solution
The relevant E°s are
• 1.23 v for O2(g) + 4H+ + 4e→ 2H2O and
• 0.77 V for the Fe3+/Fe2+ couple.
(a) The potential (with respect to the SHE, of course) is given by the Nernst equation
which works out to E = 0.82 volt. As the Le Chatelier principle predicts, the higher pH (lower [H+] compared to that at the "standard" pH of zero) reduces the electron-accepting tendency of oxygen.
(b) The Nernst equation for the reduction of Fe3+ is E = .77–.059 log Q, in which Q is the ratio [Fe2+]/[Fe3+]. With E set by the O2/H2O couple, this becomes
0.82 = 0.77 – 0.059 log Q
which gives Q = 10–0.85 or [Fe2+]/[Fe3+] = 0.14/1, so the fraction of the iron in its oxidized form is 1/1.14 = 0.88.
If we can have pH, why not pE?
As you will recall from your study of acid-base chemistry, the pH of a solution (defined as –log {H+}) is a measure of availablity (technically, the activity) of protons in the solution. As is explained in more detail here, protons tend to "fall" (in free energy) from filled donor levels (acids) to lower acceptor levels (bases.) Through the relation
$[H^+] \approx K_a \dfrac{C_a}{C_b}$
which can be rewritten as
$\dfrac{C_a}{C_b} \approx \dfrac{[H^+]}{K_a}$
in which the pH is treated as an independent variable that controls the ratio of the conjugate forms of any acid-base pairs in the solution:
$\log \left(\dfrac{C_a}{C_b}\right) \approx pH – pK_a$
In the same way, we can define the pE as the negative log of the electron activity in the solution:
$pE = –\log{e^–}$
of these particles (but not to their "concentrations") when we are considering their availability to donors and acceptors. We will not get into the details of how pE is actually calculated (it is of course related to the ordinary standard electrode potential). To get an idea of its significance, consider the following chart that shows the pE° values of some redox systems that are of immense importance in the aquatic environment.
Environmentally-important redox systems
• These pE values refer to typical environmental conditions with pH=7 and oxygen partial pressure of 0.21 atm. The scale below shows the free energy of a mole of electrons relative to their level in H2O.
• The two conjugate forms of any redox pair are present in equal concentrations when the pE is at the level at which the pair is shown. At pE's above or below this level, the reduced or oxidized form will predominate.
• The sugar glucose, denoted by the general formula for carbohydrates {CH2O}, is the source of chemical energy for most organisms. Note that it is thermodynamically stable (and thus capable of being formed by photosynthesis) only under highly reducing conditions.
• Organisms derive their metabolic free energy when electrons fall from glucose to a lower-lying acceptor on the right.
• Delivery of electrons from glucose to O2 (8) is the source of metabolic free energy for all aerobic organisms, yielding 125 kJ per mole of electrons transferred.
A few other points about this plot are worth noting:
• Anaerobic organisms must make do with electron-acceptors above oxygen. The poor bacteria that depend on reducing hydrogen ions (2) have it worst of all, gaining only a tiny amount of metabolic energy to produce a tiny puff of hydrogen gas.
• Reaction (3) is not much more efficient, but it is the vital first link in the process of natural nitrogen fixation that gets carried out in the protected electron-rich environment of the organisms that live in the root nodules of legumes.
• The consequences of (4) are readily apparent if you have ever noticed the rotten-egg odor of some poorly-aerated muddy soils.
• Reaction (6) is known as fermentation; it takes place in the anarobic soils of marshes and bogs (hence the nickname "marsh gas" for methane) and in the insides of animals from termites to cows, and sometimes to our embarrassment, in ourselves. Notice that glucose plays the double role of electron donor and acceptor here; this is a disproportionation reaction.
• Finally, if you have ever had much to do with babys' diapers, you have likely noticed the smell of ammonia produced as the anaerobic bacteria from within deliver electrons to nitrate ions (7). It's important to bear in mind that the reactions discussed above are mediated by living organisms; without the necessary enzymes to catalyze them, their rates are essentially zero.
For a more detailed chart, see Falling through the respiratory chain.
Analytical chemistry applications
A very large part of Chemistry is concerned, either directly or indirectly, with determining the concentrations of ions in solution. Any method that can accomplish such measurements using relatively simple physical techniques is bound to be widely exploited. Cell potentials are fairly easy to measure, and although the Nernst equation relates them to ionic activities rather than to concentrations, the difference between them becomes negligible in solutions where the total ionic concentration is less than about 10–3 M.
The concentrations of ions in equilibrium with a sparingly soluble salt are sufficiently low that their direct determination can be quite difficult. A far simpler and common procedure is to set up a cell in which one of the electrode reactions involves the insoluble salt, and whose net cell reaction corresponds to the dissolution of the salt. For example, to determine the Ksp for silver chloride, we could use the cell
$Ag_{(s)} | Ag^+(?\; M) || Ag^+,Cl^– | AgCl_{(s)} | Ag_{(s)}$
whose net equation corresponds to the dissolution of silver chloride:
cathode:
AgCl(s) + e → Ag(s) + Cl(aq) = +0.222 v
anode:
Ag(s) → Ag+(aq) + e = –(+0.799) v
net:
AgCl(s) → Ag+ + Cl = -0.577 v
The standard potential for the net reaction refers to a hypothetical solution in which the activities of the two ions are unity. The cell potential we actually observe corresponds to E in the Nernst equation, which is then solved for Q which gives Ksp directly.
Potentiometric titrations
In many situations, accurate determination of an ion concentration by direct measurement of a cell potential is impossible due to the presence of other ions and a lack of information about activity coefficients. In such cases it is often possible to determine the ion indirectly by titration with some other ion. For example, the initial concentration of an ion such as Fe2+ can be found by titrating with a strong oxidizing agent such as Ce4+. The titration is carried out in one side of a cell whose other half is a reference electrode:
Pt(s) | Fe2+, Fe3+ || reference electrode
Initially the left cell contains only Fe2+. As the titrant is added, the ferrous ion is oxidized to Fe3+ in a reaction that is virtually complete:
$Fe^{2+} + Ce^{4+} → Fe^{3+} + Ce^{3+}$
The cell potential is followed as the Fe2+ is added in small increments. Once the first drop of ceric ion titrant has been added, the potential of the left cell is controlled by the ratio of oxidized and reduced iron according to the Nernst equation
$E = 0.68 - 0.059 \; \log \dfrac{]Fe^{3+}]}{[Fe^{2+}]}$
which causes the potential to rise as more iron becomes oxidized.
When the equivalence point is reached, the Fe2+ will have been totally consumed (the large equilibrium constant ensures that this will be so), and the potential will then be controlled by the concentration ratio of Ce3+/Ce4+. The idea is that both species of a redox couple must be present in reasonable concentrations poise an electrode (that is, to control its potential according to the Nernst equation.) If one works out the actual cell potentials for various concentrations of all these species, the resulting titration curve looks much like the familiar acid-base titration curve. The end point is found not by measuring a particular cell voltage, but by finding what volume of titrant gives the steepest part of the curve.
Measurement of pH
Since pH is actually defined in terms of hydrogen ion activity and not its concentration, a hydrogen electrode allows a direct measure of {H+} and thus of –log {H+}, which is the pH. All you need is to measure the voltage of a cell
H2(g, 1 atm) | Pt | H+(? M) || reference electrode
In theory this is quite simple, but when it was first employed in the pre-electronics era, it required some rather formidable-looking apparatus (such as the L&N vibrating-reed electrometer setup from the 1920's shown here) and the use of explosive hydrogen gas. Although this arrangement (in which the reference electrode could be a standard hydrogen electrode) has been used for high-precision determinations since that time, it would be impractical for routine pH measurements of the kinds that are widely done, especially outside the research laboratory.
The glass electrode for pH measurements
In 1914 it was discovered that a thin glass membrane enclosing a solution of HCl can produce a potential that varies with the hydrogen ion activity {H+} in about the same way as that of the hydrogen electrode. Glass electrodes are manufactured in huge numbers for both laboratory and field measurements. They contain a built-in Ag-AgCl reference electrode in contact with the HCl solution enclosed by the membrane.
The potential of a glass electrode is given by a form of the Nernst equation very similar to that of an ordinary hydrogen electrode, but of course without the H2:
Emembrane = A + (RT/F) ln ( {H+} + B )
in which A and B are constants that depend on the particular glass membrane.
The reason a glass membrane would behave in this way was not understood until around 1970. It now appears that hydrogen ions in the external solution diffuse through the glass and push out a corresponding number of the Na+ ions which are normally present in most glasses. These sodium ions diffuse to whichever side of the membrane has the lower concentration, where they remain mostly confined to the surface of the glass, which has a porous, gelatinous nature. It is the excess charge produced by these positive ions that gives rise to the pH-dependent potential. The first commercial pH meter was developed by Arnold Beckman (1900-2004) while he was a Chemistry professor at CalTech. He was unable to interest any of the instrumentation companies in marketing it, so he founded his own company and eventually became a multi-millionaire philanthropist.
Arnold Beckman
An early Beckman pH meter
A typical modern pH meter
Ion-selective electrodes
The function of the membrane in the glass electrode is to allow hydrogen ions to pass through and thus change its potential, while preventing other cations from doing the same thing (this selectivity is never perfect; most glass electrodes will respond to moderate concentrations of sodium ions, and to high concentrations of some others.) A glass electrode is thus one form of ion-selective electrode. Since about 1970, various other membranes have been developed which show similar selectivities to certain other ions. These are widely used in industrial, biochemical, and environmental applications.
Membrane potentials and nerve conduction
You may recall the phenomena of osmosis and osmotic pressure that are observed when two solutions having different solute concentrations are separated by a thin film or membrane whose porosity allows small ions and molecules to diffuse through, but which holds back larger particles. If one solution contains a pair of oppositely-charged ionic species whose sizes are very different, the smaller ions may pass through the semipermeable membrane while the larger ones are retained. This will produce a charge imbalance between the two solutions, with the original solution having the charge sign of the larger ion. Eventually the electrical work required to bring about further separation of charges becomes too large to allow any further net diffusion to take place, and the system settles into an equilibrium state in which a constant potential difference (usually around a volt or less) is maintained. This potential difference is usually called a membrane potential or Donnan potential after the English chemist who first described this phenomenon around 1930.
Origin of a membrane potential
If the smaller ions are able to diffuse through the membrane but the larger ions cannot, a potential difference will develop between the two solutions. This membrane potential can be observed by introducing a pair of platinum electrodes.
The figure shows a simple system containing the potassium salt of a protein on one side of a membrane, and potassium chloride on the other. The proteinate anion, being too large to diffuse through the membrane, gives rise to the potential difference. The value of this potential difference can be expressed by a relation that is essentially the same as the Nernst equation, although its derivation is different. The membrane potential can be expressed in terms of the ratio of either the K+ or Cl ion activities:
The membrane surrounding most living cells contains sites or "channels" through which K+ ions are selectively transported so that the concentration of K+ inside the cell is 10-30 times that of the intracellular fluid. Taking the activity ratio as about 20, the above equation predicts that the potential difference θinside - θoutside will be
which is consistent with observed values. Transport of an ion such as K+ from a region of low concentration into the more concentrated intercellular fluid requires a source of free energy, which is supplied by ATP under enzymatic control. The metabolic processes governing this action are often referred to as "ion pumps".
Nerve conduction
Transmission of signals through the nervous system occurs not by the movement of a charge carrier through the nerve, but by waves of differential ion concentrations that travel along the length of the nerve. These concentration gradients are reduced by protein-based ion channels and ATP-activated (and energy-consuming) ion pumps specific to K+ and Ca2+ ions.
We sometimes think of our nerves as the body's wiring, but the "electricity" that they transmit is not a flow of electrons, but a rapidly-traveling wave of depolarization involving the transport of ions through the nerve membrane. The normal potential difference between the inner and outer parts of nerve cells is about –70 mv as estimated above. Transmissin of a nerve impulse is initiated by a reduction of this potential difference to about –20 mv. This has the effect of temporarily opening the Na+ channel; the influx of these ions causes the membrane potential of the adjacent portion of the nerve to collapse, leading to an effect that is transmitted along the length of the nerve. As this pulse passes, K+ and Na+ pumps restore the nerve to its resting condition.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Atmospheric oxygen is a strong oxidizing agent, but in the absence of a suitable catalyst this reaction is ordinarily too slow to be observable. In organisms that possess the appropriate enzymes, the reaction is sufficiently fast to control the availability of electrons to other redox systems.
• The pE is a useful index of electron availability, especially in natural waters and biological systems in which multiple redox systems are usually present. Although electrons tend to "fall" (in free energy) from sources to sinks, the outcome depends very strongly on the rates of the various reactions. This is quite different from proton-exchange (acid-base) reactions which are uniformly fast.
• Solubility products can often be measured by constructing a cell in which one of the electrodes involves the sparingly soluble salt, and the net cell equation corresponds to the solubility reaction.
• Potentiometric titrations are widely used to measure concentrations of species that are readily oxidized or reduced. The reaction must be fast and have a very large equilibrium constant. The equivalance point is detected by the rapid change in potential that occurs when control of the cell potential passes from the redox system of the analyte to that of the titrant.
• Measurement of pH is accomplished with an electrode consisting of a thin membrane of glass in which Na+ ions are exchanged with H+.
• Membrane (Donnan) potentials arise when passage of an ion across a semipermeable membrane is selectively facilitated or inhibited. The classic example is a sodium proteinate solution in which the protein anion is too large to pass through the membrane. In organisms, ion-specific channels or "pumps" have a similar effect.
• Nerve conduction occurs not by a flow of charge through the nerve, but by a wave of depolarization resulting from the concerted action of active channels that govern the passage of ions (mainly K+ and Na+) across the membrane enclosing the nerve. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.05%3A_Applications_of_the_Nernst_Equation.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented below.
• A battery is a galvanic cell in which some of the free energy change associated with a spontaneous electron-transfer reaction is captured in the form of electrical energy.
• A secondary or storage battery is one in which the electron-transfer reaction can be reversed by applying a charging current from an external source.
• A fuel cell is a special type of battery in which the reactants are supplied from an external source as power is produced. In most practical fuel cells, H+ions are produced at the anode (either from H2 or a hydrocarbon) and oxygen from the air is reduced to H2O at the cathode.
• The cathodic reduction of O2 is kinetically limited, necessitating the use of electrode surfaces having high catalytic activity.
• The electrodes in batteries must have very high effective surface areas, and thus be highly porous. This requirement may conflict with the other important one of efficient diffusion of reactants and products in the narrow channels within the pores.
• Batteries and fuel cells designed to power vehicles and portable devices need to have high charge-to-weight and charge-tovolume ratios.
One of the oldest and most important applications of electrochemistry is to the storage and conversion of energy. You already know that a galvanic cell converts chemical energy to work; similarly, an electrolytic cell converts electrical work into chemical free energy. Devices that carry out these conversions are called batteries. In ordinary batteries the chemical components are contained within the device itself. If the reactants are supplied from an external source as they are consumed, the device is called a fuel cell.
Introduction
The term battery derives from the older use of this word to describe physical attack or "beating"; Benjamin Franklin first applied the term to the electrical shocks that could be produced by an array of charged glass plates. In common usage, the term "call" is often used in place of battery. For portable and transportation applications especially, a battery or fuel cell should store (and be able to deliver) the maximum amount of energy at the desired rate (power level) from a device that has the smallest possible weight and volume. The following parameters are commonly used to express these attributes:
• Storage capacity or charge density, coulombs/liter or coulombs/kg;
• Energy density, J/kg or watt-hour/lb
• Power density, watts/kg
• Voltage efficiency, ratio of output voltage to E°
• Lifetime: shelf-life (resistance to self-discharge) or charge/recharge cycles
Primary and Secondary Batteries
A secondary or storage battery is capable of being recharged; its electrode reactions can proceed in either direction. During charging, electrical work is done on the cell to provide the free energy needed to force the reaction in the non-spontaneous direction. A primary cell, as expemplified by an ordinary flashlight battery, cannot be recharged with any efficiency, so the amount of energy it can deliver is limited to that obtainable from the reactants that were placed in it at the time of manufacture.
The lead-acid storage cell
The most well-known storage cell is the lead-acid cell, which was invented by Gaston Planté in 1859 and is still the most widely used device of its type. The cell is represented by
$Pb(s) | PbSO_4(s) | H_2SO_4(aq) || PbSO_4(s), PbO_2(s) | Pb(s)$
and the net cell reaction is
$Pb(s) + PbO_2(s) + 2 H_2SO_4(aq) → 2 PbSO_4(s) + 2 H_2O$
The reaction proceeds to the right during discharge and to the left during charging. The state of charge can be estimated by measuring the density of the electrolyte; sulfuric acid is about twice as dense as water, so as the cell is discharged, the density of the electrolyte decreases.
The technology of lead-acid storage batteries has undergone remarkably little change since the late 19th century. Their main drawback as power sources for electric vehicles is the weight of the lead; the maximum energy density is only about 35 Ah/kg, and actual values may be only half as much. There are also a few other problems:
• The sulfuric acid electrolyte becomes quite viscous when the temperature is low, inhibiting the flow of ions between the plates and reducing the current that can be delivered. This effect is well-known to anyone who has had difficulty starting a car in cold weather.
• These batteries tend to slowly self-discharge, so a car left idle for several weeks might be unable to start.
• Over time, PbSO4 that does not get converted to PbO2 due to lack of complete discharge gradually changes to an inert form which limits the battery capacity. Also, "fast" charging causes rapid evolution of hydrogen from the water in the electrolyte; the bubbles form on the lead surface and can tear PbO2 off the positive plate. Eventually enough solid material accumulates at the bottom of the electrolyte to short-circuit the battery, leading to its permanent demise.
The LeClanché "dry cell"
The most well-known primary battery has long been the common "dry cell" that is widely used to power flashlights and similar devices. The modern dry cell is based on the one invented by Georges Leclanché in 1866. The electrode reactions are
$Zn → Zn^{2+} + 2e^–$
$2 MnO_2 + 2H^+ + 2e^– → Mn_2O_3 + H_2O$
Despite its name, this cell is not really "dry"; the electrolyte is a wet paste containing NH4Cl to supply the hydrogen ions. The chemistry of this cell is more complicated than it would appear from these equations, and there are many side reactions and these cells have limited shelf-lifes due to self discharge. (In some of the older ones, attack by the acidic ammonium ion on the zinc would release hydrogen gas, causing the battery to swell and rupture, often ruining an unused flashlight or other device.) A more modern version, introduced in 1949, is the alkaline cell which employs a KOH electrolyte and a zinc-powder anode which permits the cell to deliver higher currents and avoids the corrosive effects of the acidic ammonium ion on the zinc.
Exercise $1$
What is the cell notation for the LeClanché dry cell?
Physical limitations of Battery Performance
The most important of these are:
• Effective surface area of the electrode. A 1-cm2 sheet of polished metal presents far less active surface than does one that contains numerous surface projections or pores. All useful batteries and fuel cells employ highly porous electrodes. Recent advances in nanotechnology are likely to greatly improve this parameter.
• Current density of electrode surface. Expressed in amperes m–2, this is essentially a measure of the catalytic ability of the electrode, that is, its ability to reduce the activation energy of the electron transfer process.
• Rate at which electroactive components can be delivered to or depart from the active electrode surface. These processes are controlled by thermal diffusion and are inhibited by the very narrow pores that are needed to produce the large active surface area.
• Side reactions and irreversible processes. The products of the discharge reaction may tend to react with the charge-storing components. Thermal diffusion can also cause self-discharge, limiting the shelf life of the battery. Recharging of some storage batteries may lead to formation of less active modifications of solid phases, thus reducing the number of charge/discharge cycles possible.
Clearly, these are all primarily kinetic and mechanistic factors which require a great deal of experimentation to understand and optimize.
The Fuel Cell
Conventional batteries supply electrical energy from the chemical reactants stored within them; when these reactants are consumed, the battery is "dead". An alternative approach would be to feed the reactants into the cell as they are required, so as to permit the cell to operate continuously. In this case the reactants can be thought of as "fuel" to drive the cell, hence the term fuel cell.
Although fuel cells were not employed for practical purposes until space exploration began in the 1960's, the principle was first demonstrated in 1839 by Sir William Grove, a Welsh lawyer and amateur chemist. At the time, it was already known that water could be decomposed into hydrogen and oxygen by electrolysis; Grove tried recombining the two gases in a simple apparatus, and discovered what he called "reverse electrolysis"— that is, the recombination of H2 and O2 into water— causing a potential difference to be generated between the two electrodes:
H2(g) → 2 H+ + 2e = 0 v
½ O2 + 2 H+ + 2e → H2O(l) = +1.23 v
H2(g) + ½ O2(g) → H2O(l) = +1.23 v
It was not until 1959 that the first working hydrogen-oxygen fuel cell was developed by Francis Thomas Bacon in England. Modern cells employ an alkaline electrolyte, so the electrode reactions differ from the one shown above by the addition of OH to both sides of the equations (note that the net reaction is the same):
H2(g) + 2 OH → 2 H2O + 2 e = 0 v
½ O2 (g) + 2 H2O + 2 e → 2 OH = +1.23 v
H2(g) + ½ O2(g) → H2O = +1.23 v
Although hydrogen has the largest energy-to-mass ratio of any fuel, it cannot be compressed to a liquid at ordinary temperatures. If it is stored as a gas, the very high pressures require heavy storage containers, greatly reducing its effective energy density. Some solid materials capable of absorbing large amount of H2 can reduce the required pressure. Other fuels such as alcohols, hydrocarbon liquids, and even coal slurries have been used; methanol appears to be an especially promising fuel.
One reason for the interest in fuel cells is that they offer a far more efficient way of utilizing chemical energy than does conventional thermal conversion. The work obtainable in the limit of reversible operation of a fuel cell is 229 kJ per mole of H2O formed. If the hydrogen were simply burned in oxygen, the heat obtainable would be ΔH = 242 kJ mol–1, but no more than about half of this heat can be converted into work so the output would not exceed 121 kJ mol–1. This limit is a consequence of the Second Law of Thermodynamics. The fraction of heat that can be converted into work ($\eta$) is a function of how far (in temperature) the heat falls as it flows through the engine and into the surroundings; this fraction is given by
$\eta=\dfrac{1 - T_{high}}{T_{low}}$
At normal environmental temperatures of around 300 K, this would have to be at least 600 K for 50% thermal efficiency.
The major limitation of present fuel cells is that the rates of the electrode reactions, especially the one in which oxygen is reduced, tend to be very small, and thus so is the output current per unit of electrode surface. Coating the electrode with a suitable catalytic material is almost always necessary to obtain usable output currents, but good catalysts are mostly very expensive substances such as platinum, so that the resulting cells are too costly for most practical uses. There is no doubt that if an efficient, low-cost catalytic electrode surface is ever developed, the fuel cell would become a mainstay of the energy economy.
Microbobial Fuel Cells
Certain types of bacteria are able to oxidize organic compounds to carbon dioxide while directly transferring electrons to electrodes. These so-called electricigen organisms may make it possible to convert renewable biomass and organic waste directly into electricity without the wasted energy and pollution produced by direct combustion. In one experiment, a graphite electrode immersed in ordinary mud (containing humic materials) was able to produce measurable amounts of electricity.
Summary
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the highlighted terms in the context of this topic.
• A battery is a galvanic cell in which some of the free energy change associated with a spontaneous electron-transfer reaction is captured in the form of electrical energy.
• A secondary or storage battery is one in which the electron-transfer reaction can be reversed by applying a charging current from an external source.
• A fuel cell is a special type of battery in which the reactants are supplied from an external source as power is produced. In most practical fuel cells, H+ ions are produced at the anode (either from H2 or a hydrocarbon) and oxygen from the air is reduced to H2O at the cathode.
• The cathodic reduction of O2 is kinetically limited, necessitating the use of electrode surfaces having high catalytic activity.
• The electrodes in batteries must have very high effective surface areas, and thus be highly porous. This requirement may conflict with the other important one of efficient diffusion of reactants and products in the narrow channels within the pores.
• Batteries and fuel cells designed to power vehicles and portable devices need to have high charge-to-weight and charge-tovolume ratios. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.06%3A_Batteries_and_Fuel_Cells.txt |
Although the development practical batteries largely paralelled the expansion of electrical technology from about the mid-19th century on, it is now thought that a very primitive kind of battery was apparently in use more than 2000 years ago. The brief popularity of electrically powered automobiles in the 1920's encouraged storage battery development. The widespread use of portable "personal" electrical devices has kept the search for better batteries very much alive.
"Baghdad Battery" - 1000 BCE?
Drawing of the three pieces. (CC-BY-SA 2.5; Ironie)
Earthenware jars containing an iron rod surrounded by a copper cylinder were discovered near Baghdad in 1938. They are believed to have been used by the Parthian civilization that occupied the region about 2000 years ago as a source of electricity to plate gold onto silver.
Allesandro Volta 1782
His "Voltaic pile", a stack of zinc and silver disks separated by a wet cloth containing a salt or a weak acid solution, was the first battery known to Western civilization.
Sir Humphry Davy 1813 Davy builds a 2000-plate battery that occupies 889 square feet in the basement of Britain's Royal Society. His earlier batteries provided power for the first public demonstration of electric lighting (carbon arc).
Michael Faraday, 1830's Faraday discovered the fundamentals of galvanic cells and electrolysis that put electrochemistry on a firm scientific basis.
1836 - Daniell cell (also known as a Crow's Foot or Gravity cell.)
John Daniell (English chemist and meterologist) developed the first modern storage cell based on Faraday's principles. This consists of a large glass jar with a copper star-shaped electrode in the bottom and a zinc "crow's foot" shaped electrode suspended near the top. The bottom of the jar was filled with a concentrated copper sulfate solution. On top of this was poured dilute sulfuric acid, whose lower density kept it on top. This was the first practical battery to find wide use to power telegraphs and railway signaling systems and home doorbells.
1839 - William Grove (Welsh)
Grove was best known in the 19th century for his "nitric acid battery" which came into wide use in early telegraphy.
Now, however, he is most famous for his "gas voltaic battery" in which discovered "reverse electrolysis": the recombination of H2 and O2 following electrolysis of water at platinum electrodes. This was the first demonstration of what we now know as the hydrogen-oxygen fuel cell (see below.)
1859 - Gaston Planté (French)
Invents the first lead-acid storage cell which consisted of two sheets of lead separated by a rubber sheet, rolled into a spiral and immersed in dilute sulfuric acid.
1866 - Georges Leclanché (French)
By 1868 twenty thousand Leclanché cells were being used in telegraph systems. The original Leclanché cells were built in porous pots which were heavy and subject to breakage. Within twenty years other inventors had modified the design into what we now know as "dry cells" which became widely used in the first flashlights (1909) and in battery-powered radios of the 1920s.
1881 - Faure and others
Development of the first practical lead-acid storage cell. The major improvement over Planté's design was the addition of a paste of PbSO4 to the positive plate.
1905 Nickel-iron cell
(Thomas Edison)
Edison, who was as much a chemist as an all-around inventor, thought that the lead in Planté-type cells made them too heavy, and that having acid in contact with any metal was an inherently bad idea. After much experimentation, he developed a successful alkaline battery. The Edison cell uses an iron anode, nickel oxide cathode, and KOH electrolyte. This cell is extremely rugged and is still used in certain industrial applications, but it was never able to displace the lead-acid cell as Edison had hoped.
1950s A similar cell, employing a nickel anode instead of iron, was the first rechargeable cell that was small enough to be used in portable consumer devices. Its main disadvantage is that it is ruined by complete discharge.
1949 - Alkaline dry cell - Lew Urry
(Eveready Battery Co.)
First commercial alkaline dry cell. These substitute KOH for the corrosive NH4Cl used in the older dry cells and last 5-8 times longer.
1947 - Mercury cell
(Ruben and Mallory, 1950's)
This was one of the first "button"-type cells which were widely used in cameras and hearing aids. The constancy of the 1.34 v output made them popular for use in sensitive instruments and cardiac pacemakers. The net cell reaction is
Zn(s) + HgO(s) → ZnO(s) + Hg(l)
Most countries have outlawed sales of these cells in order to reduce mercury contamination of the environment.
Nickel-Cadmium (NiCad) cells
The NiCad cell quickly become one of the most popular rechargeable batteries for small consumer devices. They can deliver high current and undergo hundreds of charge/discharge cycles. Because cadmium is an environmental toxin, their use is being discouraged.
1959 - Fuel cell - Francis Bacon (UK) The first practical fuel cell was developed by British engineer Francis Bacon (1904-1992). This hydrogen-oxygen cell used an alkaline electrolyte and inexpensive nickel electrodes.
Late 1960's - Nickel-metal hydride cells
The hydride ion H would be an ideal cathode material except for the fact that its oxidation product H2 is a gas. The discovery that certain compounds such as LiNi5 and ZrNi2 can act as "hydrogen sponges" made it practical to employ metal hydrides as a cathode material. One peculiarity of Ni-MH cells is that recharging them is an exothermic process, so that proper dissipation of heat must be allowed for. These batteries are widely used in cell phones, computers, and portable power tools. The electrode reactions take place in a concentrated KOH electrolyte:
Cathode (+): NiOOH + H2O + e→ Ni(OH)2 + OH
Anode (-): (1/x) MHx + OH→ (1/x) M + H2O + e
1990s - Lithium cells (Sony Corp.)
Lithium is an ideal anode material owing to its low density and high reduction potential, making Li-based cells the most compact ways of storing electrical energy. Lithium cells are used in wristwatches, cardiac pacemakers and digital cameras. Both primary (non-rechargeble) and rechargeable types have been available for some time. More recent applications are in portable power tools and— perhaps most importantly, in electric-powered or hybrid automobiles.
Modern lithium cells operate by transporting Li+ ions between electrodes into which the ions can be inserted or intercalated. Cathodes are lithium transition-metal oxides such as LiCoO3, while anodes are lithium-containing carbon, LiC6. The species that undergoes oxidation-reduction is not lithium, but the transition metal, e.g. Co(III)-Co(IV).
Lithium batteries as incendiary devices
There have been numerous reports of fires and explosions associated with lithium batteries. In 2006, the Dell Corporation had to recall 4.1 million Sony batteries that had been shipped with Dell's laptop computers and were judged to be at risk owing to a manufacturing defect. This illustrates the difficulty of concentrating a large amount of chemical energy into a small package, which is of course the goal of all battery developers eager to meet commercial demands ranging from consumer personal electronics to electrically-powered cars. The fully-charged Li+-deficient lithium cobalt oxide cathodes are inherently unstable, held in check only by a thin insulating membrane which, if accidentally breached, can lead to thermal runaway involving gaseous oxygen, carbon, organic solvents, and (in some cases) lithium chlorate— all the components necessary for a fierce fire.
Much research has gone into the development of fail-safe membranes. In one type, made by ExxonMobil and targeted at the automotive market, the pores are designed to close up and thus inhibit the passage of lithium ions when the temperature rises above a safe level.
Biological Batteries
Finally, we should mention the biological batteries that are found in a number of electric fish. The "electric organs" of these fish are modified muscle cells known as electrocytes which are arranged in long stacks. A neural signal from the brain causes all the electrocytes in a stack to become polarized simultaneously, in effect creating a battery made of series-connected cells. Most electric fish produce only a small voltage which they use for navigation, much in the way that bats use sound for echo-location of prey. The famous electric eel, however, is able to produce a 600-volt jolt that it employs to stun nearby prey. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.07%3A_Timeline_of_Battery_Development.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas. It is especially important that you know the precise meanings of all the highlighted terms in the context of this topic.
• Electrochemical corrosion of metals occurs when electrons from atoms at the surface of the metal are transferred to a suitable electron acceptor or depolarizer. Water must be present to serve as a medium for the transport of ions.
• The most common depolarizers are oxygen, acids, and the cations of less active metals.
• Because the electrons flow through the metallic object itself, the anodic and cathodic regions (the two halves of the electrochemical cell) can be at widely separated locations.
• Anodic regions tend to develop at locations where the metal is stressed or is protected from oxygen.
• Contact with a different kind of metal, either direct or indirect, can lead to corrosion of the more active one.
• Corrosion of steel can be inhibited by galvanizing, that is, by coating it with zinc, a more active metal whose dissolution leaves a negative charge on the metal which inhibits the further dissolution of Fe2+.
• Cathodic protection using an external voltage source is widely used to protect underground structures such as tanks, pipelines and piers. The source can be a sacrificial anode of zinc or aluminum, or a line-operated or photovoltaic power supply.
Corrosion can be defined as the deterioration of materials by chemical processes. Of these, the most important by far is electrochemical corrosion of metals, in which the oxidation process M → M+ + e is facilitated by the presence of a suitable electron acceptor, sometimes referred to in corrosion science as a depolarizer.
In a sense, corrosion can be viewed as the spontaneous return of metals to their ores; the huge quantities of energy that were consumed in mining, refining, and manufacturing metals into useful objects is dissipated by a variety of different routes. The economic aspects of corrosion are far greater than most people realize; the estimated cost of corrosion in the U.S. alone was $276 billion per year. Of this, about$121 billion was spent to control corrosion, leaving the difference of \$155 billion as the net loss to the economy. Utilities, especially drinking water and sewer systems, suffer the largest economic impact, with motor vehicles and transportation being a close second.
Corrosion Cells and Reactions
The special characteristic of most corrosion processes is that the oxidation and reduction steps occur at separate locations on the metal. This is possible because metals are conductive, so the electrons can flow through the metal from the anodic to the cathodic regions (Figure $1$). The presence of water is necessary in order to transport ions to and from the metal, but a thin film of adsorbed moisture can be sufficient.
A corrosion system can be regarded as a short-circuited electrochemical cell in which the anodic process is something like
$\ce{Fe(s) \rightarrow Fe^{2+}(aq) + 2 e^{-}} \label{1.7.1}$
and the cathodic steps may invove the reduction of oxygen gas
$\ce{O_2} + \ce{2 H_2O} + \ce{4e^{-}} \rightarrow \ce{4 OH^{-}} \label{1.7.2}$
or the reduction of protons
$\ce{H^{+} + e^{-}} \rightarrow \ce{1/2 H2(g)} \label{1.7.2b}$
or the reduction of a metal ion
$\ce{M^{2+} + 2 e^{–}} \rightarrow \ce{M(s)} \label{1.7.2c}$
where $\ce{M}$ is a metal.
Which parts of the metal serve as anodes and cathodes can depend on many factors, as can be seen from the irregular corrosion patterns that are commonly observed. Atoms in regions that have undergone stress, as might be produced by forming or machining, often tend to have higher free energies, and thus tend to become anodic.
If one part of a metallic object is protected from the atmosphere so that there is insufficient $\ce{O2}$ to build or maintain the oxide film, this "protected" region will often be the site at which corrosion is most active. The fact that such sites are usually hidden from view accounts for much of the difficulty in detecting and controlling corrosion.
In contrast to anodic sites, which tend to be localized to specific regions of the surface, the cathodic part of the process can occur almost anywhere. Because metallic oxides are usually semiconductors, most oxide coatings do not inhibit the flow of electrons to the surface, so almost any region that is exposed to $\ce{O2}$ or to some other electron acceptor can act as a cathode. The tendency of oxygen-deprived locations to become anodic is the cause of many commonly-observed patterns of corrosion.
Rusted-out Cars and Bathroom Stains
Anyone who has owned an older car has seen corrosion occur at joints between body parts and under paint films. You will also have noticed that once corrosion starts, it tends to feed on itself. One reason for this is that one of the products of the O2 reduction reaction is hydroxide ion. The high pH produced in these cathodic regions tends to destroy the protective oxide film, and may even soften or weaken paint films, so that these sites can become anodic. The greater supply of electrons promotes more intense cathodic action, which spawns even more anodic sites, and so on.
A very common cause of corrosion is having two dissimilar metals in contact, as might occur near a fastener or at a weld joint. Moisture collects at the junction point, acting as an electrolyte and forming a cell in which the two metals serve as electrodes. Moisture and conductive salts on the outside surfaces provide an external conductive path, effectively short-circuiting the cell and producing very rapid corrosion; this is why cars rust out so quickly in places where salt is placed on roads to melt ice.
Dissimilar-metal corrosion can occur even if the two metals are not initially in direct contact. For example, in homes where copper tubing is used for plumbing, there is always a small amount of dissolved $\ce{Cu^{2+}}$ in the water. When this water encounters steel piping or a chrome-plated bathroom sink drain, the more-noble copper will plate out on the other metal, producing a new metals-in-contact corrosion cell. In the case of chrome bathroom sink fittings, this leads to the formation of $\ce{Cr^{3+}}$ salts which precipitate as greenish stains.
Control of Corrosion
Since both the cathodic and anodic steps must take place for corrosion to occur, prevention of either one will stop corrosion. The most obvious strategy is to stop both processes by coating the object with a paint or other protective coating. Even if this is done, there are likely to be places where the coating is broken or does not penetrate, particularly if there are holes or screw threads. A more sophisticated approach is to apply a slight negative charge to the metal, thus making it more difficult for the reaction to take place:
$\ce{M -> M^{2+} + 2 e^{-}}.$
Protection Method 1: Sacrificial Coatings
One way of supplying this negative charge is to apply a coating of a more active metal. Thus a very common way of protecting steel from corrosion is to coat it with a thin layer of zinc; this process is known as galvanizing.The zinc coating, being less noble than iron, tends to corrode selectively. Dissolution of this sacrificial coating leaves behind electrons which concentrate in the iron, making it cathodic and thus inhibiting its dissolution.
The effect of plating iron with a less active metal provides an interesting contrast. The common tin-plated can (on the right) is a good example. As long as the tin coating remains intact, all is well, but exposure of even a tiny part of the underlying iron to the moist atmosphere initiates corrosion. The electrons released from the iron flow into the tin, making the iron more anodic so now the tin is actively promoting corrosion of the iron! You have probably observed how tin cans disintegrate very rapidly when left outdoors.
Protection Method 2: Cathodic Protection
A more sophisticated strategy is to maintain a continual negative electrical charge on a metal, so that its dissolution as positive ions is inhibited. Since the entire surface is forced into the cathodic condition, this method is known as cathodic protection. The source of electrons can be an external direct current power supply (commonly used to protect oil pipelines and other buried structures), or it can be the corrosion of another, more active metal such as a piece of zinc or aluminum buried in the ground nearby, as is shown in the illustration of the buried propane storage tank below.
16.09: Corrosion Gallery
Corrosion gallery
Corrosion of a nail
The nails are immersed in agar which forms a moist solid gel. The agar also contains phenolphthalein and hexacyanoiron(III) Fe(CN6) which forms a deep blue color ("prussian blue") in the presence of Fe2+. The blue colors are clearly associated with those parts of the nail that have been stressed, thus faciliting the anodic release of Fe2+ from the metal. The the pink color shows the cathodic regions that have been made alkaline by the reaction
O2 + 2 H2O + 4e → 4 OH
This clearly shows the separation between the anodic and cathodic processes in corrosion.
[Illustration from U of West Indies: link]
Water distribution main
If you live in the older part of a city where the mains are 50-100 years old, the water you drink may well have passed through a pipe in this condition! Severe corrosion like this is more common in areas where the water is acidic. Such water comes from mountain snowmelt and runoff, and usually acquires its acidity from dissolved atmospheric carbon dioxide.
Waters from rivers, lakes, and especially groundwaters from wells have usually been in sufficiently long contact with carbonate-containing sediments to have been neutralized. Water-works administrators like to make the water slightly alkaline and slightly supersaturated in calcium carbonate in order to maintain a thin coating of solid carbonate on the interior of the pipe which acts to protect it from corrosion.
Corrosion of reinforcing bars in concrete
All large concrete structures contain steel reinforcing bars ("rebars") that help ensure structural integrity under varying load conditions and especially during earthquakes. Intrusion of water, even in the form of fog or mists, can lead to serious corrosion damage, as seen in this picture of this column which supports a highway overpass.
Corrosion at metallic joints
The picture shows two steel structural members joined by cast iron flanges which have been bolted together. For some reason, one of the pieces has become more anodic than the other, leading to extensive corrosion of the upper part.
Bacterial-assisted corrosion
This gas pipe was buried in a red soil that contained iron pyrites (FeS.) The bacterium thiobacillus ferrooxidans derives its energy by oxidizing Fe2+ to the more soluble Fe3+, transferring the electrons to O2. It also oxidizes the sulfur, producing sulfuric acid. The resulting chemical cocktail has eaten a hole into the pipe.
link
These galvanized bolts were used to join wooden beams together. Subsequent movement of the beams due to varying load conditions abraded the zinc coating. A lack of oxygen near the centers of the bolts also likely contributed to the corrosion by preventing the formation of a protective oxide film.
Pitting corrosion | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.08%3A_Electrochemical_Corrosion.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas.
• In electrolysis, an external power source supplies the free energy required to drive a cell reaction in its non-spontaneous direction. An electrolytic cell is in this sense the opposite of a galvanic cell. In practice, the products of electrolysis are usually simpler than the reactants, hence the term electro-lysis.
• Transport of ions in the electrolyte in response to the potential difference between the electrodes ("drift") is largely restricted to the regions very close to the electrodes. Ionic transport in the greater part of the electrolyte is by ordinary thermal diffusion— the statistical tendency of concentrations to become uniform.
• When an aqueous solution is subjected to electrolysis, the oxidation or reduction of water can be a competing process and may dominate if the applied voltage is sufficiently great. Thus an attempt to electrolyze a solution of NaNO3 will produce only H2 and O2.
• A large number of electrolysis processes are employed by industry to refine metals and to produce both inorganic and organic products. The largest of these are the chloralkali industry (chlorine and "caustic"), and the refining of aluminum; the latter consumes approximately 5% of the electrical power generated in North America.
Electrolysis refers to the decomposition of a substance by an electric current. The electrolysis of sodium and potassium hydroxides, first carried out in 1808 by Sir Humphrey Davey, led to the discovery of these two metallic elements and showed that these two hydroxides which had previously been considered un-decomposable and thus elements, were in fact compounds.
Electrolysis of molten alkali halides is the usual industrial method of preparing the alkali metals:
Na+ + e → Na(l) = –2.71 V
Cl → ½ Cl2(g) + e = –1.36 V
Na+ + Cl → Na(l) + ½ Cl2(g) = –4.1 V
Ions in aqueous solutions can undergo similar reactions. Thus if a solution of nickel chloride undergoes electrolysis at platinum electrodes, the reactions are
Ni2+ + 2 e → Ni(s) = –0.24 V
2 Cl → Cl2(g) + 2 e = –1.36 V
Ni2+ + 2 Cl → Ni(s) + Cl2(g) = –1.60 v
Both of these processes are carried out in electrochemical cells which are forced to operate in the "reverse", or non-spontaneous direction, as indicated by the negative for the above cell reaction. The free energy is supplied in the form of electrical work done on the system by the outside world (the surroundings). This is the only fundamental difference between an electrolytic cell and the galvanic cell in which the free energy supplied by the cell reaction is extracted as work done on the surroundings.
A common misconception about electrolysis is that "ions are attracted to the oppositely-charged electrode." This is true only in the very thin interfacial region near the electrode surface. Ionic motion throughout the bulk of the solution occurs mostly by diffusion, which is the transport of molecules in response to a concentration gradient. Migration— the motion of a charged particle due to an applied electric field, is only a minor player, producing only about one non-random jump out of around 100,000 random ones for a 1 volt cm–1 electric field. Only those ions that are near the interfacial region are likely to undergo migration.
Electrolysis in aqueous solutions
Water is capable of undergoing both oxidation
$H_2O \rightarrow O_{2(g)} + 4 H^+ + 2 e^– \;\;\; E^o = -1.23 V$
and reduction
$2 H_2O + 2 e^– \rightarrow H_{2(g)} + 2 OH^– \;\;\; E^o = -0.83 \;V$
Thus if an aqueous solution is subjected to electrolysis, one or both of the above reactions may be able to compete with the electrolysis of the solute. For example, if we try to electrolyze a solution of sodium chloride, hydrogen is produced at the cathode instead of sodium:
H2O + 2 e → H2(g) + 2 OH E =+0.41 V ([OH] = 10-7 M)
Cl → ½ Cl2(g) + e = –1.36 V
Cl + H2O → 2 H2(g) + ½ Cl2(g) + 2 OH E = –0.95 V
[This illustration is taken from the excellent Purdue University Chemistry site]Reduction of Na+ (E° = –2.7 v) is energetically more difficult than the reduction of water (–1.23 V), so in aqueous solution the latter will prevail.
Electrolysis of salt ("brine") is carried out on a huge scale and is the basis of the chloralkali industry.
Electrolysis of water
Pure water is an insulator and cannot undergo significant electrolysis without adding an electrolyte. If the object is to produce hydrogen and oxygen, the electrolyte must be energetically more difficult to oxidize or reduce than water itself. Electrolysis of a solution of sulfuric acid or of a salt such as NaNO3 results in the decomposition of water at both electrodes:
H2O + 2 e → H2(g) + 2 OH E =+0.41 V ([OH] = 10-7 M)
2 H2O → O2(g) + 4 H+ + 2 e E° = -0.82 V
2 H2O(l) → 2 H2(g) + O2(g) E = -1.23 V
Electrolytic production of hydrogen is usually carried out with a dilute solution of sulfuric acid. This process is generally too expensive for industrial production unless highly pure hydrogen is required. However, it becomes more efficient at higher temperatures, where thermal energy reduces the amount of electrical energy required, so there is now some interest in developing high-temperature electrolytic processes. Most hydrogen gas is manufactured by the steam reforming of natural gas.
Faraday's laws of electrolysis
One mole of electric charge (96,500 coulombs), when passed through a cell, will discharge half a mole of a divalent metal ion such as Cu2+. This relation was first formulated by Faraday in 1832 in the form of two laws of electrolysis:
1. The weights of substances formed at an electrode during electrolysis are directly proportional to the quantity of electricity that passes through the electrolyte.
2. The weights of different substances formed by the passage of the same quantity of electricity are proportional to the equivalent weight of each substance.
The equivalent weight of a substance is defined as the molar mass, divided by the number of electrons required to oxidize or reduce each unit of the substance. Thus one mole of V3+ corresponds to three equivalents of this species, and will require three faradays of charge to deposit it as metallic vanadium.
Most stoichiometric problems involving electrolysis can be solved without explicit use of Faraday's laws. The "chemistry" in these problems is usually very elementary; the major difficulties usually stem from unfamiliarity with the basic electrical units:
• current (amperes) is the rate of charge transport; 1 amp = 1 C/sec.
• power (watts) is the rate of energy production or consumption; $1 W = 1 J/sec = 1 volt-amp; 1 watt-sec = 1 J, 1 kW-h = 3600 J.$
Example $1$
A metallic object to be plated with copper is placed in a solution of CuSO4.
1. To which electrode of a direct current power supply should the object be connected?
2. What mass of copper will be deposited if a current of 0.22 amp flows through the cell for 1.5 hours?
Solution
1. Since Cu2+ ions are being reduced, the object acts as a cathode and must be connected to the negative terminal (where the electrons come from!)
2. The amount of charge passing through the cell is
(0.22 amp) × (5400 sec) = 1200 C
or
(1200 C) ÷ (96500 c F–1) = 0.012 F
Since the reduction of one mole of Cu2+ ion requires the addition of two moles of electrons, the mass of Cu deposited will be
(63.54 g mol–1) (0.5 mol Cu/F) (.012 F) = 0.39 g of copper
Example $2$
How much electric power is required to produce 1 metric ton (1000 kg) of chlorine from brine, assuming the cells operate at 2.0 volts and assuming 100 % efficiency?
Solution
• moles of Cl2 produced: (106 g) ÷ 70 g mol–1 = 14300 mol Cl2
• faradays of charge: (2 F/mol) × (14300 mol) = 28600 F
• charge in coulombs: (96500 C/F) × (28600 F) = 2.76 × 109 C
• duration of electrolysis: (3600 s/h) x (24 h) = 86400 s
• current (rate of charge delivery): (2.76 × 109 amp-sec) ÷ (86400 sec) = 32300 amps
• power (volt-amps): (2.0 V) × (32300 amp) = 64.6 kW
• energy in kW-h: (64.6 kW) × (24 h) = 1550 kW-h
• energy in joules: (1550 kW-h) × (3.6 MJ/kW-h) = 5580 MJ (megajoules)
(In the last step, recall that 1 W = 1 J/s, so 1 kW-h = 3.6 MJ)
Industrial Electrolytic Processes
For many industrial-scale operations involving the oxidation or reduction of both inorganic and organic substances, and especially for the production of the more active metals such as sodium, calcium, magnesium, and aluminum, the most cost-effective reducing agent is electrons supplied by an external power source. The two most economically important of these processes are described below.
The chloralkali industry
The electrolysis of brine is carried out on a huge scale for the industrial production of chlorine and caustic soda (sodium hydroxide). Because the reduction potential of Na+ is much higher than that of water, the latter substance undergoes decomposition at the cathode, yielding hydrogen gas and OH.
anode
reactions
2 Cl → Cl2(g) + 2 e -1.36 V i
4 OH → O2(g) + 2 H2O + 4 e
-0.40 V ii
cathode
reactions
Na+ + e → Na(s) -2.7 V iii
H2O + 2 e → H2(g) + 2 OH
+.41 V iv
A comparison of the s would lead us to predict that the reduction (ii) would be favored over that of (i). This is certainly the case from a purely energetic standpoint, but as was mentioned in the section on fuel cells, electrode reactions involving O2 are notoriously slow (that is, they are kinetically hindered), so the anodic process here is under kinetic rather than thermodynamic control. The reduction of water (iv) is energetically favored over that of Na+ (iii), so the net result of the electrolysis of brine is the production of Cl2 and NaOH ("caustic"), both of which are of immense industrial importance:
$\ce{2 NaCl + 2 H2O -> 2 NaOH + Cl2(g) + H2(g)} \nonumber$
Since chlorine reacts with both OH and H2, it is necessary to physically separate the anode and cathode compartments. In modern plants this is accomplished by means of an ion-selective polymer membrane, but prior to 1970 a more complicated cell was used that employed a pool of mercury as the cathode. A small amount of this mercury would normally find its way into the plant's waste stream, and this has resulted in serious pollution of many major river systems and estuaries and devastation of their fisheries.
Electrolytic refining of aluminum
Aluminum is present in most rocks and is the most abundant metallic element in the earth's crust (eight percent by weight.) However, its isolation is very difficult and expensive to accomplish by purely chemical means, as evidenced by the high E° (–1.66 V) of the Al3+/Al couple. For the same reason, aluminum cannot be isolated by electrolysis of aqueous solutions of its compounds, since the water would be electrolyzed preferentially. And if you have ever tried to melt a rock, you will appreciate the difficulty of electrolyzing a molten aluminum ore! Aluminum was in fact considered an exotic and costly metal until 1886, when Charles Hall (U.S.A) and Paul Hérault (France) independently developed a practical electrolytic reduction process.
The Hall-Hérault process takes advantage of the principle that the melting point of a substance is reduced by admixture with another substance with which it forms a homogeneous phase. Instead of using the pure alumina ore Al2O3 which melts at 2050°C, it is mixed with cryolite, which is a natural mixture of NaF and AlF3, thus reducing the temperature required to a more manageable 1000°C. The anodes of the cell are made of carbon (actually a mixture of pitch and coal), and this plays a direct role in the process; the carbon gets oxidized (by the oxide ions left over from the reduction of Al3+ to CO, and the free energy of this reaction helps drive the aluminum reduction, lowering the voltage that must be applied and thus reducing the power consumption. This is important, because aluminum refining is the largest consumer of industrial electricity, accounting for about 5% of all electricity generated in North America. Since aluminum cells commonly operated at about 100,000 amperes, even a slight reduction in voltage can result in a large saving of power.
The net reaction is
$\ce{2 Al_2O_3 + 3 C \rightarrow 4 Al + 3 CO_2} \nonumber$
However, large quantities of CO and of HF (from the cryolite), and hydrocarbons (from the electrodes) are formed in various side reactions, and these can be serious sources of environmental pollution. | textbooks/chem/General_Chemistry/Chem1_(Lower)/16%3A_Electrochemistry/16.10%3A_Electrolytic_Cells_and_Electrolysis.txt |
The chapters in this section deal with the rates and mechanisms of chemical change. These topics stand in contrast to the subjects of equilibrium and thermodynamics that control the direction of chemical change. Chemical change is guided and driven by energetics (thermodynamics), but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products.
• 17.1: Rates of reactions and rate laws
Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products. The energetic aspects of change are governed by the laws of thermodynamics.
• 17.2: Reaction Rates Typically Change with Time
On this page we extend the concept of differential rate laws introduced on the previous page to integral rate laws and reaction half-life that are of great importance in most practical applications of kinetics.
• 17.3: Collision and activation- the Arrhenius law
Why are some reactions so much faster than others, and why are reaction rates independent of the thermodynamic tendency of the reaction to take place? These are the central questions we address in this unit. In doing so, we open the door to the important topic of reaction mechanisms: what happens at the microscopic level when chemical reactions take place? To keep things as simple as possible, we will restrict ourselves to reactions that take place in the gas phase.
• 17.4: Reaction Mechanisms
The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collision") or as the breaking-up of a molecule ("dissociation") into simpler units.
• 17.5: Kinetics of Reactions in Solution
The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions.
• 17.6: Catalysts and Catalysis
Catalysts play an essential role in our modern industrial economy, in our stewardship of the environment, and in all biological processes. This lesson will give you a glimpse into the wonderful world of catalysts, helping you to understand what they are and how they work.
• 17.7: Experimental methods of chemical kinetics
Studies of the dynamics of chemical processes impinge on almost every area of chemistry and biochemistry. It is useful for students even at the general chemistry level to have some understanding of the experimental techniques that have informed what you have already learned about kinetics.
17: Chemical Kinetics and Dynamics
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Describe the contrasting roles of thermodynamics and kinetics in understanding chemical change.
• Given a balanced net equation, write an expression for the rate of a reaction.
• Sketch a curve showing how the instantaneous rate of a reaction might change with time.
• Determine the order of a reaction of the form A → B + C from experimental data for the concentrations of its products at successive times.
• Describe the initial rate and isolation methods of determining the orders of the individual reactants in a reaction involving multiple reactants.
• Explain the difference between differential and integral rate laws.
• Sketch out a plot showing how the concentration of a component ([A] or ln [A]) that follows first-order kinetics will change with time. Indicate how the magnitude of the rate constant affects this plot
• Define the half-life of a reaction.
• Given the half-life for a first-order reaction A → products along with the initial value of [A]o, find [A]t at a subsequent time an integral number of half-lifes later.
• Describe the conditions under which a reaction can appear to have an order of zero.
Chemical change is guided and driven by energetics, but the actual route it takes and the speed with which it occurs is the subject of "dynamics". Dynamics is itself divided into two general areas: kinetics, which deals with the rate of change and is the subject of this lesson. Mechanistics, introduced in a later lesson, is an exploration of the "road map" that links reactants to products.
The energetic aspects of change are governed by the laws of thermodynamics (the "dynamics" part of this word is related to the historical origins of the field and is not a part of dynamics in the sense of these lessons.)
Energetics + dynamics = chemical change
Chemical change is driven by the tendency of atoms and molecules to rearrange themselves in a way that results in the maximum possible dispersion of thermal energy into the world. The observable quantity that measures this spreading and sharing of energy is the free energy of the system. As a chemical change takes place, the quantities of reactants and products change in a way that leads to a more negative free energy. When the free energy reaches its minimum possible value, there is no more net change and the system is said to be in equilibrium.
The beauty of thermodynamics is that it enables us to unfailingly predict the net direction of a reaction and the composition of the equilibrium state even without conducting the experiment; the standard free energies of the reactants and products, which can be independently measured or obtained from tables, are all we need.
Half the Story
Thermodynamics points the way and makes it possible...but it says nothing about how long it will take to get there!
It is worth noting that the concept of "time" plays no role whatsover in thermodynamics. But kinetics is all about time. The "speed" of a reaction — how long it takes to reach equilibrium — bears no relation at all to how spontaneous it is (as given by the sign and value of ΔG°) or whether it is exothermic or endothermic (given by the sign of ΔH°). Moreover, there is no way that reaction rates can be predicted in advance; each reaction must be studied individually. One reason for this is that
The stoichiometric equation for the reaction says nothing about its mechanism. By mechanism, we mean, basically, "who does what to whom". Think of a reaction mechanism as something that goes on in a "black box" that joins reactants to products:
The inner workings of the black box are ordinarily hidden from us, are highly unpredictable and can only be inferred by indirect means. Consider, for example, the gas-phase formation reactions of the hydrogen halides from the elements.
The Balanced equation says nothing about a Reaction's Mechanism
The thermodynamics of these reactions are all similar (they are all highly exothermic), but their dynamics (their kinetics and mechanisms) could not be more different.
$H_{2(g)} + I_{2(g)} → 2 HI_{(g)}$
$H_{2(g)} + Br_{2(g)} → 2 HBr_{(g)}$
$H_{2(g)} + Cl_{2(g)} → 2 HCl_{(g)}$
Careful experiments, carried out over many years, are consistent with the simplest mechanism: a single collision between the two reactant molecules results in a rearrangement of the bonds:
One might be tempted to suppose that this would proceed in a similar way, but experiments reveal that the mechanism of this reaction is far more complex. The reaction takes place in a succession of steps, some of which involve atomic H and Br.
The mechanism of this reaction is different again. Although the first two reactions reach equilibrium in minutes to an hour or so at temperatures of 300 to 600 K, a mixture of hydrogen and chlorine will not react at all in the dark, but if you shine a light on the mixture, it goes off with a bang as the instantaneous reaction releases heat and expands the gas explosively.
What is particularly noteworthy is that these striking differences cannot be reliably predicted from theory; they were revealed only by experimentation.
The rates of chemical reactions
Chemical reactions vary greatly in the speed at which they occur. Some are essentially instantaneous, while others may take years to reach equilibrium. The speed of a chemical reaction may be defined as the change in concentration of a substance divided by the time interval during which this change is observed:
$\color{red} \text{rate} = \dfrac{\Delta \text{concentration}}{\Delta \text{time}} \label{2-1}$
For a reaction of the form A + B → C, the rate can be expressed in terms of the change in concentration of any of its components:
$\text{rate} = \dfrac{-\Delta[A]}{\Delta t} \label{Eq1a}$
$\text{rate} = \dfrac{-\Delta[B]}{\Delta t} \label{Eq1b}$
$\text{rate} = \dfrac{\Delta[C]}{\Delta t} \label{Eq1c}$
in which Δ[A] is the difference between the concentration of $A$ over the time interval $\Delta t = t_2 – t_1$:
$Δ[A] = [A]_2 – [A]_1 \label{2-2}$
Notice the minus signs in Equations $\ref{Eq1a}$ and $\ref{Eq1a}$; the concentration of a reactant always decreases with time, so Δ[A] and Δ[B] are both negative. Since negative rates does not make much sense, rates expressed in terms of a reactant concentration are always preceded by a minus sign in order to make the rate come out positive.
Consider now a reaction in which the coefficients are different:
$A + 3B → 2D$
It is clear that [B] decreases three times as rapidly as [A], so in order to avoid ambiguity when expressing the rate in terms of different components, it is customary to divide each change in concentration by the appropriate coefficient:
$\text{rate} = \dfrac{-\Delta[A]}{\Delta t}= \dfrac{-\Delta[B]}{3 \Delta t} = \dfrac{+\Delta[D]}{2\Delta t}$
Each of the above quotients is a legitimate expression of the rate of this particular reaction; they all yield the same number. Which one you employ when doing a calculation is largely a matter of convenience.
Example $1$: Oxidation of Ammonia
For the oxidation of ammonia
$4 NH_3 + 3O_2 \rightarrow 2 N_2 + 6 H_2O \nonumber$
it was found that the rate of formation of N2 was 0.27 mol L–1 s–1.
1. At what rate was water being formed?
2. At what rate was ammonia being consumed?
Solution:
1. From the equation stoichiometry, Δ[H2O] = 6/2 Δ[N2], so the rate of formation of H2O is 3 × (0.27 mol L–1 s–1) = 0.81 mol L–1 s–1.
2. 4 moles of NH3 are consumed for every 2 moles of N2 formed, so the rate of disappearance of ammonia is 2 × (0.27 mol L–1 s–1) = 0.54 mol L–1 s–1.
Comment: Because of the way this question is formulated, it would be acceptable to express this last value as a negative number.
Instantaneous rates
Most reactions slow down as the reactants are consumed. Consequently, the rates given by the expressions shown above tend to lose their meaning when measured over longer time intervals Δt. Thus for the reaction whose progress is plotted here, the actual rate (as measured by the increasing concentration of product) varies continuously, being greatest at time zero. The instantaneous rate of a reaction is given by the slope of a tangent to the concentration-vs.-time curve. Three such rates have been identified in this plot.
An instantaneous rate taken near the beginning of the reaction (t = 0) is known as an initial rate (label (1) here). As we shall soon see, initial rates play an important role in the study of reaction kinetics. If you have studied differential calculus, you will know that these tangent slopes are derivatives whose values can very at each point on the curve, so that these instantaneous rates are really limiting rates defined as
$rate = \lim_{\Delta t \rightarrow 0} \dfrac{-\Delta[A]}{\Delta t}$
However, if you does not know calculus, just bear in mind that the larger the time interval Δt, the smaller will be the precision of the instantaneous rate.
Instantaneous rates are also known as differential rates.
Rate Laws and Reaction Order
The relation between the rate of a reaction and the concentrations of reactants is expressed by its rate law. For example, the rate of the gas-phase decomposition of dinitrogen pentoxide
$2N_2O_5 → 4NO_2 + O_2$
has been found to be directly proportional to the concentration of $N_2O_5$:
$\text{rate} = k [N_2O_5]$
Be very careful about confusing equilibrium constant expressions with those for rate laws. The expression for $K_{eq}$ can always be written by inspecting the reaction equation, and it contains a term for each component (raised to the appropriate power) whose concentration changes during the reaction. For this reaction it is given by
$K_{eq} = \dfrac{[NO_2]^2 [O_2]}{[N_2O_5]^2}$
In contrast, the expression for the rate law generally bears no necessary relation to the reaction equation, and must be determined experimentally. More generally, for a reaction of the form
$n_A A + n_B B + ... → products$
the rate law will be
$rate = [A]^a[B]^b ...$
, nB.
Since the rate of a reaction has the dimensions of (concentration/time), the dimensions of the rate constant k will depend on the exponents of the concentration terms in the rate law. To make this work out properly, if we let $p$ be the sum of the exponents of the concentration terms in the rate law
$p = a + b + ...$
then k will have the dimensions (concentration1–p/time).
Reaction order
The order of a rate law is the sum of the exponents in its concentration terms. For the N2O5 decomposition with the rate law k[N2O5], this exponent is 1 (and thus is not explicitly shown); this reaction is therefore a first order reaction. We can also say that the reaction is "first order in N2O5". For more complicated rate laws, we can speak of the overall reaction order and also the orders with respect to each component. As an example, consider a reaction
$A + 3B + 2C → \text{products}$
whose experimental rate law is
$rate = k[A] [B]^2$
We would describe this reaction as
• third-order overall,
• first-order in A,
• second-order in B, and
• zero-order in C.
Zero-order means that the rate is independent of the concentration of a particular reactant. However, enough C must be present to allow the equilibrium mixture to form.
Example $2$
The rate of oxidation of bromide ions by bromate in an acidic aqueous solution
$6H^+ + BrO_3^– + 5Br^– → 3 Br_2 + 3 H_2O \nonumber$
is found to follow the rate law
$rate = k[Br^–][BrO_3^–][H^+]^2$
What happens to the rate if, in separate experiments,
1. [BrO3] is doubled;
2. the pH is increased by one unit;
3. the solution is diluted to twice its volume, with the pH kept constant by use of a buffer?
Solution
1. Since the rate is first-order in bromate, doubling its concentration will double the reaction rate.
2. Increasing the pH by one unit will decrease the [H+] by a factor of 10. Since the reaction is second-order in [H+], this will decrease the rate by a factor of 100.
3. Dilution reduces the concentrations of both Br2 and BrO3 to half their original values. Doing this to each concentration alone would reduce the rate by a factor of 2, so reducing both concentration will reduce the rate by a factor of 4, to (½)×(½) = ¼ of its initial value.
How reaction orders are observed
Observing rate-vs.-concentration proportionality
In order to determine the value of the exponent in a rate equation term, we need to see how the rate varies with the concentration of the substance. For a single-reactant decomposition reaction of the form
A → products
in which the rate is –d[A]/dt, we simply plot [A] as function of time, draw tangents at various intervals, and see how the slopes of these tangents (the instantaneous rates) depend on [A].
• If doubling the concentration of A doubles the rate, then the reaction is first-order in A.
• If doubling the concentration results in a four-fold rate increase, the reaction is second-order in A.
Example $3$
Use the tabulated experimental data to determine the order of the reaction
$2 N_2O_5 → 4 NO_2 + O_2 \nonumber$
Time (min) p(N2O5) [N2O5] mol L-1
Rate (mol L-1 min-1)
0 301.6 0.0152
10 224.8 0.0113 3.4 × 10–4
20 166.7 0.0084 2.5
30 123.2 0.0062 1.8
40 92.2 0.0046 1.3
69.1 69.1 0.0035 1.0
Solution
The ideal gas law can be used to convert the partial pressures of $N_2O_5$ to molar concentrations. These are then plotted (left) to follow their decrease with time. The rates are computed from the slopes of the tangents (blue lines) and their values plotted as a function of $[N_2O_5]$ and $[N)2O)5]^2$. It is apparent that the rates are directly proportional to $[N)2O)5]^1$, indicating that this is a first-order reaction.
Initial rate method
When there is more than one reactant, the method described above is rarely practical, since the concentrations of the different reactants will generally fall at different rates, depending on the stoichiometry. Instead, we measure only the rate near the beginning of the reaction, before the concentrations have had time to change significantly. The experiment is then repeated with a different starting concentration of the reactant in question, but keeping the concentrations of any others the same. After the order with respect to one component is found, another series of trials is conducted in which the order of another component is found.
Using the same dinitrogen pentoxide decomposition as in the previous example, we conduct a series of runs using five different initial concentrations of $N_2O_5$. Note that we use only the tangents at the beginning of each plot — that is, at times t=0.
We then plot the five initial rates of consumption of $N_2O_5$ as a function of its molar concentration. As before, we see that these rates are directly proportional to $[N_2O_5]$. The slope of this plot gives the value of the rate constant.
rate = (5.2 × 10–3) [N2O5] mol L–1 s–1
Example $4$
A study of the gas-phase reduction of nitric oxide by hydrogen
$2 NO + 2 H_2 → N_2 + 2 H_2O \nonumber$
yielded the following initial-rate data (all pressures in torr):
experiment P(NO) P(H2)
initial rate (torr s–1)
1
359 300 1.50
2
300 300 1.03
3
152 300 0.25
4
300 289 1.00
5
300 205 0.71
6
200 147 0.51
Find the order of the reaction with respect to each component. In looking over this data, take note of the following:
• The six runs recorded here fall into two groups, in which the initial pressures of H2 and of NO, respectively, are held constant.
• All the data are expressed in pressures, rather than in concentrations. We can do this because the reactants are gases, whose concentrations are directly proportional to their partial pressures when T and V are held constant. And since we are only interested in comparing the ratios of pressures and rates, the units cancel out and does notmatter. It is far easier experimentally to adjust and measure pressures than concentrations.
Solution
Experiments 2 and 3: Reduction of the initial partial pressure of NO by a factor of about 2 (300/152) results in a reduction of the initial rate by a factor of about 4, so the reaction is second-order in nitric oxide.
Experiments 4 and 6: Reducing the initial partial pressure of hydrogen by a factor of approximately 2 (289/147) causes a similar reduction in the initial rate, so the reaction is first-order in hydrogen.
The rate law is thus
$rate = k[NO]^2[H_2]$
Dealing with multiple reactants: the isolation method
It is not always practical to determine orders of two or more reactants by the method illustrated in the preceding example. Fortunately, there is another way to accomplish the same task: we can use excess concentrations of all the reactants except the one we wish to investigate. For example, suppose the reaction is
$A + B + C → \text{products}$
and we need to find the order with respect to [B] in the rate law. If we set [B]o to 0.020 M and let [A]o = [C]o = 2.00M, then if the reaction goes to completion, the change in [A] and [C] will also be 0.020 M which is only 1 percent of their original values. This will often be smaller than the experimental error in determining the rates, so it can be neglected. By "flooding" the reaction mixture with one or more reactants, we are effectively isolating the one in which we are interested. | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.01%3A_Rates_of_reactions_and_rate_laws.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
• Describe the contrasting roles of thermodynamics and kinetics in understanding chemical change.
• Given a balanced net equation, write an expression for the rate of a reaction.
• Sketch a curve showing how the instantaneous rate of a reaction might change with time.
• Determine the order of a reaction of the form A → B + C from experimental data for the concentrations of its products at successive times.
• Describe the initial rate and isolation methods of determining the orders of the individual reactants in a reaction involving multiple reactants.
• Explain the difference between differential and integral rate laws.
• Sketch out a plot showing how the concentration of a component ([A] or ln [A]) that follows first-order kinetics will change with time. Indicate how the magnitude of the rate constant affects this plot
• Define the half-life of a reaction.
• Given the half-life for a first-order reaction A → products along with the initial value of [A]o, find [A]t at a subsequent time an integral number of half-lifes later.
• Describe the conditions under which a reaction can appear to have an order of zero.
On this page we extend the concept of differential rate laws introduced on the previous page to integral rate laws and reaction half-life that are of great importance in most practical applications of kinetics.
Differential and integral rate laws
Measuring instantaneous rates as we have described on the previous page of this unit is the most direct way of determining the rate law of a reaction, but is not always convenient, and it may not even be possible to do so with any precision.
• If the reaction is very fast, its rate may change more rapidly than the time required to measure it; the reaction may be finished before even an initial rate can be observed.
• In the case of very slow reactions, observable changes in concentratons occur so slowly that the observation of a truly "instantaneous" rate becomes impractical.
The ordinary rate law (more precisely known as the instantaneous or differential rate law) tells us how the rate of a reaction depends on the concentrations of the reactants. But for many practical purposes, it is more important to know how the concentrations of reactants (and of products) change with time. For example, if you are carrying out a reaction on an industrial scale, you would want to know how long it will take for, say, 95% of the reactants to be converted into products.
This is the purpose of an integrated rate law.
Integrating the rate law
This is easy to do, but only some courses expect you to know how to do it. For a quick run-through, click here. If you have had even a bit of calculus, here is an opportunity to put it to use!
Expressing the "speed" of a reaction: the half-life
How long does it take for a chemical reaction to occur under a given set of conditions? As with many "simple" questions, no meaningful answer can be given without being more precise. In this case,
How do we define the point at which the reaction is "completed"?
A reaction is "completed" when it has reached equilibrium — that is, when concentrations of the reactants and products are no longer changing. If the equilibrium constant is quite large, then the answer reduces to a simpler form: the reaction is completed when the concentration of a reactant falls to zero. In the interest of simplicity, we will assume that this is the case in the remainder of this discussion.
"How long?" may be too long
If the reaction takes place very slowly, the time it takes for every last reactant molecule to disappear may be too long for the answer to be practical. In this case, it might make more sense to define "completed" when a reactant concentration has fallen to some arbitrary fraction of its initial value — 90%, 70%, or even only 20%.
The particular fraction one selects depends on the cost of the reactants in relation to the value of the products, balanced against the cost of operating the process for a longer time or the inconvenience of waiting for more product. This kind of consideration is especially imporant in industrial processes in which the balances of these costs affect the profitability of the operation.
The half-life of a reaction
Instead of trying to identify the time required for the reaction to become completed, it is far more practical to specify the the time required for the concentration of a reactant to fall to half of its initial value. This is known as the half-life (or half-time) of the reaction.
First-order reactions
The law of exponential change
The rate at which a reactant is consumed in a first-order process is proportional to its concentration at that time. This general relationship, in which a quantity changes at a rate that depends on its instantaneous value, is said to follow an exponential law.
Exponential relations are widespread in science and in many other fields. Consumption of a chemical reactant or the decay of a radioactive isotope follow the exponential decay law. Its inverse, the law of exponential growth, describes the manner in which the money in a continuously-compounding bank account grows with time, or the population growth of a colony of reproducing organisms.
The reason that the exponential function y=ex so efficiently describes such changes stems from the remarkable property that dy/dx = ex; that is, ex is its own derivative, so the rate of change of y is identical to its value at any point. A nice discussion of this property can be found here .
The first-order integrated rate law
The integrated rate law for a first-order reaction
A → products
is a common example of the law of exponential change. For a reactant A, its concentration [A]t at time t is given by
[A]t = [A]o × ekt
in which [A]o is its initial concentration and k is the first-order rate constant.
The "e" in the exponential term is of course the base of the natural logarithms, and the negative sign in its exponent means that the value of this term diminishes as t increases, as we would expect for any kind of a decay process.
A more convenient form of the integrated rate law is obtained by taking the natural logarithm of both sides:
ln [A] = –kt + ln [A]o(4-1)
This has the form of an equation for a straight line
y = mx + b
in which the slope m corresponds to the rate constant k. This means that, for a first-order reaction, a plot of ln [A] as a function of time gives a straight line with a slope of –k.
Half-life
After a period of one half-life, t = t½ and we can write
$\dfrac{[\mathrm{A}]_{1 / 2}}{[\mathrm{~A}]_0}=\dfrac{1}{2}=\mathrm{e}^{-k t_{1 / 2}} \nonumber$
(in which we express the exponential as a function in order to make it stand out more prominently.) Taking logarithms of both sides (remember that
ln ex = x) yields
$\ln 0.5=-k t \nonumber$
Solving for the half-life, we obtain the simple relation
$t_{1 / 2}=\dfrac{0.693}{k} \nonumber$
which tells us that the half-life of a first-order reaction is a constant. This means that 100,000 molecules of a reactant will be reduced to 50,000 in the same time interval needed for ten molcules to be reduced to five.
It should be clear that the rate constant and the half life of a first-order process are inversely related.
Example $1$
The half-life of a first-order reaction was found to be 10 min at a certain temperature. What is its rate constant in reciprocal seconds?
Solution:
From the above equation, k = –0.693/(600 s) = 0.00115 s–1
The decay of radioactive nuclei is always a first-order process.
Example $2$
The mass-241 isotope of americium, widely used as an ionizing source in smoke detectors, has a half-life of 432 years.
a) What fraction of the Am241 in a smoke detector will have decayed after 50 years?
b) How long will it take for the activity to decline to 80% of its initial value?
c) What would be the "seventh-life" of Am241?
Second-order reactions
Integration of the second-order rate law
$-\dfrac{d[\mathrm{~A}]}{d t}=-k|\mathrm{~A}|^2 \nonumber$
yields
$-\dfrac{1}{\mid \mathrm{A}]}=\dfrac{1}{\left|\mathrm{~A}_{\mathrm{o}}\right|}+k t \nonumber$
which is easily rearranged into a form of the equation for a straight line (try showing this yourself!) and yields plots similar to the one shown on the left below.
The half-life is given by
$t_{1/2}=\dfrac{1}{k\left|\mathrm{~A}_0\right|} \nonumber$
(see here for details)
Notice that the half-life of a second-order reaction depends on the initial concentration, in contrast to its constancy for a first-order reaction. For this reason, the concept of half-life for a second-order reaction is far less useful.
Zero-order processes
In some reactions, the rate is apparently independent of the reactant concentraton, in which case
Integration of the differential rate law yields [A]t = [A]0k
$rate =k[\mathrm{~A}]^0=k \nonumber$
Note the word "apparently" in the preceding sentence; zero-order kinetics is always an artifact of the conditions under which the reaction is carried out. For this reason, reactions that follow zero-order kinetics are often referred to as pseudo-zero-order reactions.
Clearly, a zero-order process cannot continue after a reactant has been exhausted. Just before this point is reached, the reaction will revert to another rate law instead of falling directly to zero as depicted at the upper left.
There are two general conditions that can give rise to zero-order rates:
This situation commonly occurs when a reaction is catalyzed by attachment to a solid surface (heterogeneous catalysis) or to an enzyme.
For example, the decomposition of nitrous oxide
N2O (g)→ N2(g) + ½ O2(g)
in the presence of a hot platinum wire (which acts as a catalyst) is zero-order, but it follows more conventional kinetics when carried out entirely in the gas phase.
In this case, the N2O molecules that react are limited to those that have attached themselves to the surface of the solid catalyst. Once all of the sites on the limited surface of the catalyst have been occupied, additional gas-phase molecules must wait until the decomposition of one of the adsorbed molecules frees up a surface site.
Enzyme-catalyzed reactions in organisms begin with the attachment of the substrate to the active site on the enzyme, leading to the formation of an enzyme-substrate complex. If the number of enzyme molecules is limited in relation to substrate molecules, then the reaction may appear to be zero-order.
This is most often seen when two or more reactants are involved. Thus if the reaction
A + B → products
is first-order in both reactants so that
rate = k [A][B]
then if B is present in great excess, the reaction will appear to be zero order in B (and first order overall). This commonly happens when B is H2O and the reaction is carried out in aqueous solution.
1. Only a small fraction of the reactant molecules are in a location or state in which they are able to react, and this fraction is continually replenished from the larger pool.
2. When two or more reactants are involved, the concentrations of some are much greater than those of others
Summary
The following table compares the rate parameters of zero-, first-, and second-order reactions of the form aA → products. See also this excellent page from Purdue University that shows plots of [A], ln [A], and 1/[A] for each reaction order.
zero order first order second order
differential rate law rate = k rate = k[A] rate = k[A]2
integrated rate law $[A]=[A]_0=-a k t$ $\ln \dfrac{[\mathrm{A}]_0}{[\mathrm{~A}]}=a k t$ $\dfrac{1}{[\mathrm{~A}]}-\dfrac{1}{[\mathrm{~A}]_0}=a k t$
half-life $\dfrac{[\mathrm{A}]_0}{2 a k}$ $\frac{\ln 2}{a k}=\frac{.693}{a k}$ $\frac{1}{a k[\mathrm{~A}]_0}$
plot that gives straight line [A] vs. t ln [A] vs. t 1/[A] vs. t
plot, showing interpretation of slope and intercept | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.02%3A_Reaction_Rates_Typically_Change_with_Time.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
• Explain the meaning of a reaction mechanism and define elementary step and intermediate.
• Describe the role of collisions in reaction mechanisms, and explain why not all collisions lead to the formation of products.
• Sketch out activation energy diagrams for simple reactions that are endothermic or exothermic,
• Explain how an activated complex differs from an intermediate.
• Define catalyst, and sketch out an activation energy diagram that illustrates how catalysts work.
• Explain the significance of the various terms that appear in the Arrhenius Law.
• Sketch out a typical Arrhenius Law plot for a hypothetical reaction at higher and lower temperatures.
• Explain how the activation energy of a reaction can be determined experimentally.
• Explain the significance of the various terms that appear in the pre-exponential factor of the Arrhenius equation.
Why are some reactions so much faster than others, and why are reaction rates independent of the thermodynamic tendency of the reaction to take place? These are the central questions we address in this unit. In doing so, we open the door to the important topic of reaction mechanisms: what happens at the microscopic level when chemical reactions take place? We can thank Prof. Svante Arrhenius for unlocking this door! To keep things as simple as possible, we will restrict ourselves to reactions that take place in the gas phase.The same principles will apply to reactions in liquids and solids, but with added complications that we will discuss in a later unit.
Reaction mechanisms
The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collison") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an intermediate — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation.
A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called elementary reactions) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as
unimolecular A → by far the most common
bimolecular A + B →
termolecular A + B + C → very rare
Collision theory of chemical change: Molecules must collide before they can react
This fundamental rule must guide any analysis of an ordinary chemical reaction mechanism.
This explains why termolecular processes are so uncommon. The kinetic theory of gases tells us that for every 1000 binary collisions, there will be only one event in which three molecules simultaneously come together. Four-way collisions are so improbable that this process has never been demonstrated in an elementary reaction. Consider a simple bimolecular step
$\ce{A + B -> products} \nonumber$
Clearly, if two molecules A and B are to react, they must approach closely enough to disrupt some of their existing bonds and to permit the creation of any new ones that are needed in the products. We call such an encounter a collision.
The frequency of collisions between A and B in a gas will be proportional to the concentration of each; if we double [A], the frequency of A-B collisions will double, and doubling [B] will have the same effect. So if all collisions lead to products, than the rate of a bimolecular process will be first-order in A and B, or second-order overall:
$\text{rate} = k[\ce{A}][\ce{B}] \nonumber$
However, not all collisions are equal. In a gas at room temperature and normal atmospheric pressure, there will be about 1033 collisions in each cubic centimetre every second. If every collision between two reactant molecules yielded products, all reactions would be complete in a fraction of a second.
When two billiard balls collide, they simply bounce off of each other. This is also the most likely outcome if the reaction between A and B requires a significant disruption or rearrangement of the bonds between their atoms. In order to effectively initiate a reaction, collisions must be sufficiently energetic (kinetic energy) to bring about this bond disruption. More about this further on.
And there is often one additional requirement. In many reactions, especially those involving more complex molecules, the reacting species must be oriented in a manner that is appropriate for the particular process. For example, in the gas-phase reaction of dinitrogen oxide with nitric oxide, the oxygen end of N2O must hit the nitrogen end of NO; reversing the orientation of either molecule prevents the reaction.
Owing to the extensive randomization of molecular motions in a gas or liquid, there are always enough correctly-oriented molecules for some of the molecules to react. But of course, the more critical this orientational requirement is, the fewer collisions will be effective.
Anatomy of a collision
Energetic collisions between molecules cause interatomic bonds to stretch and bend farther, temporarily weakening them so that they become more susceptible to cleavage. Distortion of the bonds can expose their associated electron clouds to interactions with other reactants that might lead to the formation of new bonds. Chemical bonds have some of the properties of mechanical springs, whose potential energy depends on the extent to which they are stretched or compressed. Each atom-to-atom bond can be described by a potential energy diagram that shows how its energy changes with its length. When the bond absorbs energy (either from heating or through a collision), it is elevated to a higher quantized vibrational state (indicated by the horizontal lines) that weakens the bond as its length oscillates between the extended limits corresponding to the curve.
A particular collision will typically excite a number of bonds in this way. Within about 10–13 second this excitation gets distributed among the other bonds in the molecule in rather complex and unpredictable ways that can concentrate the added energy at a particularly vulnerable point. The affected bond can stretch and bend farther, making it more susceptible to cleavage. Even if the bond does not break by pure stretching, it can become distorted or twisted so as to expose nearby electron clouds to interactions with other reactants that might encourage a reaction.
Consider, for example, the isomerization of cyclopropane to propene which takes place at fairly high temperatures in the gas phase.
We can imagine the collision-to-product sequence in the following [grossly oversimplified] way:
Note that
• To keep things simple, we do not show the hydrogen atoms here. This is reasonable because C–C bonds are weaker then C–H bonds and thus less likely to be affected.
• The collision at will usually be with another cyclopropane molecule, but because no part of the colliding molecule gets incorporated into the product, it can in principle be a noble gas or some other non-reacting species;
• Although the C–C bonds in cyclopropane are all identicial, the instantaneous localization of the collisional energy can distort the molecule in various ways (), leading to a configuration sufficiently unstable to initiate the rearrangement to the product.
Unimolecular processes also begin with a collision
The cyclopropane isomerization described above is typical of many decomposition reactions that are found to follow first-order kinetics, implying that the process is unimolecular. Until about 1921, chemists did not understand the role of collisions in unimolecular processes. It turns out that the mechanisms of such reactions are really rather complicated, and that at very low pressures they do follow second-order kinetics. Such reactions are more properly described as pseudounimolecular.
Activation energy
Higher temperatures, faster reactions
The chemical reactions associated with most food spoilage are catalyzed by enzymes produced by the bacteria which mediate these processes. It is common knowledge that chemical reactions occur more rapidly at higher temperatures. Everyone knows that milk turns sour much more rapidly if stored at room temperature rather than in a refrigerator, butter goes rancid more quickly in the summer than in the winter, and eggs hard-boil more quickly at sea level than in the mountains. For the same reason, cold-blooded animals such as reptiles and insects tend to be noticeably more lethargic on cold days.
It is not hard to understand why this should be. Thermal energy relates direction to motion at the molecular level. As the temperature rises, molecules move faster and collide more vigorously, greatly increasing the likelyhood of bond cleavages and rearrangements as described above.
Activation energy diagrams
Most reactions involving neutral molecules cannot take place at all until they have acquired the energy needed to stretch, bend, or otherwise distort one or more bonds. This critical energy is known as the activation energy of the reaction. Activation energy diagrams of the kind shown below plot the total energy input to a reaction system as it proceeds from reactants to products.
In examining such diagrams, take special note of the following:
• The "reaction coordinate" plotted along the abscissa represents the changes in atomic coordinates as the system progresses from reactants to products. In the very simplest elementary reactions it might correspond to the stretching or twisting of a particular bond, and be shown to a scale. In general, however, the reaction coordinate is a rather abstract concept that cannot be tied to any single measurable and scaleable quantity.
• The activated complex (also known as the transition state) represents the structure of the system as it exists at the peak of the activation energy curve. It does not correpond to an identifiable intermediate structure (which would more properly be considered the product of a separate elementary process), but rather to whatever configuration of atoms exists during the collision, which lasts for only about 0.1 picosecond.
• Activation energy diagrams always incorporate the energetics (ΔU or ΔH) of the net reaction, but it is important to understand that the latter quantities depend solely on the thermodynamics of the process which are always independent of the reaction pathway. This means that the same reaction can exhibit different activation energies if it can follow alternative pathways.
• With a few exceptions for very simple processes, activation energy diagrams are largely conceptual constructs based on our standard collision model for chemical reactions. It would be unwise to read too much into them.
Gallery of activation energy plots
Activation energy diagrams can describe both exothermic and endothermic reactions:
... and the activation energies of the forward reaction can be large, small, or zero (independently, of course, of the value of ΔH):
Processes with zero activation energy most commonly involve the combination of oppositely-charged ions or the pairing up of electrons in free radicals, as in the dimerization of nitric oxide (which is an odd-electron molecule).
In this plot for the dissociation of bromine, the Ea is just the enthalpy of atomization
Br2(g) → 2 Br· (g)
and the reaction coordinate corresponds roughly to the stretching of the vibrationally-excited bond. The "activated complex", if it is considered to exist, is just the last, longest "stretch". The reverse reaction, being the recombination of two radicals, occurs immediately on contact.
Where does the activation energy come from?
In most cases, the activation energy is supplied by thermal energy, either through intermolecular collisions or (in the case of thermal dissocation) by thermal excitation of a bond-stretching vibration to a sufficiently high quantum level.
As products are formed, the activation energy is returned in the form of vibrational energy which is quickly degraded to heat. It's worth noting, however, that other sources of activation energy are sometimes applicable:
• Absorption of light by a molecule (photoexcitation) can be a very clean and efficient, but it doesn't always work. It's not enough that the wavelength of the light correspond to the activation energy; it must also fall within the absorption spectrum of the molecule, and (in a complex molecule) enough of it must end up in the right part of the molecule, such as in a particular bond.
• Electrochemical activation. Molecules capable of losing or gaining electrons at the surface of an electrode can undergo activation from an extra potential (known as the overvoltage) between the electrode and the solution. The electrode surface often plays an active role, so the process is also known as electrocatalysis.
Catalysts can reduce activation energy
A catalyst is usually defined as a substance that speeds up a reaction without being consumed by it. More specifically, a catalyst provides an alternative, lower activation energy pathway between reactants and products. As such, they are vitally important to chemical technology; approximately 95% of industrial chemical processes involve catalysts of various kind. In addition, most biochemical processes that occur in living organisms are mediated by enzymes, which are catalysts made of proteins.
It is important to understand that a catalyst affects only the kinetics of a reaction; it does not alter the thermodynamic tendency for the reaction to occur. Thus there is a single value of ΔH for the two pathways depicted in the plot on the right
Temperature and kinetic energy
A review of the principles of gas molecular velocities and the Boltzmann distribution can be found on the "KMT-classic" page.
In the vast majority of cases, we depend on thermal actvation, so the major factor we need to consider is what fraction of the molecules possess enough kinetic energy to react at a given temperature. According to kinetic molecular theory, a opulation of molecules at a given temperature is distributed over a variety of kinetic energies that is described by the Maxwell-Boltzman distribution law.
The two distribution plots shown here are for a lower temperature T1 and a higher temperature T2. The area under each curve represents the total number of molecules whose energies fall within particular range. The shaded regions indicate the number of molecules which are sufficiently energetic to meet the requirements dictated by the two values of Ea that are shown.
It is clear from these plots that the fraction of molecules whose kinetic energy exceeds the activation energy increases quite rapidly as the temperature is raised. This the reason that virtually all chemical reactions (and all elementary reactions) are more rapid at higher temperatures.
2 The Arrhenius law
By 1890 it was common knowledge that higher temperatures speed up reactions, often doubling the rate for a 10-degree rise, but the reasons for this were not clear. Finally, in 1899, the Swedish chemist Svante Arrhenius (1859-1927) combined the concepts of activation energy and the Boltzmann disribution law into one of the most important relationships in physical chemistry:
Take a moment to focus on the meaning of this equation, neglecting the A factor for the time being.
First, note that this is another form of the exponential decay law we discussed in the previous section of this series. What is "decaying" here is not the concentration of a reactant as a function of time, but the magnitude of the rate constant as a function of the exponent –Ea /RT. And what is the significance of this quantity? If you recall that RT is the average kinetic energy, it will be apparent that the exponent is just the ratio of the activation energy Ea to the average kinetic energy. The larger this ratio, the smaller the rate (hence the negative sign.) This means that high temperature and low activation energy favor larger rate constants, and thus speed up the reaction. And because these terms occur in an exponent, their effects on the rate are quite substantial.
The two plots below show the effects of the activation energy (denoted here by E) on the rate constant. Even a modest activation energy of 50 kJ/mol reduces the rate by a factor of 108.
The logarithmic scale in the right-hand plot leads to nice straight lines, as described under the next heading below.
Looking at the role of temperature, we see a similar effect. (If the x-axis were in "kilodegrees" the slopes would be more comparable in magnitude with those of the kilojoule plot at the above right.)
Determining the activation energy
The Arrhenius equation
$k=A \mathrm{e}^{-E_{a} / R T}$
can be written in a non-exponential form which is often more convenient to use and to interpret graphically. Taking the logarithms of both sides and separating the exponential and pre-exponential terms yields
$\ln k=\ln \left(A \mathrm{e}^{-E_{\mathrm{a}} / R T}\right)=\ln A+\ln \left(\mathrm{e}^{-E_{\mathrm{a}} / R T}\right)$
$\ln k=\ln A-\dfrac{E_{a}}{R T}$
which is the equation of a straight line whose slope is $–E_a /R$. This affords a simple way of determining the activation energy from values of $k$ observed at different temperatures; we just plot $\ln k$ as a function of $1/T$.
Example $1$: Isomerization of cyclopropane
Thus for the isomerization of cyclopropane to propene
the following data were obtained (calculated values shaded in pink):
T, °C 477 523 577 623
1/T, K–1 × 103 1.33 1.25 1.18 1.11
k, s–1 0.00018 0.0027 0.030 0.26
ln k –8.62 –5.92 –3.51 –1.35
From the calculated slope, we have
– (Ea/R) = –3.27 × 104 K
Ea=– (8.314 J mol–1 K–1) (–3.27 × 104 K) = 273 kJ mol–1
Comment: This activation energy is rather high, which is not surprising because a carbon-carbon bond must be broken in order to open the cyclopropane ring. (C–C bond energies are typically around 350 kJ/mol.) This is why the reaction must be carried out at high temperature.
You don't always need a plot
(... if you are willing to live a bit dangerously!) Since the ln k-vs.-1/T plot yields a straight line, it is often convenient to estimate the activation energy from experiments at only two temperatures. To see how this is done, consider that
$\ln k_{2}-\ln k_{1}=\left(\ln A-\frac{E_{a}}{R T_{2}}\right)-\left(\ln A-\frac{E_{a}}{R T_{1}}\right)=\frac{E_{a}}{R}\left(\frac{1}{T_{1}}-\frac{1}{T_{2}}\right) \nonumber$
(... in which we have made the ln-A term disappear by subtracting the expressions for the two ln-k terms.) Solving the expression on the right for the activation energy yields
$E_{a}=\dfrac{R \ln \dfrac{k_{2}}{k_{1}}}{\dfrac{1}{T_{1}}-\dfrac{1}{T_{2}}} \nonumber$
Example $2$
A widely used rule-of-thumb for the temperature dependence of a reaction rate is that a ten-C° rise in the temperature approximately doubles the rate. (This is obviously not generally true, especially when a strong covalent bond must be broken.) But for a reaction that does show this behavior, what would the activation energy be?
Solution
We will center our ten-degree interval at 300 K. Substituting into the above expression yields
= (8.314)(0.693) / (.00339 - 0.00328)
= (5.76 J mol–1 K–1) / (0.00011 K–1) = 52400 J mol–1 = 52.4 kJ mol–1
Example $3$
It takes about 3.0 minutes to cook a hard-boiled egg in Los Angeles, but at the higher altitude of Denver, where water boils at 92°C, the cooking time is 4.5 minutes. Use this information to estimate the activation energy for the coagulation of egg albumin protein.
Solution:
The ratio of the rate constants at the elevations of LA and Denver is 4.5/3.0 = 1.5, and the respective temperatures are 373K and 365K. With the subscripts 2 and 1 referring to LA and Denver respectively, we have
Ea = (8.314)(ln 1.5) / (1/365 – 1/273) = (8.314)(.405) / (0.00274 – 0.00366)
= (3.37 J mol–1 K–1) / (0.000923 K–1) = 3650 J mol–1 = 3.65 kJ mol–1
Comment: This rather low value seems reasonable because protein denaturation involves the disruption of relatively weak hydrogen bonds; no covalent bonds are broken
Crickets and popcorn
• Many biological processes exhibit a temperature dependence that follows the Arrhenius law, and can thus be characterized by an activation energy. See this interesting Dartmouth U. page that looks at the kinetics of cricket chirps.
• In an article on the Kinetics of Popping of Popcorn (Cereal Chemisty 82(1): 53-59), J. Byrd and M. Perona found that popping follows a first-order rate law with an activation energy of 53.8 kJ/mol.
The pre-exponential factor
It is now time to focus in on the pre-exponential term A in the Arrhenius equation. We have been neglecting it because it is not directly involved in relating temperature and activation energy, which is the main practical use of the equation. But since A multiplies the exponential term, its value clearly contributes to the value of the rate constant and thus of the rate.
and of the temperature.
If this fraction were unity, the Arrhenius law would reduce to
$k = A$
In other words, A is the fraction of molecules that would react if either the activation energy were zero, or if the kinetic energy of all molecules exceeded Ea — admittedly, an uncommon scenario.
It's all about collisions
So what would limit the rate constant if there were no activation energy requirements? The most obvious factor would be the rate at which reactant molecules come into contact. This can be calculated from kinetic molecular theory and is known as the frequency- or collision factor Z. In some reactions, the relative orientation of the molecules at the point of collision is important, so we can also define a geometrical or steric factor (commonly denoted by ρ (Greek lower case rho).
In general, we can express A as the product of these two factors:
$A = Zρ$
Values of $ρ$ are generally very difficult to assess; they are sometime estimated by comparing the observed rate constant with the one in which A is assumed to be the same as Z.
Direction makes a difference
The more complicated the structures of the reactants, the more likely that the value of the rate constant will depend on the trajectories at which the reactants approach each other. We showed one example of this near the top of the page, but for another, consider the addition of a hydrogen halide such as HCl to the double bond of an alkene, converting it to a chloroalkane. This kind of electrophilic addition reaction is well-known to all students of organic chemistry.
Experiments have shown that the reaction only takes place when the HCl molecule approaches the alkene with its hydrogen-end, and in a direction that is approximately perpendicular to the double bond, as shown at below.
The reason for this becomes apparent when we recall that HCl is highly polar owing to the high electronegativity of chlorine, so that the hydrogen end of the molecule is slightly positive. The double bond of ethene consists of two clouds of negative charge corresponding to the σ (sigma) and π (pi) molecular orbitals. The latter, which extends above and below the plane of the C2H4 molecule, interacts with and attracts the HCl molecule.
If, instead, the HCl approaches with its chlorine end leading as in , electrostatic repulsion between the like charges causes the two molecules to bounce away from each other before any reaction can take place. The same thing happens in ; the electronegativity difference between carbon and hydrogen is too small to make the C–H bond sufficiently polar to attract the incoming chlorine atom.
The lesson you should take from this example is that once you start combining a variety of chemical principles, you gradually develop what might be called "chemical intuition" which you can apply to a wide variety of problems. This is far more important than memorizing specific examples. Now that you know what it takes to get a reaction started, you are ready for the next lesson that describes their actual mechanisms. | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.03%3A_Collision_and_activation-_the_Arrhenius_law.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially important that you know the precise meanings of all the green-highlighted terms in the context of this topic.
• Explain what is meant by the mechanism of a reaction.
• Define an elementary reaction, and state how it differs from an ordinary net chemical reaction.
• Sketch out an activation energy diagram for a multistep mechanism involving a rate-determining step, and relate this to the activation energy of the overall reaction.
• Write the rate law expression for a two-step mechanism in which the rate constants have significantly different magnitudes.
• Write the rate law expression for a three-step reaction in which one step is a rapid equilibrium, and the other two steps have significantly different magnitudes.
• Define a chain reaction, and list some of the different kinds of steps such a reaction will involve.
• Define a branching chain reaction, and explain how such reactions can lead to explosions.
We are now ready to open up the "black box" that lies between the reactants and products of a net chemical reaction. What we find inside may not be very pretty, but it is always interesting because it provides us with a blow-by-blow description of how chemical reactions take place.
The mechanism of a chemical reaction is the sequence of actual events that take place as reactant molecules are converted into products. Each of these events constitutes an elementary step that can be represented as a coming-together of discrete particles ("collision") or as the breaking-up of a molecule ("dissociation") into simpler units. The molecular entity that emerges from each step may be a final product of the reaction, or it might be an intermediate — a species that is created in one elementary step and destroyed in a subsequent step, and therefore does not appear in the net reaction equation.
For an example of a mechanism, consider the decomposition of nitrogen dioxide into nitric oxide and oxygen. The net balanced equation is
$\ce{2 NO2(g) → 2 NO(g) + O2(g)} \nonumber$
The mechanism of this reaction is believed to involve the following two elementary steps:
\begin{align*} \ce{2 NO2} &→ \ce{NO3 + NO} \label{step 1} \[4pt] \ce{NO3} &→ \ce{NO + O2} \label{step 2} \end{align*}
Note that the intermediate species $\ce{NO3}$ has only a transient existence and does not appear in the net equation.
Properties of a Mechanism
A useful reaction mechanism
• consists of a series of elementary steps (defined below) that can be written as chemical equations, and whose sum gives the net balanced reaction equation;
• must agree with the experimental rate law;
• can rarely if ever be proved absolutely.
It is important to understand that the mechanism of a given net reaction may be different under different conditions. For example, the dissociation of hydrogen bromide
$\ce{2 HBr(g) → H2(g) + Br2(g)} \nonumber$
proceeds by different mechanisms (and follows different rate laws) when carried out in the dark (thermal decomposition) and in the light (photochemical decomposition). Similarly, the presence of a catalyst can enable an alternative mechanism that greatly speeds up the rate of a reaction.
Elementary Steps
A reaction mechanism must ultimately be understood as a "blow-by-blow" description of the molecular-level events whose sequence leads from reactants to products. These elementary steps (also called elementary reactions) are almost always very simple ones involving one, two, or [rarely] three chemical species which are classified, respectively, as
unimolecular $A →$ by far the most common
bimolecular $A + B →$ common
termolecular $A + B + C →$ very rare
Elementary reactions differ from ordinary net chemical reactions in two important ways:
• The rate law of an elementary reaction can be written by inspection. For example, a bimolecular process always follows the second-order rate law $rate=k[A][B]$.
• Elementary steps often involve unstable or reactive species that do not appear in the net reaction equation.
Some net reactions do proceed in a single elementary step, at least under certain conditions. However, without careful experimentation, one can never be sure. The gas-phase formation of $\ce{HI}$ from its elements was long thought to be a simple bimolecular combination of $\ce{H2}$ and $\ce{I2}$, but it was later found that under certain conditions, it follows a more complicated rate law.
Multi-step (Consecutive) Reactions
Mechanisms in which one elementary step is followed by another are very common.
$\ce{ A + B → \cancel{Q} } \tag{step 1}$
$\ce{B + \cancel{Q} → C} \tag{step 2}$
$\ce{A + 2B → C} \tag{net reaction}$
(As must always be the case, the net reaction is just the sum of its elementary steps.)
In this example, the species $Q$ is an intermediate, usually an unstable or highly reactive species. If both steps proceed at similar rates, rate law experiments on the net reaction would not reveal that two separate steps are involved here. The rate law for the reaction would be
$rate = k[A][B]^2 \nonumber$
(Bear in mind that intermediates such as $Q$ cannot appear in the rate law of a net reaction.)
When the rates are quite different, things can get interesting and lead to quite varied kinetics as well as some simplifying approximations. When the rate constants of a series of consecutive reactions are quite different, a number of relationships can come into play that greatly simplify our understanding of the observed reaction kinetics.
Approximation 1: The Rate-Determining Step Approximation
The rate-determining step is also known as the rate-limiting step. We can generally expect that one of the elementary reactions in a sequence of consecutive steps will have a rate constant that is smaller than the others. The effect is to slow the rates of all the reactions — very much in the way that a line of cars creeps slowly up a hill behind a slow truck.
The three-step reaction depicted here involves two intermediate species I1 and I2, and three activated complexes numbered X1-3. Although the step I 2 → products has the smallest individual activation energy Ea3, the energy of X3 with respect to the reactants determines the activation energy of the overall reaction, denoted by the leftmost vertical arrow . Thus the rate-determining step is
$X_1 → X_2. \nonumber$
Chemists often refer to elementary reactions whose forward rate constants have large magnitudes as "fast", and those with forward small rate constants as "slow". Always bear in mind, however, that as long as the steps proceed in single file (no short-cuts!), all of them will proceed at the same rate. So even the "fastest" members of a consecutive series of reactions will proceed as slowly as the "slowest" one.
Approximation 2: The Rapid Equilibrium Approximation
In many multi-step processes, the forward and reverse rate constants for the formation of an intermediate $Q$ are of similar magnitudes and sufficiently large to make the reaction in each direction quite rapid. Decomposition of the intermediate to product is a slower process:
$\ce{A <=>[k_1][k_{-1}] Q ->[k_2] B} \nonumber$
This is often described as a rapid equilibrium in which the concentration of $Q$ can be related to the equilibrium constant
$K = \dfrac{k_1}{k_{–1}} \nonumber$
This is just the Law of Mass Action. It should be understood, however, that true equilibrium is never achieved because $Q$ is continually being consumed; that is, the rate of formation of $Q$ always exceeds its rate of decomposition. For this reason, the steady-state approximation described below is generally preferred to treat processes of this kind.
Approximation 3: The Steady-State Approximation
Consider a mechanism consisting of two sequential reactions
$\ce{A ->[k_1] Q ->[k_2] B} \nonumber$
in which $Q$ is an intermediate. The time-vs-concentration profiles of these three substances will depend on the relative magnitudes of $k_1$ and $k_2$, as shown in the following diagrams. Construction of these diagrams requires the solution of sets of simultaneous differential equations, which is [fortunately!] beyond the scope of this course.
The steady-state approximation is usually not covered in introductory courses, although it is not particularly complicated mathematically.
In the left-hand diagram, the rate-determining step is clearly the conversion of the rapidly-formed intermediate into the product, so there is no need to formulate a rate law that involves $Q$. But on the right side, the formation of $Q$ is rate-determining, but its conversion into $B$ is so rapid that $[Q]$ never builds up to a substantial value. (Notice how the plots for $[A]$ and $[B]$ are almost mutually inverse.) The effect is to maintain the concentration of $Q$ at an approximately constant value. This steady-state approximation can greatly simplify the analysis of many reaction mechanisms, especially those that are mediated by enzymes in organisms.
Converting Mechanisms into Rate Laws
We are unable to look directly at the elementary steps hidden within the "black box" of the reaction mechanism, we are limited to proposing a sequence that would be consistent with the reaction order which we can observe. Chemical intuition can guide us in this, for example by guessing the magnitudes of some of the activation energies. In the end, however, the best we can do is to work out a mechanism that is plausible; we can never "prove" that what we come up with is the actual mechanism.
Example $\PageIndex{1A}$: A + B → C
Consider the following reaction:
$A + B → C\nonumber$
One possible mechanism might involve two intermediates $Q$ and $R$:
Step 1 $\ce{A + B ->[k_1] Q}$ (slow, rate-determining)
Step 2 $\ce{Q + A ->[k_2] R }$ (fast)
Step 3 $\ce{ R + B ->[k_3] C }$ (fast)
The rate law corresponding to this mechanism would be that of the rate-determining step:
$\text{rate} = k_1[A][B]. \nonumber$
If the first step in a mechanism is rate-determining, it is easy to find the rate law for the overall expression from the mechanism. If the second or a later step is rate-determining, determining the rate law is slightly more complicated and often requires either of the two approximation above to identify.
Example $\PageIndex{1B}$: A + B → C
An alternative mechanism for the following reaction:
$A + B → C \nonumber$
in which the rate-determining step involves one of the intermediates would display third-order kinetics:
Step 1 $\ce{A + B <=>[k_1][k_{-1}] Q}$ (rapid equilibrium)
Step 2 $\ce{Q + A ->[k_2] R }$ (slow, rate-determining)
Step 3 $\ce{ R + B ->[k_3] C }$ (fast)
Since intermediates cannot appear in rate law expressions, we must express $[Q]$ in the rate-determining step in terms of the other reactants. To do this, we make use of the fact that Step 1 involves an equilibrium constant $K_1$:
$K_1 = \dfrac{k_1}{k_{-1}} = \dfrac{[Q]}{[A][B]} \nonumber$
Solving this for $[Q]$, we obtain
$[Q] = K_1[A][B]. \nonumber$
We can now express the rate law for 2 as
\begin{align*} \text{rate} &= k_2 K_1[A][B][A] \[4pt] &= k[A]^2[B] \end{align*}
in which the constants $k_2$ and $K_1$ have been combined into a single constant $k$.
Example $2$: F2 + 2 NO2 → 2 NO2F
Consider the following reaction:
$\ce{ F2 + 2 NO2 → 2 NO2F }\nonumber$
Application of the "chemical intuition" mentioned in the above box would lead us to suspect that any process that involves breaking of the strong F–F bond would likely be slow enough to be rate limiting, and that the resulting atomic fluorines would be very fast to react with another odd-electron species:
Step 1 $\ce{F2 + 2 NO2 ->[k_1] NO2F + F + NO2}$ (slow, rate-determining)
Step 2 $\ce{F + NO2 ->[k_2] NO2F}$ (very fast)
If this mechanism is correct, then the rate law of the net reaction would be that of the rate-determining step:
$\text{rate} = k_1[F_2][NO_2] \nonumber$
Example $3$: Decomposition of ozone
Ozone is an unstable allotrope of oxygen that decomposes back into ordinary dioxygen according to the net reaction
$\ce{2 O3 → 3 O2} \nonumber$
A possible mechanism would be the simple one-step bimolecular collision suggested by the reaction equation, but this would lead to a second-order rate law which is not observed. Instead, experiment reveals a more complicated rate law:
$\text{rate} = [O_3]^2[O_2]^{–1} \nonumber$
What's this? It looks as if O2 actually inhibits the reaction in some way. The generally-accepted mechanism for this reaction is:
Step 1 $\ce{O3 <=>[k_1][k_{-1}] O2+ O}$ (rapid equilibrium)
Step 2 $\ce{O + O3 ->[k_2] 2 O2}$ (rate-determining)
Does this seem reasonable? Note that
• The equilibrium in Step 1 should be quite rapid because ozone's instability predicts a low activation energy for the forward process. The same can be said for the reverse process which involves the highly reactive oxygen atom which we would expect to rapidly gobble up one of the nonbonding electron pairs on the O2 molecule.
• The inhibitory effect of O2 can be explained by the Le Chatelier effect — being a product of the equilbrium in 1, buildup of oxygen forces it back to the left.
• In Step 2, one might expect the O atom to react with ozone as rapidly as it does with O2 in the reverse of Step 1, but other studies show that this is not the case. The rate constant k2 is probably fairly large. Given the strong covalent bond in O2, we not expect any significant reverse reaction.
To translate this mechanism into a rate law, we first write the equilibrium constant for Step 1 and solve it for the concentration of the intermediate:
$K =\dfrac{[O_2][O]}{[O_3]} \nonumber$
$[O] = \dfrac{k_1[O_3]}{k_{-1}[O_2]} \nonumber$
We substitute this value of $\ce{[O]}$ into the rate expression $\ce{[O][O3]}$ for Step 2, which yields the experimentally-obtained rate law
$rate = k_1K \dfrac{[O_3]^2}{[O_2]} \nonumber$
Example $4$: A reaction having an apparently Negative Ea
Consider the gas-phase oxidation of nitric oxide:
$\ce{2 NO + O_2 → 2 NO_2} \nonumber$
This reaction, like most third-order reactions, is not termolecular but rather a combination of an equilibrium followed by a subsequent bimolecular step:
Step 1 $\ce{ NO + NO <=>[k_1][k_{-1}] N2O2}$ (equilibrium, eq. const. K)
Step 2 $\ce{ N2O2 + O2 ->[k_2] 2 NO2}$ (rate-limiting)
Since the intermediate $\ce{N2O2}$ may not appear in the rate equation, we need to express its concentration in terms of the reactant $NO$. As in the previous example, we do this through the equilibrium constant of Step 1:
$K = \dfrac{[N_2O_2]}{[NO]^2} \nonumber$
$[N_2O_2] = K [NO]^2 \nonumber$
\begin{align*} \text{rate} &= k_2 [N_2O_2 ][O_2] \[4pt] &= k_2 K [NO]^2 [O_2] \end{align*}
The unusual feature of this net reaction is that its rate diminishes as the temperature increases, suggesting that the activation energy is negative. Reaction 1 involves bond formation and is exothermic, so as the temperature rises, $K$ decreases (Le Chatelier effect). At the same time, k2 increases, but not sufficiently to overcome the decrease in $K$. So the apparently negative activation energy of the overall process is simply an artifact of the magnitudes of the opposing temperature coefficients of k2 and $K$.
Chain Reactions (Positive Feedback Mechanisms)
Many important reaction mechanisms, particularly in the gas phase, involve intermediates having unpaired electrons, commonly known as free radicals. Free radicals are often fairly stable thermodynamically and may be quite long-lived by themselves, but they are highly reactive, and hence kinetically labile. The dot ·, representing the unpaired electron, is not really a part of the formula, and is usually shown only when we want to emphasize the radical character of a species. The "atomic" forms of many elements that normally form diatomic molecules are free radicals; H·, O·, and Br· are common examples. The simplest and most stable (ΔG = +87 kJ/mol) molecular free radical or "odd-electron molecule" is nitric oxide, NO·.
The most important chemical property of a free radical is its ability to pass the odd electron along to another species with which it reacts. This chain-propagation process creates a new radical which becomes capable of initiating another reaction. Radicals can, of course, also react with each other, destroying both ("chain termination") while creating a new covalent-bonded species.
Max Bodenstein
Much of the pioneering work in this field, of which the $\ce{HBr}$ synthesis is a classic example, was done by the German chemist Max Bodenstein (1871-1942)
The synthesis of hydrogen bromide from its elements illustrates the major features of a chain reaction. The figures in the right-hand column are the activation energies per mole.
Step 1 Br2 + M → 2 Br· chain initiation (ΔH° = +188 kJ)
Step 2 Br· + H2 → H· + HBr chain propagaton (+75 kJ)
Step 3 H· + Br2 → HBr + Br· chain propagation (–176 kJ)
Step 4 H· + HBr → H2 + Br· chain inhibition (–75 kJ)
Step 5 2 Br· + M → M* + Br2 chain termination (–188 kJ)
Note the following points:
• To start the reaction, a free radical must be formed (1). If the temperature of the gas is fairly high, then Br· can be formed from the more energetic collisions of Br2 molecules with some other molecule M (most likely a second Br2). This is known as thermal activation; Another way of creating free radicals is photochemical activation.
• Reactions 2 and 3 consume a free radical but form another, thus "propagating" the chain.
• In reaction 4, a molecule of the product is destroyed, thus partially un-doing the net process.
• If the only the first four reactions were active, then the cycle would continue indefinitely. But reaction 5 consumes the Br· chain carrier without producing new radicals, thus terminating the chain. The function of the molecule M is to absorb some of the kinetic energy of the collision so that the two bromine atoms do not simply bounce apart. The product M* represents a thermally-excited M which quickly dissipates its energy to other molecules.
• Other reactions, such as the recombination of hydrogen atoms, also take place, but their contribution to the overall kinetics is usually very small.
The rate laws for chain reactions tend to be very complex, and often have non-integral orders.
Branching Chain Reactions: Explosions
The gas-phase oxidation of hydrogen has been extensively studied over a wide range of temperatures and pressures.
$\ce{H2(g) + 1/2 O2(g) → H2O(g)}\quad ΔH^o = –242\, kJ/mol \nonumber$
This reaction does not take place at all when the two gases are simply mixed at room temperature. At temperatures around 500-600°C it proceeds quite smoothly, but when heated above 700° or ignited with a spark, the mixture explodes. As with all combustion reactions, the mechanism of this reaction is extremely complex (you do not want to see the rate law!) and varies somewhat with the conditions. Some of the major radical formation steps are
Step 1 H2 + O2 → HO2· + H· chain initiation
Step 2 H2 + HO2· → HO· + H2O chain propagaton
Step 3 H· + O2 → HO· + O· chain propagaton + branching
Step 4 O· + H2 → HO· + H· chain propagation + branching
Step 5 HO· + H2 → H2O + H· chain propagation
The species HO2· is known as perhydroxyl; HO· is hydroxyl (not to be confused with the hydroxyl ion HO:)
Reactions 3 and 4 give birth to more radicals than they consume, so when these are active, each one effectively initiates a new chain process causing the rate overall rate to increase exponentially, producing an explosion.
Explosion Limits
An explosive reaction is a highly exothermic process that once initiated, goes to completion very rapidly and cannot be stopped. The destructive force of an explosion arises from the rapid expansion of the gaseous products as they absorb the heat of the reaction. There are two basic kinds of chemical explosions:
• Thermal explosions occur when heat is released by a reaction more rapidly than it can escape from the reaction space. This causes the reaction to proceed more rapidly, releasing even more heat, resulting in an ever-accelerating runaway rate.
• Chain-branching explosions occur when the number of chain carriers increases exponentially, effectively seeding new reaction centers in the mixture. The process then transforms into a thermal explosion.
Whether or not a reaction proceeds explosively depends on the balance between formation and destruction of the chain-carrying species. This balance depends on the temperature and pressure, as illustrated here for the hydrogen-oxygen reaction.
• Direct recombination of chain carriers generally requires a three-body collision with another molecule to absorb some of the kinetic energy; such ternary processes are unlikely at very low pressures. Thus below the lower explosion limit , the radicals (including those produced by a spark) are usually able to reach the walls of the container and combine there — or in the case of an open-air explosion, simply combine with other molecules as they exit the active volume of the reaction.
• Within the explosion zone between and (which, for most gases, are known as the lower- and upper explosion limits), the propagation and branching processes operate efficiently and explosively, even when the mixture is heated homogeneously.
• Above , the concentration of gas molecules is sufficient to enable ternary collisions which allow chain-termination processes to operate efficiently, thus suppressing branching. Above this upper limit, reactions involving most gases proceed smoothly.
• Hydrogen is unusual in that it exhibits a third explosion limit . The cause of this was something of a mystery for some time, but it is now believed that explosions in this region do not involve branching, but are thermally induced by the reaction
HO2· + H2 → H2O2 + H· .
The lower explosion limit of gas mixtures varies with the size, shape, and composition of the enclosing container. Needless to say, experimental determination of explosion limits requires some care and creativity. Upper and lower explosion limits for several common fuel gases are shown below. | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.04%3A_Reaction_Mechanisms.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas:
• Describe some of the major differences between the kinetics of reactions in the gas phase, compared with those in liquid solutions.
• What role do solvent cages play in solution kinetics?
• Explain the distinction between diffusion-control and activation-control of reaction rates in solutions.
• How can the polarity of a solvent affect the energetics of a reaction mechanism?
The kinetics fundamentals we covered in the earlier sections of this lesson group relate to processes that take place in the gas phase. But chemists and biochemists are generally much more concerned with solutions. This lesson will take you through some of the extensions of basic kinetics that you need in order to understand the major changes that occur when reactions take place in liquid solutions.
What's different about kinetics in liquid solutions?
Most of the added complications of kinetics and rate processes in liquid solutions arise from the much higher density of the liquid phase. In a typical gas at atmospheric pressure, the molecules occupy only about 0.2 per cent of the volume; the other 99.8 percent is empty space. In a liquid, molecules may take up more than half the volume, and the "empty" spaces are irregular and ever-changing as the solvent molecules undergo thermal motions of their own.
In a typical liquid solution, the solvent molecules massively outnumber the reactant solute molecules, which tend to find themselves momentarily (~10–11 sec) confined to a "hole" within the liquid. This trapping becomes especially important when the solvent is strongly hydrogen-bonded as is the case with water or alcohol.
When thermal motions occasionally release a solute molecule from this trap, it will jump to a new location. The jumps are very fast (10–12 - 10–13 sec) and short (usually a few solvent-molecule diameters), and follow an entirely random pattern, very much as in Brownian motion. Consider a simple bimolecular process A + B → products. The reactant molecules will generally be jumping from hole to hole in the solvent matrix, only occasionally finding themselves in the same solvent cage where thermal motions are likely to bring them into contact.
Table $1$: Solvent cages and encounter pairs
A pair of reactants end up in the same solvent cage, where they bounce around randomly and exchange kinetic energy with the solvent molecules. Eventually the two reactants form an encounter pair. If they fail to react the first time, they have many more opportunities during the lifetime of the cage. The products form and begin to move away from each other. Finally, after about 10–11 sec, the solvent cage breaks up and the products diffuse away.
The process can be represented as
$A + B \rightarrow \{AB\} → \text{products}$
in which the $\{AB\}$ term represents the caged reactants including the encounter pair and the activated complex.
Contrast this scenario with a similar reaction taking place in the gas phase; the molecules involved in the reaction will often be the only ones present, so a significant proportion of the collisions will be $A$-$B$ encounters. However, if the collision should fail to be energetically or geometrically viable, the reactant molecules fly apart and are unlikely to meet again anytime soon. In a liquid, however, the solute molecules are effectively in a constant state of collision — if not with other reactants, then with solvent molecules which can exchange kinetic energy with the reactants. So once an A-B encounter pair forms, the two reactants get multiple whacks at each other, greatly increasing the probability that they will obtain the kinetic energy needed to kick them over the activation hump before the encounter pair disintegrates.
Limiting Cases: Diffusion-Controlled and Activation-Controlled Reactions
The encounter pair model introduces some new rate parameters:
$\ce{A + B <=>[k1][k_{-1}] {AB} -> products}$
The first step is an equilibrium between the reactants outside and inside the solvent cage. The rate constants $k_1$ and $k_2$ reflect those relating to diffusion of molecules through the solvent; their values are strongly dependent on the viscosity (and thus the temperature) of the solvent. (Note that $k_1$ is a second-order rate constant, while $k_2$ is first-order.)
Diffusion is the transport of a substance through a concentration gradient; that is, from a region of higher concentration to one of lower concentration. Think of the way the color of tea spreads out when a tea bag is immersed in hot water. Diffusion occurs because random thermal motions are statistically more likely to move molecules out of a region of higher concentration than in the reverse direction, simply because in the latter case fewer molecules are available to make the reverse trip. Eventually the concentrations become uniform and equilibrium is attained.
As molecules diffuse through a liquid, they must nudge neighboring molecules out of the way. The work required to do this amounts to an activation energy, so diffusion can be thought of as a kinetic process with its own rate constant kd and activation energy. These parameters depend on the sizes of the solute and solvent molecules and on how strongly the latter interact with each other. This suggests two important limiting cases for reactions in solution.
For water at room temperature, k1 is typically 109-1010 dm–3 mol–1 s–1 and k2 is around 10–9-10–10 dm–3 mol–1 s–1. Given these values, k3 > 1012 s–1 implies diffusion control, while values < 109 s–1 are indicative of activation control.
• Diffusion Controlled ($k_3 \gg k_2$): If the activation energy of the A+B reaction is very small or if escape of molecules from the {AB} cage is difficult, the kinetics will be dominated by k1, and thus by the activation energy of diffusion. Such a process is said to be diffusion controlled. Reactions in aqueous solution in which Ea > 20 kJ/mol are likely to fall into this category.
• Activation Controlled ($k_3 \ll k_2$): Alternatively, if the activation energy of the A+B reaction dominates the kinetics, and the reaction is activation-controlled.
Several general kinds of reactions are consistently very "fast" and thus are commonly found to be diffusion-controlled in most solvents:
Units Matter
Gas-phase rate constants are normally expressed in units of mol s–1, but rate constants of reactions in solution are conventionally given in mol/L units, or dm3 mol–1 s–1. Conversion between them depends on a number of assumptions and is non-trivial.
• Recombination of atoms and radicals
For example the formation of I2 from I atoms in hexane at 298 K has k3 = 1.3×1012 dm3 mol–1 s–1.
• Acid-base reactions which involve the transport of H+ and OH ions tend to be very fast.
The most famous of these is one of the fastest reactions known:$H^+ + OH^– → H_2O \nonumber$ for which k3 = 1.4×1011 dm3 mol–1 s–1 at 298 K.
Polar solvents such as water and alcohols interact with ions and polar molecules through attractive dipole-dipole and ion-dipole interactions, leading to lower-energy solvated forms which stabilize these species. In this way, a polar solvent can alter both the thermodynamics and kinetics (rate) of a reaction.
Solvent Thermodynamic Effect
If the products of the reaction are markedly more or less polar than the reactants, solvent polarity can change the overal thermodynamics (equilibrium constant) of the reaction. Nowhere is this more apparent than when an ionic solid such as salt dissolves in water. The Na+ and Cl ions are bound together in the solid through strong coulombic forces; pulling the solid apart in a vacuum or in a nonpolar solvent is a highly endothermic process. In contrast, dissolution of NaCl in water is slightly exothermic and proceeds spontaneously.
The water facilitates this process in two important ways. First, its high dielectric constant of 80 reduces the force between the separated ions to 1/80 of its normal value. Secondly, the water molecules form a solvation shell around the ions (lower left), rendering them energetically (thermodynamically) more stable than they were in the NaCl solid.
Solvent Kinetic Effect
In the same way, a reaction whose mechanism involves the formation of an intermediate or activated complex having a polar or ionic character will have its activation energy, and thus its rate, subject to change as the solvent polarity is altered. As an example we will consider an important class of reactions that you will hear much about if you take a course in organic chemistry. When an aqueous solution of a strong base such as KOH is added to a solution of tertiary-butyl chloride in ethanol, the chlorine is replaced by a hydroxyl group, leaving t-butyl alcohol as a product:
This reaction is one of a large and important class known as SN1 nucleophilic substitution processes that are discussed in most organic chemistry courses. In these reactions, a species that possesses a pair of non-bonding electrons (also called a nucleophile or Lewis base) uses them to form a new bond with an electrophile — a compound in which a carbon atom has a partial positive charge owing to its bonds to electron-withdrawing groups. In the example here, other nucleophiles such as NH3 or even H2O would serve as well.
In order to reflect the generality of this process and to focus on the major changes that take place, we will represent this reaction as
Extensive studies of this class of reactions in the 1930's revealed that it proceeds in two activation energy-controlled steps, followed by a simple dissociation into the products:
In step , which is rate-determining, the chlorine leaves the alkyl chloride which becomes an intermediate known as a carbocation ("cat-ion"). These ions, in which the central carbon atom lacks a complete octet, are highly reactive, and in step the carbocation is attacked by the hydroxide ion which supplies the missing electron. The immediate product is another cation in which the positive charge is on the oxygen atom. This oxonium ion is unstable and rapidly dissociates ()into the alcohol and a hydrogen ion.
The reaction coordinate diagram helps us understand the effect of solvent polarity on this reaction. Polar solvent molecules interact most strongly with species in which the electric charge is concentrated in one spot. Thus the carbocation is stabilized to a greater degree than are the activated complexes in which the charge is spread out between the positive and negative ends. As the heavy green arrows indicate, a more polar solvent will stabilize the carbocation more than it will either of the activated complexes; the effect is to materially reduce the activation energy of the rate-determining step, and thus speed up the reaction. Because neither the alkyl chloride nor the alcohol is charged, the change in solvent polarity has no effect on the equilibrium constant of the reaction. This is dramatically illustrated by observing the rate of the reaction in solvents composed of ethanol and water in varying amounts:
Table $2$: Data
% water 10 20 30 40 50 60
$k_1 \times 10^6$
1.7 9.1 40.3 126 367 1294 | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.05%3A_Kinetics_of_Reactions_in_Solution.txt |
Learning Objectives
Make sure you thoroughly understand the following essential ideas which have been presented above. It is especially imortant that you know the precise meanings of all the green-highlighted terms in the context of this topic.
• What are catalysts, and how do they work in terms altering the parameters of a reaction?
• Describe the similarities and differences between the three principal classes of catalysts.
• Define physisorption and chemisorption, and explain the role of the latter in initiating a catalytic event.
• What is the meaning and significance of the Sabatier Principle?
• What are the principal differences between the Langmuir-Hinshelwood and Eley-Rideal mechanisms of heterogeneous catalysis?
• Describe the role of the enzyme-substrate complex.
• How might the particular properties of the amino-acids surrounding the active site of an enzyme contribute to its catalytic effect?
• Describe the lock-and-key and induced-fit models of enzyme action.
• Explain the function and significance of allosteric sites on an enzyme.
It almost seems like magic! A mixture of gaseous H2 and O2 can coexist indefinitely without any detectable reaction, but if a platinum wire is heated and then immersed in the gaseous mixture, the reaction proceeds gently as the heat liberated by the reaction makes the wire glow red-hot. Catalysts play an essential role in our modern industrial economy, in our stewardship of the environment, and in all biological processes. This lesson will give you a glimpse into the wonderful world of catalysts, helping you to understand what they are and how they work.
What are Catalysts?
Catalysts have no effect on the equilibrium constant and thus on the equilibrium composition. Catalysts are substances that speed up a reaction but which are not consumed by it and do not appear in the net reaction equation. Also — and this is very important — catalysts affect the forward and reverse rates equally; this means that catalysts have no effect on the equilibrium constant and thus on the composition of the equilibrium state. Thus a catalyst (in this case, sulfuric acid) can be used to speed up a reversible reaction such as ester formation or its reverse, ester hydrolysis:
The catalyst has no effect on the equilibrium constant or the direction of the reaction. The direction can be controlled by adding or removing water (Le Chatelier principle).
Catalysts function by allowing the reaction to take place through an alternative mechanism that requires a smaller activation energy. This change is brought about by a specific interaction between the catalyst and the reaction components. You will recall that the rate constant of a reaction is an exponential function of the activation energy, so even a modest reduction of $E_a$ can yield an impressive increase in the rate.
Catalysts provide alternative reaction pathways
Catalysts are conventionally divided into two categories: homogeneous and heterogeneous. Enzymes, natural biological catalysts, are often included in the former group, but because they share some properties of both but exhibit some very special properties of their own, we will treat them here as a third category.
Some common examples of catalysis
How to burn a Sugar Cube
When heated by itself, a sugar cube (sucrose) melts at 185°C but does not burn. But if the cube is rubbed in cigarette ashes, it burns before melting owing to the catalytic action of trace metal compounds in the ashes.
Platinum as an Oxidation Catalyst
The surface of metallic platinum is an efficient catalyst for the oxidation of many fuel vapors. This property is exploited in flameless camping stoves (left). The image at the right shows a glowing platinum wire heated by the slow combustion of ammonia on its surface. However, if you dip a heated Pt wire into liquid ammonia, you get a miniature explosion: see video below.
This is oxidation of ammonia by oxygen catalysed by warmed platinum wire. Oxygen is bubbled through ammonia solution, in which it mixes with ammonia gas present. The reaction causes the platinum wire to glow, and the hot wire ignites a mixture of ammonia and oxygen.
Decomposition of Hydrogen Peroxide
Hydrogen peroxide is thermodynamically unstable according to the reaction
$\ce{2 H2O2 → 2 H2O + O2 } \quad \quad ΔG^o = –210\, kJ\, mol^{–1}$
In the absence of contaminants this reaction is very slow, but a variety of substances, ranging from iodine, metal oxides, trace amount of metals, greatly accelerate the reaction, in some cases almost explosively owing to the rapid release of heat. The most effective catalyst of all is the enzyme catalase, present in blood and intracellular fluids; adding a drop of blood to a solution of 30% hydrogen peroxide induces a vigorous reaction.
Rapid liberation of O2 can result in a spectacular bubble bath if some soap is added. [This same reaction has been used to power a racing car!
Potassium iodide efficiently catalyzes H2O2 deposition. This short video shows what happens when some colored soap, H2O2, and KI are combined. But "don't try this at [your] home"!
Each kind of catalyst facilitates a different pathway with its own activation energy. Because the rate is an exponential function of Ea (Arrhenius equation), even relatively small differences in Ea's can have dramatic effects on reaction rates. Note especially the values for catalase; the chemist is still a rank amateur compared to what Nature can accomplish through natural selection!
catalyst
Ea
kJ/mol
relative rate
no catalyst 75 1
iodide ion 56 2140
colloidal platinum 50 24,000
catalase (enzyme) 21 2,900,000,000
How catalytic activity is expressed
Changes in the rate constant or of the activation energy are obvious ways of measuring the efficacy of a catalyst. But two other terms have come into use that have special relevance in industrial applications.
Turnover number
The turnover number (TON) is an average number of cycles a catalyst can undergo before its performance deteriorates (see below). Reported TONs for common industrial catalysts span a very wide range from perhaps 10 to well over 105, which approaches the limits of diffusion transport.
Turnover Frequency
This term, which was originally applied to enzyme-catalyzed reactions, has come into more general use. It is simply the number of times the overall catalyzed reaction takes place per catalyst (or per active site on an enzyme or heterogeneous catalyst) per unit time:is defined as
The number of active sites S on a heterogeneous catalyst is often difficult to estimate, so it is often replaced by the total area of the exposed catalyst, which is usually experimentally measurable. TOFs for heterogeneous reactions generally fall between 10–2 to 102 s–1.
Homogeneous Catalysis
As the name implies, homogeneous catalysts are present in the same phase (gas or liquid solution) as the reactants. Homogeneous catalysts generally enter directly into the chemical reaction (by forming a new compound or complex with a reactant), but are released in their initial form after the reaction is complete, so that they do not appear in the net reaction equation.
Iodine-catalyzed cis-trans isomerization
Unless you are taking an organic chemistry course in which your instructor indicates otherwise, don't try to memorize these mechanisms. They are presented here for the purpose of convincing you that catalysis is not black magic, and to familiarize you with some of the features of catalyzed mechanisms. It should be sufficient for you to merely convince yourself that the individual steps make chemical sense.
You will recall that cis-trans isomerism is possible when atoms connected to each of two doubly-bonded carbons can be on the same (cis) or opposite (trans) sides of the bond. This reflects the fact that rotation about a double bond is not possible.
Conversion of an alkene between its cis- and trans forms can only occur if the double bond is temporarily broken, thus freeing the two ends to rotate. Processes that cleave covalent bonds have high activation energies, so cis-trans isomerization reactions tend to be slow even at high temperatures. Iodine is one of several catalysts that greatly accelerate this process, so the isomerization of butene serves as a good introductory example of homogeneous catalysis.
The mechanism of the iodine-catalyzed reaction is believed to involve the attack of iodine atoms (formed by the dissociation equilibrium on one of the carbons in Step :
During its brief existence, the free-redical activated complex can undergo rotation about the C—C bond, so that when it decomposes by releasing the iodine (), a portion of the reconstituted butene will be in its trans form. Finally, the iodine atom recombine into diiodine. Since processes and cancel out, iodine does not appear in the net reaction equation — a requirement for a true catalyst.
Acid-base catalysis
Many reactions are catalyzed by the presence of an acid or a base; in many cases, both acids and bases will catalyze the same reaction. As one might expect, the mechanism involves the addition or removal of a proton, changing the reactant into a more kinetically labile form. A simple example is the addition of iodine to propanone
I2 + (CH3)2–C=O → (CH2I)(CH3)–C=O
The mechanism for the acid-catalyzed process involves several steps. The role of the acid is to provide a proton that attaches to the carbonyl oxygen, forming an unstable oxonium ion . The latter rapidly rearranges into an enol (i.e., a carbon connected to both a double bond (ene) and a hydroxyl (ol) group.) This completes the catalytic part of the process, which is basically an acid-base (proton-transfer) reaction in which the role of the proton is to withdraw an electron from the ketone oxygen.
In the second stage, the enol reacts with the iodine. The curved arrows indicate shifts in electron locations. In below, an electron is withdrawn from the π orbital of the double bond by one of the atoms of the I2 molecule. This induces a shift of electrons in the latter, causing half of this molecule to be expelled as an iodide ion. The other half of the iodine is now an iodonium ion I+ which displaces a proton from one of the methyl groups. The resultant carbonium ion then expels the -OH proton to yield the final neutral product.
Perhaps the most well-known acid-catalyzed reaction is the hydrolysis (or formation) of an ester — a reaction that most students encounter in an organic chemistry laboratory course. This is a more complicated process involving five steps; its mechanism is discussed here. See also this U. Calgary site, which describes both the acid- and base-catalyzed reaction.
Oxidation-reduction catalysis
Many oxidation-reduction (electron-transfer) reactions, including direct oxidation by molecular oxygen, are quite slow. Ions of transition metals capable of existing in two oxidation states can often materially increase the rate. An example would be the reduction of Fe3+ by the vanadium ion V3+:
$\ce{V^{3+} + Fe^{3+} → V^{4+} + Fe^{2+}}$
This reaction is catalyzed by either Cu+or Cu3+, and the rate is proportional to the concentration of V3+and of the copper ion, but independent of the Fe3+ concentration. The mechanism is believed to involve two steps:
1 V3+ + Cu2+ V4+ + Cu+ (rate-determining)
2 Fe3+ + Cu+ Fe2+ + Cu2+ (very fast)
(If Cu+ is used as the catalyst, it is first oxidized to Cu2+ by step 2.)
Hydrogen Peroxide, again
Ions capable of being oxidized by an oxidizing agent such as H2O2 can serve as catalysts for its decomposition. Thus H2O2 oxidizes iodide ion to iodate, when then reduces another H2O2 molecule, returning an I ion to start the cycle over again:
$H_2O_2 + I^– → H_2O + IO^–$
$H_2O_2 + IO^– → H_2O + O_2 + I^–$
Iron(II) can do the same thing. Even traces of metallic iron can yield enough Fe2+ to decompose solutions of hydrogen peroxide.
$H_2O_2 + Fe^{2+} → H_2O + Fe^{3+}$
$H_2O_2 + Fe^{3+} + 2H^+ → H_2O + O_2 + Fe^{2+} + 2H^+$
Heterogeneous catalysts
As its name implies, a heterogeneous catalyst exists as a separate phase (almost always a solid) from the one (most commonly a gas) in which the reaction takes place. The catalytic affect arises from disruption (often leading to dissociation) of the reactant molecules brought about by their interaction with the surface of the catalyst.
Model of a catalyst consisting of clusters of 8-10 platinum atoms (blue) deposited on an aluminum oxide surface. This catalyst efficiently removes hydrogen atoms from propane, converting it into the industrially-important propylene. [source]
Unbalanced forces at surfaces
You will recall that one universal property of matter is the weak attractive forces that arise when two particles closely approach each other. (See here for a quick review.) When the particles have opposite electric charges or enter into covalent bonding, these far stronger attraction dominate and define the "chemistry" of the interacting species.
The molecular units within the bulk of a solid are bound to their neighbors through these forces which act in opposing directions to keep each under a kind of "tension" that restricts its movement and contributes to the cohesiveness and rigidity of the solid.
At the surface of any kind of condensed matter, things are quite different. The atoms or molecules that reside on the surface experience unbalanced forces which prevents them from assuming the same low potential energies that characterize the interior units. (The same thing happens in liquids, and gives rise to a variety of interfacial effects such as surface tension.)
But in the case of a solid, in which the attractive forces tend to be stronger, something much more significant happens. The molecular units that reside on the surface can be thought of as partially buried in it, with their protruding parts (and the intermolecular attractions that emerge from them) exposed to the outer world. The strength of the attractive force field which emanates from a solid surface varies in strength depending on the nature of the atoms or molecules that make up the solid.
Don't confuse adsorption with absorption; the latter refers to the bulk uptake of a substance into the interior of a porous material. At the microscopic level, of course, absorption also involves adsorption. The process in which molecules in a gas or a liquid come into contact with and attach themselves to a solid surface is known as adsorption. Adsorption is almost always an exothermic process and its strength is conventionally expressed by the enthalpy or "heat" of adsorption ΔHads.
Chemisorption and Physisorption
Two general categories of adsorption are commonly recognized, depending on the extent to which the electronic- or bonding structure of the attached molecule is affected. When the attractive forces arise from relatively weak van der Waals interactions, there is little such effect and ΔHads tends to be small. This condition is described as physical adsorption (physisorption). Physisorption of a gas to a surface is energetically similar to the condensation of the gas to a liquid, it usually builds up multiple layers of adsorbed molecules, and it proceeds with zero activation energy.
Of more relevance to catalytic phenomena is chemisorption, in which the adsorbate is bound to the surface by what amounts to a chemical bond. The resulting disruption of the electron structure of the adsorbed species "activates" it and makes it amenable to a chemical reaction (often dissociation) that could not be readily achieved through thermal activation in the gas or liquid phase. In contrast to physisorption, chemisorption generally involves an activation energy (supplied by ΔHads) and the adorbed species is always a monolayer.
Mechanisms of reactions on surfaces
Dissociative adsorption
The simplest heterogeneous process is chemisorption followed by bond-breaking as described above. The most common and thoroughly-studied of these is the dissociation of hydrogen which takes place on the surface of most transition metals. The single 1s electron of each hydrogen atom coordinates with the d orbitals of the metal, forming a pair of chemisorption bonds (indicated by the red dashed lines). Although these new bonds are more stable than the single covalent bond they replace, the resulting hydrogen atoms are able to migrate along the surface owing to the continuous extent of the d-orbital conduction band.
The Langmuir-Hinshelwood mechanism
Although the adsorbed atoms ("adatoms") are not free radicals, they are nevertheless highly reactive, so if a second, different molecular species adsorbs onto the same surface, an interchange of atoms may be possible. Thus carbon monoxide can be oxidized to CO2 by the process illustrated below:
In this example, only the O2 molecule undergoes dissociation . The CO molecule adsorbs without dissociation , configured perpendicular to the surface with the chemisorption bond centered over a hollow space between the metal atoms. After the two adsorbed species have migrated near each other , the oxygen atom switches its attachment from the metal surface to form a more stable C=O bond with the carbon , followed by release of the product molecule.
The Eley-Rideal mechanism
An alternative mechanism eliminates the second chemisorption step; the oxygen adatoms react directly with the gaseous CO molecules by replacing the chemisorption bond with a new C–O bond as they swoop over the surface:
Examples of both mechanisms are known, but the Langmuir-Hinshelwood mechanism is more importantin that it exploits the activation of the adsorbed reactant. In the case of carbon monoxide oxdation, studies involving molecular beam experiments support this scheme. A key piece of evidence is the observation of a short time lag between contact of a CO molecule with the surface and release of the CO2, suggesting that CO remains chemisorbed during the interval.
The Sabatier Principle
To be effective, these processes of adsorption, reaction, and desorption must be orchestrated in a way that depends critically on the properties of the catalyst in relation to the chemisorption properties (ΔHads) of the reactants and products.
• Adsorption of the reactant onto the catalytic surface (2) must be strong enough to perturb the bonding within the species to dissociate or activate it;
• If the resulting fragments must migrate to other locations on the surface (3-4), their chemisorption must be weak enough to allow this movement but not so small that they escape before they have a chance to react;
• The product species must have sufficiently small ΔHads values to ensure their rapid desorption from the catalyst (5) so that surface is freed up to repeat the cycle
The importance of choosing a catalyst that achieves the proper balance of the heats of adsorption of the various reaction components is known as the Sabatier Principle, but is sometimes referred to as the "just-right" or "Goldilocks principle". Remember the story of Goldilocks and the Three Bears? ... or see this UTube video.
In its application to catalysis, this principle is frequently illustrated by a "volcano diagram" in which the rate of the catalyzed reaction is plotted as a function of ΔHads of a substrate such as H2 on a transition metal surface.
The plot at the left shows the relative effectiveness of various metals in catalyzing the decomposition of formic acid HCOOH. The vertical axis is plotted as temperature, the idea being that the better the catalyst, the lower the temperature required to maintain a given rate.
The catalytic cycle
This term refers to the idealized sequence of steps between the adsorption of a reactant onto the catalyst and the desorption of the product, culminating in restoration of the catalyst to its original condition. A typical catalytic cycle for the hydrogenation of propene is illustrated below.
This particular reaction
H3C–CH=CH2 + H2 → H3C–CH2–CH3
takes place spontaneously only in the reverse direction, but it is representative of the process used to hydrogenate the carbon-carbon double bonds in vegetable oils to produce solid saturated fats such as margarine.
Catalyst poisoning and breakdown
Catalyst poisoning, brought about by irreversible binding of a substance to its surface, can be permanent or temporary. In the latter case the catalyst can be regenerated, usually by heating to a high temperature. In organisms, many of the substances we know as "poisons" act as catalytic poisons on enzymes. If catalysts truly remain unchanged, they should last forever, but in actual practice, various events can occur that limit the useful lifetime of many catalysts.
• Impurities in the feedstock or the products of side reactions can bind permanently to a sufficient number of active sites to reduce catalytic efficiency over time, "poisoning" the catalyst.
• Physical deterioration of the catalyst or of its support, often brought about by the high temperatures sometimes used in industrial processes, can reduce the effective surface area or the accessibility of reactants to the active sites.
Catalysts tend to be rather expensive, so it is advantageous if they can be reprocessed or regenerated to restore their activity. It is a common industrial practice to periodically shut down process units to replace spent catalysts.
How heterogeneous catalysts work
The actual mechanisms by which adsorption of a molecule onto a catalytic surface facilitates the cleavage of a bond vary greatly from case to case.
We give here only one example, that of the dissociation of dixoygen O2 on the surface of a catalyst capable of temporarily donating an electron which enters an oxygen antibonding molecular orbital that will clearly destabilize the O–O bond. (Once the bond has been broken, the electron is given back to the catalyst.)
Types of catalytically active surfaces
Heterogeneous catalysts mostly depend on one or more of the followng kinds of surfaces:
Active surface type Remarks
Atoms at surfaces or crystal edges in macromolecular 2- and 3-D networks such as graphite or quartz may have free valences
Surfaces and edges are sites of intense electric fields able to interact with ions and polar molecules
Oxides can acquire H+ and/or OH groups able to act as acid- or base catalyts
"Electron gas" at metal surfaces can perturb bonding in substrate molecules
Vacant d orbitals can provide a variety of coordination sites for activation.
Semiconductors (including many oxides) can supply electrons, thermally excited through reasonably small (<50 kJ) band gaps.
Some factors affecting catalyst efficacy
Since heterogeneous catalysis requires direct contact between the reactants and the catalytic surface, the area of active surface goes at the top of the list. In the case of a metallic film, this is not the same as the nominal area of the film as measured by a ruler; at the microscopic level, even apparently smooth surfaces are highly irregular, and some cavities may be too small to accommodate reactant molecules.
Consider, for example, that a 1-cm cube of platinum (costing roughly \$1000) has a nominal surface area of only 6 cm2. If this is broken up into 1012 smaller cubes whose sides are 10–6 m, the total surface area would be 60,000 cm2, capable in principle of increasing the rate of a Pt-catalyzed reaction by a factor of 104. These very finely-divided (and often very expensive) metals are typically attached to an inert supporting surface to maximize their exposure to the reactants.
Surface topography. At the microscopic level, even an apparently smooth surface is pitted and uneven, and some sites will be more active than others. Penetration of molecules into and out of some of the smaller channels of a porous surface may become rate-limiting.
An otherwise smooth surface will always possess a variety of defects such as steps and corners which offer greater exposure and may be either the only active sites on the surface, or overly active so as to permanently bind to a reactant, reducing the active area of the surface. In one study, it was determined that kink defects constituting just 5 percent of platinum surface were responsible for over 90% of the catalytic activity in a certain reaction.
Steric factors
When chemisorbtion occurs at two or more locations on the reactant, efficient catalysis requires that the spacing of the active centers on the catalytic surface be such that surface bonds can be formed without significant angular distortion.
Thus activation of the ethylene double bond on a nickel surface proceeds efficiently because the angle between the C—Ni bonds and the C—C is close to the tetrahedral value of 109.5° required for carbon sp3 hybrid orbital formation. Similarly, we can expect that the hydrogenation of benzene should proceed efficiently on a surface in which the active sites are spaced in the range of 150 and 250 pm.
This is one reason why many metallic catalysts exhibit different catalytic activity on different crystal faces.
Some special types of heterogeneous catalysts
Metal-cluster catalysts
As the particle size of a catalyst is reduced, the fraction of more highly exposed step, edge, and corner atoms increases. An extreme case occurs with nano-sized (1-2 nm) metal cluster structures composed typically of 10-50 atoms.
[link] →
Metallic gold, well known for its chemical inertness, exhibits very high catalytic activity when it is deposited as metallic clusters on an oxide support. For example, O2 dissociates readily on Au55 clusters which have been found to efficiently catalyze the oxidation of hydrocarbons [article].
Zeolite catalysts
Zeolites are clay-like aluminosilicate solids that form open-framework microporous structures that may contain linked cages, cavities or channels whose dimensions can be tailored to the sizes of the reactants and products. To those molecules able to diffuse through these spaces, zeolites are in effect "all surface", making them highly efficient. This size-selectivity makes them important for adsorption, separation, ion-exchange, and catalytic applications. Many zeolites occur as minerals, but others are made synthetically in order to optimize their properties.
As catalyts, zeolites offer a number of advantages that has made them especially important in "green chemistry" operations in which the number of processing steps, unwanted byproducts, and waste stream volumes are minimized.
Enzymes and biocatalysis
"A Jug of Wine, a Loaf of Bread, and . . ." Enzymes!
This distortion of Robert FitzGerald's already-distorted translation of the famous quatrain from the wonderful Rubaiyat of Omar Khayyam underlines the central role that enzymes and their technology have played in civilization since ancient times.
Illustration by Edmund Dulac (1882-1953)
Fermentation and wine-making have been a part of human history and culture for at least 8000 years, but recognition of the role of catalysis in these processes had to wait until the late nineteenth century. By the 1830's, numerous similar agents, such as those that facilitate protein digestion in the stomach, had been discovered. The term "enzyme", meaning "from yeast", was coined by the German physiologist Wilhelm Kühne in 1876. In 1900, Eduard Buchner (1860-1917, 1907 Nobel Prize in Chemistry) showed that fermentation, previously believed to depend on a mysterious "life force" contained in living organisms such as yeast, could be achieved by a cell-free "press juice" that he squeezed out of yeast.
By this time it was recognized that enzymes are a form of catalyst (a term introduced by Berzelius in 1835), but their exact chemical nature remained in question. They appeared to be associated with proteins, but the general realization that enzymes are proteins began only in the 1930s when the first pure enzyme was crystallized, and did not become generally accepted until the 1950s. It is now clear that nearly all enzymes are proteins, the major exception being a small but important class of RNA-based enzymes known as ribozymes.
Proteins are composed of long sequences of amino acids strung together by amide bonds; this sequence defines the primary structure of the protein.. Their huge size (typically 200-2000 amino acid units, with total molecular weights 20,000 - 200,000) allows them to fold in complicated ways (known as secondary and tertiary structures) whose configurations are essential to their catalytic function.
Because enzymes are generally very much larger than the reactant molecules they act upon (known in biochemistry as substrates), enzymatic catalysis is in some ways similar to heterogeneous catalysis. The main difference is that the binding of a subtrate to the enzyme is much more selective.
Precursors and cofactors
Most enzymes come into being as inactive precursors (zymogens) which are converted to their active forms at the time and place they are needed.
• For example, the enzymes that lead to the clotting of blood are supposed to remain inactive until bleeding actually begins; a major activating factor is exposure of the blood to proteins in the damaged vessel wall.
• The enzyme that catalyzes the formation of lactose (milk sugar) in the mammary gland is formed during pregnancy, but it remains inactive until the time of birth, when hormonal changes cause a modifier unit to bind to and activate the enzyme.
Conversion to the active form may involve a simple breaking up of the protein by hydrolysis of an appropriate peptide bond or the addition of a phosphate or similar group to one of the amino acid residues.
Many enzyme proteins also require "helper" molecules, known as cofactors, to make them catalytically active. These may be simple metal ions (many of the trace nutrient ions of Cu, Mn, Mo, V, etc.) or they may be more complex organic molecules which are called coenzymes. Many of the latter are what we commonly refer to as vitamins. Other molecules, known as inhibitors, decrease enzyme activity; many drugs and poisons act in this way.
The enzyme-substrate complex
The standard model of enzyme kinetics consists of a two-step process in which an enzyme binds reversibly to its substrate S (the reactant) to form an enzyme-substrate complex ES:
The enzyme-substrate complex plays a role similar to that of the activated complex in conventional kinetics, but the main function of the enzyme is to stabilize the transition state.
In the second, essentially irreversible step, the product and the enzyme are released:
The overall kinetics of the process were worked out in 1913 by the German biochemist Leonor Michaelis and his Canadian student Maud Menten.
The basic kinetic treatment of this process involves the assumption that the concentrations [E] and [ES] reach steady-state values which do not change with time. (The detailed treatment, which is beyond the scope of this course, can be found here.)
The overall process is described by the Michaelis-Menten equation which is plotted here. The Michaelis constant kM is defined as shown, but can be simplified to the ES dissociation constant k–1/k1 in cases when dissociation of the complex is the rate-limiting step. The quantity vmax is not observed directly, but can be determined from kM as shown here.
Enzymes are proteins
In order to understand enzymes and how they catalyze reactions, it is first necessary to review a few basic concepts relating to proteins and the amino acids of which they are composed.
Amino acids: polar, nonpolar, positive, negative.
The 21 amino acids that make up proteins all possess the basic structure shown here, where R represents either hydrogen or a side chain which may itself contain additional –NH2 or –COOH groups. Both kinds of groups can hydrogen-bond with water and with the polar parts of substrates, and therefore contribute to the amino acid's polarity and hydophilic nature. Side chains that contain longer carbon chains and especially benzene rings have the opposite effect, and tend to render the amino acid non-polar and hydrophobic.
The pKa of an ionizable group is the pH at which half of these groups will be ionized. For example, if a –COOH group has a pKa of 2.5, it will exist almost entirely in this form at a pH of about 1.5 or less, and as the negative ion –COO at pH above about 3.5.
Both the –NH2 and –COOH groups are ionizable (i.e., they can act as proton donors or acceptors) and when they are in their ionic forms, they will have an electric charge. The –COOH groups have pKa's in the range 1.8-2.8, and will therefore be in their ionized forms –COO at ordinary cellular pH values of around 7.4. The amino group pKa's are around 8.8-10.6, so these will also normally be in their ionized forms NH3+.
This means that at ordinary cellular pH, both the carboxyl and amino groups will be ionized. But because the charges have opposite signs, an amino acid that has no extra ionizable groups in its side chain will have a net charge of zero. But if the side chain contains an extra amino or carboxyl group, the amino acid can carry a net electric charge. The following diagram illustrates typical amino acids that fall into each of the groups described above.
For clarity, only the charges in the side chains are shown here, but of course the carboxyl and amino groups at the head end of each amino acid will also be ionized. The small green numbers are pKa values. See here for a complete list of amino acids that occur in proteins.
Proteins
Proteins are made up of one or more chains of amino acids linked to each other through peptide bonds by elimination of a water molecule.
The product shown above is called a peptide, specifically it is a dipeptide because it contains two amino acid residues (what's left after the water has been removed.) Proteins are simply very long polypeptide chains, or combinations of them. (The distinction between a long polypeptide and a small protein is rather fuzzy!)
Globular proteins
Most enzymes fall into the category of globular proteins. In contrast to the fibrous proteins that form the structural components of tissues, globular proteins are soluble in water and rarely have any systematic tertiary structures. They are made up of one or more amino-acid ("peptide") chains which fold into various shapes that can roughly be described as spherical — hence the term "globular", and the suffix "globin" that is frequently appended to their names, as in "hemoglobin".
Protein folding is a spontaneous process that is influenced by a number of factors. One of these is obviously their primary amino-acid sequence that facilitates formation of intramolecular bonds between amino acids in different parts of the chain. These consist mostly of hydrogen bonds, although disulfide bonds S—S between sulfur-containing amino acids are not uncommon.
In addition to these intramolecular forces, interactions with the surroundings play an important role. The most import of these is the hydrophobic effect, which favors folding conformations in which polar amino acids (which form hydrogen bonds with water) are on the outside, while the so-called hydrophobic amino acids remain in protected locations within the folds.
How enzymes work: the active site
The catalytic process mediated by an enzyme takes place in a depression or cleft that exposes the substrate to only a few of the hundreds-to-thousands of amino acid residues in the protein chain. The high specificity and activity of enzyme catalysis is sensitively dependent on the shape of this cavity and on the properties of the surrounding amino acids
In 1894, long before it was clear that enzymes are proteins, the German chemist Emil Fischer suggested the so-called lock-and-key model as a way of understanding how a given enzyme can act specificilly on only one kind of substrate molecule. This model is essentially an elaboration of the one we still use for explaining heterogeneous catalysis.
Although the basic lock-and-key model continues to be useful, it has been modified into what is now called the induced-fit model. This assumes that when the substrate enters the active site and interacts with the surrounding parts of the amino acid chain, it reshapes the active site (and perhaps other parts of the enzyme) so that it can engage more fully with the substrate.
One important step in this process is to squeeze out any water molecules that are bound to the substrate and which would interfere with its optimal positioning.
Within the active site, specific interactions between the substrate and appropriately charged, hydrophlic and hydrophobic amino acids of the active site then stabilize the transition state by distorting the substrate molecule in such a way as to lead to a transition state having a substantially lower activation energy than can be achieved by ordinay non-enzymatic catalysis. Beyond this point, the basic catalytic steps are fairly conventional, with acid/base and nucleophilic catalysis being the most common.
For a very clear and instructive illustration of a typical multi-step sequence of a typical enzyme-catalyzed reaction, see this page from Mark Bishop's online textbook, from which this illustration is taken.
Enzyme regulation and inhibition
If all the enzymes in an organism were active all the time, the result would be runaway chaos. Most cellular processes such as the production and utilization of energy, cell division, and the breakdown of metabolic products must operate in an exquisitely choreographed, finely-tuned manner, much like a large symphony orchestra; no place for jazz-improv here!
Nature has devised various ways of achieving this; we described the action of precursors and coenzymes above. Here we focus on one of the most important (and chemically-interesting) regulatory mechanisms.
Allosteric regulation: tweaking the active site
There is an important class of enzymes that possess special sites (distinct from the catalytically active sites) to which certain external molecules can reversibly bind. Although these allosteric sites, as they are called, may be quite far removed from the catalytic sites, the effect of binding or release of these molecules is to trigger a rapid change in the folding pattern of the enzyme that alters the shape of the active site. The effect is to enable a signalling or regulatory molecule (often a very small one such as NO) to modulate the catalytic activity of the active site, effectively turning the enzyme on or off.
In some instances, the product of an enzyme-catalyzed reaction can itself bind to an allosteric site, decreasing the activity of the enzyme and thus providing negative feedback that helps keep the product at the desired concentration. It is believed that concentrations of plasma ions such as calcium, and of energy-supplying ATP are, are regulated in this way.
Allosteric enzymes are more than catalysts: they act as control points in metabolic and cellular signalling networks.Allosteric enzymes frequently stand at the beginning of a sequence of enzymes in a metabolic chain, or at branch points where two such chains diverge, acting very much like traffic signals at congested intersections.
Enzyme inhibition
As is the case with heterogeneous catalysts, certain molecules other than the normal substrate may be able to enter and be retained in the active site so as to competitively inhibit an enzyme's activity. This is how penicillin and related antibiotics work; these molecules covalently bond to amino acid residues in the active site of an enzyme that catalyzes the formation of an essential component of bacterial cell walls. When the cell divides, the newly-formed progeny essentially fall apart.
Enzymes outside the cell
Artist Mike Perkins' conception of an immobilized enzyme within a functionalized nanoporous silica pore. The enzyme is shown as green with positively charged regions shown in red. The blue structures inside the pores represent the negatively charged functional groups added to the mesoporous silica to create a favorable chemical environment for the enzyme. The favorable environment in each pore attracts the enzyme molecule to move into the unoccupied pore. This environment stabilizes and increases the chemical reactivity of the enzyme, allowing it to convert harmful substrate materials (purple particles) into useful or harmless products (yellow and red particles). [link]
Enzymes have been widely employed in the food, pulp-and-paper, and detergent industries for a very long time, but mostly as impure whole-cell extracts.
In recent years, developments in biotechnology and the gradual move of industry from reliance on petroleum-based feedstocks and solvents to so-called "green" chemistry have made enzymes more attractive as industrial catalysts. Compared to the latter, purified enzymes tend to be expensive, difficult to recycle, and unstable outside of rather narrow ranges of temperature, pH, and solvent composition.
Immobilized enzymes
Many of the problems connected with the use of free enzymes can be overcome by immobilizing the enzyme. This can be accomplished in several ways:
Binding the enzyme to an inert solid supporting material
This was first done in 1916, using charcoal or aluminum hydroxide. since the 1960's when this method became more popular, synthetic polymers, often engineered for a specific application, have come into wide use. Some natural biopolymers such as cellulose or starch, and inorganic solids such as silica and alumina have also been employed.
Trapping the enzyme in a porous material
Polymerization of acrylamide in the presence of an enzyme yields a porous gel which alllows free diffusion of substrate and products, Other materils used for this purpose include natural gels such as alginate (obtained from seaweed), porous silica or ceramic solids, and zeolites.
Cross-linked enzymes
If the enzyme is chemically linked to another substance (usually a very small molecule), the resulting solid, known as a cross-linked enzyme aggregate, can retain high catalytic efficiency over a very wide range of conditions. A more recent related development involves precipitating a less-purified form of the enzyme (by addition of ethylene glycol or ammonium sulfate, for example) and then cross-linking the resulting aggregates.
[images↑]
References: more to explore
Catalysis
1. Gadi Rothenberg- Catalysis: Concepts and Green Applications (Wiley-VCH, 2008)
2. R.A. van Santan, M. Neurock - Molecular Heterogeneous Catalysis: A conceptual and computational approach (Wiley-VCH, 2006)
3. History of Catalysis: List of Web pages, including Fifty Years of Catalysis - A list of major advances in the field 1949-1999. More detailed History pages from UK
4. Catalysis and its applications - a very readable and interesting account of the applications of catalysis to various industrially-important fields. | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.06%3A_Catalysts_and_Catalysis.txt |
Studies of the dynamics of chemical processes impinge on almost every area of chemistry and biochemistry. It is useful for students even at the general chemistry level to have some understanding of the experimental techniques that have informed what you have already learned about kinetics.
The time domain of chemical kinetics
Kinetics deals with rates of change, so as a preface to this section, it is useful to consider the range of times within which the study of chemical processes takes place. Prior to about 1945, the practical range for kinetic investigations began at about 103 sec and extended out to around 108 sec, this upper limit being governed by the length of time that a graduate student (who usually does the actual work) can expect to get a Ph.D. degree. The development of flash photolysis and flow methods in the mid-1940s moved the lower time limit down to the millisecond range. The most dramatic changes began to occur in the 1970s as the availablity of lasers capable of very short-lived pulses and fast electronics gradually opened up the study of reactions that are completed in nanoseconds, picoseconds, and even femtoseconds (10–15 sec).
The basic way of obtaining the information needed to determine rate constants and reaction orders is to bring the reactants together and then measure successive changes in concentration of one of the components as a function of time. Two important requirements are
• The time required to take a measurement must be very short compared to the time the reaction takes to run to completion;
• The temperature must be held constant — something than can pose a problem if the reaction is highly exothermic.
Measuring the concentration of a reactant or product directly — that is by chemical analysis, is awkward and seldom necessary. When it cannot be avoided, the reaction sample must usually be quenched in some way in order to stop any further change until its composition can be analyzed. This may be accomplished in various ways, depending on the particular reaction. For reactions carried out in solution, especially enzyme-catalyzed ones, it is sometimes practical to add a known quantity of acid or base to change the pH, or to add some other inhibitory agent. More commonly, however, the preferred approach is to observe some physical property whose magnitude is proportional to the extent of the reaction.
Classical and laboratory-class methods
These methods are applicable to reactions that are not excessively fast, typically requiring a few minutes or hours to run to completion. They were about the only methods available before 1950. Most first-year laboratory courses students will include at least one experiment based on one of these methods.
Optical methods
Light absorption is perhaps the most widely-used technique. If either a reactant or a product is colored, the reaction is easily followed by recording the change in transmission of an appropriate wavelength after the reaction is started.
When a beam of light passes through a solution containing a colored substance, the fraction that is absorbed is directly proportional to the concentration of that substance and to the length of the light's path through the solution. The latter can be controlled by employing a cell or cuvette having a fixed path length.
If I0 is the intensity of the light incident on the cell and I is the intensity that emerges on the other side, then the percent absorption is just 100 × I / I0. Because a limited 1-100 scale of light absorption is often inadequate to express the many orders of magnitude frequently encountered, a logarithmic term optical density is often employed. The relation between this, the cell path length, concentration, and innate absorption ability of the colored substance is expressed by Beer's law.
The simplest absorbance measuring device is a colorimeter in which a beam of white light from an incandescent lamp is passed through a cell (often just a test tube) and onto a photodetector whose electrical output is directly proportional to the light intensity. Before beginning the experiment, a zeroing control sets the meter to zero when the light path is blocked off, and a sensitivity control sets it to 100 when a cell containing an uncolored solution (a "blank") is inserted.
The sensitivity and selectivity of such an arrangement is greatly enhanced by adjusting the wavelength of the light to match the absorption spectrum of the substance being measured. Thus if the substance has a yellow color, it is because blue light is being absorbed, so a blue color filter is placed in the light path.
More sophisticated absorption spectrophotometers employ two cells (one for the sample and another reference cell for the blank. They also allow one to select the particular wavelength range that is most strongly absorbed by the substance under investigation, which can often extend into the near-ultraviolet region.
Light scattering measurements (made with a nephthelometer) can be useful for reactions that lead to the formation a fine precipitate/ A very simple student laboratory experiment of this kind can be carried out by placing a conical flask containing the reaction mixture on top of a marked piece of paper. The effects of changing the temperature or reactant concentrations can be made by observing how long it takes for precipitate formation to obscure the mark on the paper.
Other optical methods such a fluorescence and polarimetry (measurement of the degree to which a solution rotates the plane of polarized light) are also employed when applicable.
Other methods
• Measurement of gases: The classical method of following reactions that produce changes in the number of moles of gases is to observe changes in pressure or volume. An alternative method for following the loss of a gas is to place the reaction container on an electronic balance and monitor the loss in weight. This is generally less accurate than pressure measurements, but is sometimes used in student experiments.
• pH measurements: Many reactions yield or consume hydrogen ions and are conveniently followed by means of a pH meter.
• Electrical Conductance: Reactions that yield or consume ionic substances are often studied by measuring the electrical conductance of the solution. Conductimetry is usually carried out by balancing the solution conductance with a known resistance in a bridge arrangement. An audio-frequency alternating current is used in order to avoid electrolysis
The study of rapid reactions
Rapid mixing
The traditional experimental methods described above all assume that we can follow the reaction after its components have combined into a homogeneous mixture of known concentrations. But what can we do if the time required to complete the mixing process is comparable to or greater than the time needed for the reaction to run to completion? For reactions that take place in milliseconds, the standard approach since the 1950s has been to employ a flow technique of some kind. An early example was used to study fast gas-phase reactions in which one of the reactants is a free radical such as OH that can be produced by an intense microwave discharge acting on a suitable source gas mixture. This gas, along with the other reactant being investigated, is made to flow through a narrow tube at a known velocity.
If the distance between the point at which the reaction is initiated and the product detector is known, then the time interval can be found from the flow rate. By varying this distance, the time required to obtain the maximum yield can then be determined.
Although this method is very simple in principle, it can be rather complicated in practice, as the illustration shows. Owing to the rather large volumes required, his method is more practical for the study of gas-phase reactions than for solutions, for which the stopped-flow method described below is generally preferred.
Stopped-flow and Quenched-flow methods
These are by far the most common means of studying fast solution-phase reactions over time intervals of down to a fraction of a millisecond. The use of reasonably simple devices is now practical even in student laboratory experiments. These techniques make it possible to follow not only changes in the concentrations of reactants and products, but also the buildup and decay of reaction intermediates.
The basic stopped-flow apparatus consists of two or more coupled syringes that rapidly inject the reactants into a small mixing chamber and then through an observation cell that can be coupled to instruments that measure absorption, fluorescence, light scattering, or other optical or electrical properties of the solution. As the solution flows through the cell, it empties into a stopping syringe that, when filled, strikes a backstop that abruptly stops the flow. The volume that the stopping syringe can accept is adjusted so that the mixture in the cell has just become uniform and has reached a steady state; at this point, recording of the cell measurement begins and its change is followed.
Of course, there are many reactions that cannot be followed by changes in light absorption or other physical properties that are conveniently monitored. In such cases, it is often practical to quench (stop) the reaction after a desired interval by adding an appropriate quenching agent. For example, an enzyme-catalyzed reaction can be stopped by adding an acid, base, or salt solution that denatures (destroys the actvity of) the protein enzyme. Once the reaction has been stopped, the mixture is withdrawn and analyzed in an appropriate manner.
The quenched-flow technique works something like the stopped-flow method described above, with a slightly altered plumbing arrangement. The reactants A and B are mixed and fed directly through the diverter valve to the measuring cell, which is not shown in this diagram. After a set interval that can vary from a few milliseconds to 200 sec or more, the controller activates the quenching syringe and diverter valve, flooding the cell with the quenching solution.
Too fast to mix? No problem!
In order to investigate reactions that are complete in less than a millisecond, one can start with a pre-mixed sample in which one of active reactants is generated in situ. Alternatively, a rapid change in pressure or temperature can alter the composition of a reaction that has already achieved equilibrium.
Flash photolysis
Many reactions are known which do not take place in the absence of light whose wavelength is sufficiently short to supply the activation energy needed to break a bond, often leading to the creation of a highly reactive radical. A good example is the combination of gaseous Cl2 with H2, which procedes explosively when the system is illuminated with visible light. In flash photolysis, a short pulse of light is used to initiate a reaction whose progress can be observed by optical or other means.
Photolysis refers to the use of light to decompose a molecule into simpler units, often ions or free radicals. In contrast to thermolysis (decomposition induced by high temperature), photolysis is able to inject energy into a molecule almost instantaneously and can be much "cleaner", meaning that there are fewer side reactions that often lead to complex mixtures of products. Photolysis can also be highly specific; the wavelength of the light that triggers the reaction can often be adjusted to activate one particular kind of molecule without affecting others that might be present.
All this had been known for a very long time, but until the mid-1940's there was no practical way of studying the kinetics of the reactions involving the highly reactive species producd by photolysis. In 1945, Ronald Norrish of Cambridge University and his graduate student George Porter conceived the idea of using a short-duration flash lamp to generate gas-phase CH2 radicals, and then following the progress of the reaction of these radicals with other species by means of absorption spectroscopy.
Norrish and Porter shared the 1967 Nobel Prize in Chemistry for this work.Nanosecond flash photolysis setup.
In a flash photolysis experiment, recording of the absorbance of the sample cell contents is timed to follow the flash by an interval that can be varied in order to capture the effects produced by the product or intermediate as it is formed or decays. Flash durations of around 1 millisecond permitted one to follow processes having lifetimes in the microsecond range, but the advent of fast lasers gradually extended this to picoseconds and femtoseconds.
Flash photolysis revolutionized the study of organic photochemistry, especially that relating to the chemistry of free radicals and other reactive species that cannot be isolated or stored, but which can easily be produced by photolysis of a suitable precursor. It has proven invaluable for understanding the complicated kinetics relating to atmospheric chemistry and smog formation. More recently, flash photolysis has become an important tool in biochemistry and cellular physiology.
Perturbation-relaxation methods
Many reactions, especially those that take place in solution, occur too rapidly to follow by flow techniques, and can therefore only be observed when they are already at equilibrium. The classical examples of such reactions are two of the fastest ones ever observed, the dissociation of water
\[2 H_2O → H_3O^+ + OH^–\]
and the formation of the triiodide ion in aqueous solution
\[I^– + I_2 → I_3^–\]
Reactions of these kinds could not be studied until the mid-1950s when techniques were developed to shift the equilibrium by imposing an abrupt physical change on the system.
For example, if the reaction A B is endothermic, then according to the Le Chatelier principle, subjecting the system to a rapid jump in temperature will shift the equilibrium state to one in which the product B has a higher concentration. The composition of the system will than begin to shift toward the new equilibrium composition at a rate determined by the kinetics of the process.
For the general case illustrated here, the quantity "x" being plotted is a measurable quantity such as light absorption or electrical conductivity that varies linearly with the composition of the system. In a first-order process, x will vary with time according to
xt = xo ekt
After the abrupt perturbation at time to, the relaxation time t* is defined as the half-time for the return to equilibrium — that is, as the time required for xo to decrease by Δx/e = Δx/2.718. The derivation of t* and the relations highlighted in yellow can be found in most standard kinetics textbooks. Temperature jumps are probably most commonly used. They can be brought about in various ways:
This is the method that Manfred Eigen pioneered when, in the early 1960's, he measured the rate constant of what was then the fastest reaction ever observed:
H+ + OH → H2O k = 1.3 × 1011 M–1 sec–1
A detailed description of the kinetics of this process can be seen here. Eigen shared the 1967 Nobel Prize in Chemistry with Porter and Norrish, who developed flash photolysis. Eigen's Nobel lecture, in keeping with his legendary sense of humor, was titled "immeasurably fast reactions". His later work has focused on self-organizing systems and the origin of life, molecular genetics, and neurology.
• high-voltage electric discharge: A capacitor, charged to 5-10 kV, is discharged through a solution to which an electrolyte has been added to provide a conductive path.
• laser irradiation: The sample is irradiated with a laser whose wavelength corresponds to an absorption peak in the sample. Infrared lasers are often used for this purpose.
• mixing of two pre-equilibrated solutions: Two solutions, otherwise identical but at different temperatures, are rapidly mixed in a stopped-flow type of apparatus. Although this method is not as fast, it has the advantage of allowing both negative and positive T-jumps. The device shown here uses 0.1-mL samples and provides jumps of up to ±40 C° over a few microseconds. Observation times, however, are limited to 1-2 milliseconds owing to thermal dissipation.
Pressure jumps
According to the Le Chatelier principle, a change in the applied pressure will shift the equilibrium state of any reaction which involves a change in the volume of a system. Aside from the obvious examples associated with changes in the number of moles of gases, there are many more subtle cases involving formation of complexes, hydration shells and surface adsorption, and phase changes. One area of considerable interest is the study of protein folding, which has implications in diseases such as Parkinson's and Alzheimer's.
The pressure-jump is applied to the cell through a flexible membrane that is activated by a high-pressure gas supply, or through an electrically-actuated piezoelectric crystal. The latter method is employed in the device shown here, which can produce P-jumps of around 1GPa over sub-millisecond time intervals.
More on P-jumps: this 1999 review covers many non-biochemical applications. See also this article on protein folding and aggregation.
Shock tubes: extreme jumping
When a change in pressure propagates through a gas at a rate greater than the ordinary compressions and rarefactions associated with the travel of sound, a moving front (a shock wave) of very high pressure forms. This in turn generates an almost instantaneous rise in the temperature that can approach several thousand degrees in magnitude.
A shock tube is an apparatus in which shock waves can be generated and used to study the kinetics of gas-phase reactions that are otherwise inaccessible to kinetic measurements. Since all molecules tend to dissociate at high temperatures, shock tubes are widely used to study dissociation processes and the chemistry of the resulting fragments. For example, the shock-induced decomposition of carbon suboxide provides an efficient means of investigating carbon atom reactions:
\[C_3O_2 → C + CO\]
Shock tube techniques are also useful for studying combustion reactions, including those that proceed explosively.
The shock tube itself consists of two sections separated by a breakable diaphragm of metal or plastic. One section is filled with a "driver" gas at a very high pressure, commonly helium, but often mixed with other inert gases to adjust the properties of the shock. The other, longer section of the tube contains the "driven" gas — the reactants — at a low pressure, usually less than 1 atmosphere.
The reaction is initiated by causing the diaphragm to rupture, either by means of a mechanical plunger or by raising the pressure beyond its bursting point. The kinetics of the reaction are monitored by means of an absorption or other optical monitoring device that is positioned at a location along the reaction tube that is appropriate to the time course of the reaction. | textbooks/chem/General_Chemistry/Chem1_(Lower)/17%3A_Chemical_Kinetics_and_Dynamics/17.07%3A_Experimental_methods_of_chemical_kinetics.txt |
The science of chemistry is concerned with the composition, properties, and structure of matter and with the ways in which substances can change from one form to another. Since anything that has mass and occupies space can be classified as matter, this means that chemistry is involved with almost everything in the universe. But this definition is too broad to be useful. Chemistry isn't the only science that deals with the composition and transformations of matter. Some matter is composed of cells, which transform by meiosis and other processes that biologists study. Matter is also composed of subatomic particles called leptons, which transform by processes like annihilation studied by physicists. Chemists are unique because they understand or explain everything, from our bodies to our universe, in terms of the properties of just over 100 kinds of atoms found in all matter and the amazing variety of molecules and other atomic-scale structures that are created by forming and breaking bonds between atoms.
• 1.1: Prelude to Chemistry
The science of chemistry is concerned with the composition, properties, and structure of matter and with the ways in which substances can change from one form to another. Since anything that has mass and occupies space can be classified as matter, this means that chemistry is involved with almost everything in the universe. But this definition is too broad to be useful. Chemistry isn't the only science that deals with the composition and transformations of matter.
• 1.2: What Chemists Do
What are some of the things that chemists do? Like most scientists, they observe and measure components of the natural world. Based on these observations they try to place things into useful, appropriate categories and to formulate scientific laws which summarize the results of a great many observations. Like other scientists, chemists try to explain their observations and laws by means of theories or models. They constantly make use of atoms, molecules, and other very small particles.
• 1.3: Handling Large and Small Numbers
Results often involve very large numbers or very small fractions. Such numbers are inconvenient to write and hard to read correctly. One approach involves what is called scientific notation or exponential notation.
• 1.4: The International System of Units (SI)
The metric system has undergone continuous evolution and improvement since its original adoption by France. Beginning in 1899, a series of international conferences have been held for the purpose of redefining and regularizing the system of units. In 1960 the Eleventh Conference on Weights and Measures proposed major changes in the metric system and suggested a new name — the International System of Units — for the revised metric system.
• 1.5: SI Prefixes
The SI base units are not always of convenient size for a particular measurement. For example, the meter would be too big for reporting the thickness of this page, but rather small for the distance from Chicago to Detroit. To overcome this obstacle the SI includes a series of prefixes, each of which represents a power of 10. These allow us to reduce or enlarge the SI base units to convenient sizes.
• 1.6: Measurements, Quantities, and Unity Factors
Let us assume that you are faced with a specific problem. Then we can see how scientific thinking might help solve it. Suppose that you live near a large plant which manufactures cement. Smoke from the plant settles on your car and house, causing small pits in the paint. You would like to stop this air-pollution problem—but how?
• 1.7: Errors in Measurement
Scientific measurements are of no value (or at least, they're not really scientific) unless they are given with some statement of the errors they contain. If a poll reports that one candidate leads another by 5%, that may be politically useful for the winning candidate to point out. But all respectable polls are scientific, and report errors.
• 1.8: Volume
Volume is the amount of 3D space a substance or object occupies. The most commonly used derived units are those of volume. As we have already seen, calculation of the volume of an object requires that all 3 dimensions are multiplied together (length, width, and height). Thus the SI unit of volume is the cubic meter. This is rather large for use in the chemical laboratory, and so the cubic decimeter or cubic centimeter are more commonly used.
• 1.9: Density
The terms heavy and light are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. However, something else is alluded to when we say that oak is heavier than balsa wood. A small shaving of oak would obviously weigh less than a roomful of balsa wood, but oak is heavier in the sense that a piece of given size weighs more than the same-size piece of balsa. What we are actually comparing is the mass per unit volume, that is, the density.
• 1.10: Conversion Factors and Functions
The "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions. When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, that is based on a mathematical equation that relates parameters.
01: Introduction - The Ambit of Chemistry
The explosive decomposition of nitrogen triiodide shown in the video is an exciting (and forceful) example of chemistry in action. For many of us, this explosion is the work of chemists who work magic at the lab bench. In reality, chemistry is much broader. The transformations of matter can be as fast and explosive as the video shows, or involve the slow decay of living organisms into fossil fuels. The compounds around us in everyday life have many properties that would have been conceived as magic only a few centuries ago. Could you imagine describing to previous generations neon signs or electricity from batteries? Could you imagine what they would say seeing the decomposition of nitrogen triiodide?
Nitrogen Triiodide is a contact explosive. Here I have some on five filter papers stacked on a ring stand and the original filter paper is just off to the left of the screen (if you watch when it detonates you can see a secondary purple cloud from the left).
The science of chemistry is concerned with the composition, properties, and structure of matter and with the ways in which substances can change from one form to another. Since anything that has mass and occupies space can be classified as matter, this means that chemistry is involved with almost everything in the universe. But this definition is too broad to be useful. Chemistry isn't the only science that deals with the composition and transformations of matter. Some matter is composed of cells, which transform by meiosis and other processes that biologists study. Matter is also composed of subatomic particles called leptons, which transform by processes like annihilation studied by physicists. Chemists are unique because they understand or explain everything, from our bodies to our universe, in terms of the properties of just over 100 kinds of atoms found in all matter and the amazing variety of molecules and other atomic-scale structures that are created by forming and breaking bonds between atoms. In the video explosion, bonds break in nitrogen triiodide (NI3) molecules and the atoms recombine to form nitrogen (N2) and iodine (I2). So chemistry is defined by its approach, not its subject matter. Chemistry explains or understands any subject in terms of the properties of atoms and molecules. Chemistry, in other words, is not just something that happens in laboratories. It is a unique perspective, or way of understanding, all that is around us, and even inside us. Chemistry is going on in places as diverse as the smallest bacterium, a field of ripening wheat, a modem manufacturing plant, the biospheres of planets such as Earth, the vast reaches of interstellar space, and even your eyes and brain as you read these words.
ChemPRIME recognizes that the chemical perspective can add to our understanding of anything that catches our interest. Our goal will be to add another dimension--understanding in terms of the properties of atoms and molecules and how they interact--to many subjects, thereby making it clear how the study of chemistry is of importance to a wide variety of people. Biologists, for example, have examined smaller and smaller organisms, cells, and cell components, until, in the study of viruses and genes, they joined forces with chemists who were interested in larger and larger molecules. The result was a new inter-disciplinary field called molecular biology, and a reinforcement of the idea that living organisms are complicated, highly organized chemical systems. Chemists interact in similar ways with scientists in areas such as chemical physics, geochemistry, pharmacology, toxicology, ecology, meteorology, oceanography, and many others. Current practice in these fields is such that a person lacking basic chemical knowledge is at a severe disadvantage, because the perspective of the molecular level has become so important.
Chemistry also underlies a great deal of modern technology. The manufacture of such basic commodities as steel, aluminum, glass, plastics, paper, textiles, and gasoline all involve chemists and chemistry. Without the abundant cheap supply of these and other substances that chemistry has helped to produce, our lives would be much less comfortable. However, as we are now beginning to discover, callous and indiscriminate use of technology can produce disadvantages as well as benefits. Automobiles, power plants, and industrial processes spew into the air harmful substances that are not always easy or cheap to eliminate. Rivers and lakes are also more easily contaminated than we once thought - substances once believed harmless now have proven to be the opposite.
Issues involving the effects of technology on the environment affect everyone - not just scientists. Decisions about them are political, at least in part, and require some chemical knowledge on the part of voters as well as their elected representatives. At the very least, a citizen needs to be able to distinguish valid and invalid arguments put forward by scientific “experts” regarding such issues. (In some cases such “expertise” may be mainly a willingness to speak out rather than a command of the scientific and political issues.) It is to be hoped also that more persons will follow the example of Russell Peterson, formerly a researcher in a large chemical company, who served as Governor of Delaware and Chairman of the President’s Council on Environmental Quality. Only by a combination of scientists willing to leave their laboratories and citizens willing to master some of the basics of science can intelligent political decisions be made in a democratic society. Indeed, such a combination may be a necessity if democracy is not to degenerate into an oligarchy ruled by those who control the experts.
Given the universality of chemistry, its central role among the sciences, and its importance in modern life, how is it possible to learn much about it in a short time? If everything has a chemical aspect, because atoms and molecules can aid in understanding everything, is the field of chemistry so broad and all-encompassing that one cannot master enough to make its study worth-while? We think the answer to this second question is a resounding no! This entire book has been designed to help you learn a good deal of chemistry in a short time. If it is successful, the first question will have been answered as well.
An important and valuable technique of science and scientists is that of subdividing large, seemingly unsolvable problems into smaller, simpler parts. If the latter are chosen carefully, each can be mastered. Individual small advances can then be combined to yield an important, more complicated result. Thus chemists have not acted on the assumption that because all the world may be understood in terms of chemicals and chemistry, they should try to study it all at once. Rather, many chemists do much of their work under controlled, laboratory conditions, advancing in small steps toward more general, useful, and exciting results.
Since people’s process of studying and understanding chemistry is far from complete, we can narrow our area of interest considerably by redefining chemistry as those things that chemists can explain on the molecular level, or are attempting to explain in terms of atoms and molecules and their properties. This more restricted view constitutes the main theme of this book. We hope that when you have finished with it, you will have a solid background in the facts, laws, and theories of chemistry, as well as those modes of behavior and thinking that chemists have found useful in solving the problems they have faced. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.01%3A_Prelude_to_Chemistry.txt |
What are some of the things that chemists do? Like most scientists, they observe and measure components of the natural world. Based on these observations they try to place things into useful, appropriate categories and to formulate scientific laws which summarize the results of a great many observations. Indeed, it is a fundamental belief of all science that natural events do not occur in a completely unpredictable fashion. Instead, they obey natural laws. Therefore observations and measurements made on one occasion can be duplicated by the same or another person on another occasion. Communication of such results, another important activity, affords an opportunity for the entire scientific community to test an individual’s work. Eventually a consensus is reached, and there is agreement on a new law.
Like other scientists, chemists try to explain their observations and laws by means of theories or models. They constantly make use of atoms, molecules, and other very small particles. Using such theories as their guides, chemists synthesize new materials. Well over 3 million compounds are now known, and more than 9000 are in large-scale commercial production. Even a backpacker going “back to nature” takes along synthetic materials such as nylon, aluminum, and aspirin.
Chemists also analyze the substances they make and those found in nature. Determining the composition of a substance is the first step in understanding its chemical properties, and detection of very small quantities of some materials in the natural world is essential in controlling air and water pollution. Another role that chemists play is in studying the processes (chemical reactions) by which one substance can be transformed into another. Will the reactions occur without prodding? If so, how quickly? Is energy given off? Can the reactions be controlled - made to occur only when we want them to?
These and many other problems that interest chemists are the subject matter of this online textbook. We will return to each several times and in increasing detail. Do not forget, however, that chemistry is more than just what chemists do. Many persons in other sciences as well as in daily life are constantly doing chemistry, whether they call it by name or not. Indeed, each of us is an intricate combination of chemicals, and everything we do depends on chemical reactions. Although space limitations will prevent us from exploring all but a fraction of the applications of chemistry to other fields, we have included such excursions as often as seemed appropriate. From them we hope that you will be able to learn how to apply chemical facts and principles to the problems you will face in the future. Many of the problems are probably not even known yet, and we could not possibly anticipate them. If you have learned how to think “chemically” or “scientifically,” however, we believe that you will be better prepared to face them.
Elements, Compounds and Mixtures
Chemists have a unique way of characterizing the objects of their study. To a chemist, matter containing one kind of atom or molecule is considered a pure substance, so water (H2O) and aluminum metal (Al) are pure substances. To a chemist, aluminum is an element, the simplest kind of pure substance which contains only one kind of atom (a physicist might consider aluminum a very complex "mixture" of things like leptons, protons, quarks).
To a chemist, the pure substance water is a compound (it contains two kinds of atoms bound to one another in just one kind of molecule). An environmentalist might consider water "pure" even if it contains the normal amount of dissolved oxygen and carbon dioxide, but no other "pollutants". To a chemist, water containing oxygen is no longer a pure substance, but a mixture. In pure water, the ratio of hydrogen to oxygen atoms is always 2:1, but as soon as oxygen is dissolved in the water, the ratio is no longer fixed (because oxygen does not form a new molecule by reacting with H2O), a sure sign of impurity to a chemist.
Chemical Reactions
Because the dissolved oxygen does not react with water to form a new compound, we consider "dissolving" to be a physical process (one that is not understood in terms of atoms or molecules and bond formation or breaking). Changes like dissolution, evaporation, and condensation are considered "physical changes" because, while they are clearly interesting changes, they might be studied in physics with no reference to atomic recombinations. If, however, the dissolved oxygen is used by a fish to convert food to carbon dioxide (CO2) and water (H2O), a chemical reaction has occurred, because the process is best understood at the level of atoms, molecules, and bonds between them. The process by which the dissolved oxygen is absorbed by fish might be considered a "biological process" because knowledge of the cellular structure of the fish gills is important in its understanding. Once more, the mode of understanding is more important than the subject of study in determining whether a process is considered "chemical", "physical", or "biological". Later in this book, we will use atomic and molecular concepts like non-covalent bond formation to understand dissolution, and then it should probably be considered a chemical process, and it will become clear that no absolute distinction between physical and chemical processes is possible (or necessary).
Homogeneous and Heterogeneous
It is interesting that water containing ice cubes is a pure substance to a chemist (only H2O molecules are present), but it is heterogeneous. To a chemist, "heterogeneous" does not indicate "impurity", nor does "homogeneous" indicate purity. Oxygen (or salt) dissolved in water is a homogeneous mixture, because every part of it looks like every other part. Of course, when we say "homogeneous" we usually assume that the substance is not being observed with anything with higher magnification than a crude microscope. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.02%3A_What_Chemists_Do.txt |
Example 3 from Measurements, Quantities and Unity Factors(opens in new window) illustrates a common occurrence in science—results often involve very large numbers or very small fractions. The United States used 66 500 000 000 000 000 000 J (joules) of energy in 1971, and the mass of a water molecule is 0.000 000 000 000 000 000 000 029 9 g. Such numbers are inconvenient to write and hard to read correctly. (We have divided the digits into groups of three to make it easier to locate the decimal point. Spaces are used instead of commas because many countries use a comma to indicate the decimal.)
There are two ways of handling this problem. We can express a quantity in larger or smaller units, as in Example 4 from Measurements, Quantities and Unity Factors(opens in new window), or we can use a better way to write small and large numbers. The latter approach involves what is called scientific notation or exponential notation. The position of the decimal point is indicated by a power (or exponent) of 10. For example,
$138= 13.8\cdot 10= 1.38\cdot 10\cdot 10= 1.38\cdot 10^2 \nonumber$
$0.004\ 83= \frac{4.83}{10\cdot10\cdot10} = 4.83\cdot\frac{1}{10^3} = 4.83\cdot10^{-3} \nonumber$
A number with a negative exponent is simply the reciprocal of (one divided by) the same number with the equivalent positive exponent. Therefore decimal fractions (numbers between zero and one) may be expressed using negative powers of 10. Numbers between 1 and 10 require no exponential part, and those larger than 10 involve positive exponents. By convention the power of 10 is chosen so that there is one digit to the left of the decimal point in the ordinary number. That is, we would usually write 5280 as 5.28 × 103 not as 0.528 × 104 or 52.8 × 102.
To convert a number from ordinary to scientific notation, count how many places the decimal point must be shifted to arrive at a number between 1 and 10. If these shifts are to the left, the number was large to begin with and we multiply by a large (that is, positive) power of 10. If the shift is to the right, a reciprocal (negative) power of 10 must be used.
Example $1$ : Scientific Notation
Express the following numbers in scientific notation: (a) 7563; (b) 0.0156.
Solution
a) In this case the decimal point must be shifted left three places:
Therefore we use an exponent of +3:
$7563= 7.563\cdot10^3 \nonumber$
b) Shifting the decimal point two places to the right yields a number between 1 and 10:
Therefore the exponent is –2:
$0.0156= 1.56\cdot10^{-2} \nonumber$
When working with exponential notation, it is often necessary to add, subtract, multiply, or divide numbers. When multiplying and dividing, you must remember that multiplication corresponds to addition of exponents, and division to their subtraction.
Multiplication: $\text{10}^a\cdot\text{10}^b = \text{10}^{\small(a + b\small)}$
Division: $\frac{10^a}{10^b} = 10^{(a - b)}$
Hence $(3.0\cdot10^5)\cdot(5.0\cdot10^3) = 15.0\cdot10^{(5+3)}= 15.0\cdot10^8= 1.50\cdot10^9$ and $\frac{3.0\cdot10^5}{5.0\cdot10^3} =0.6\cdot10^{(5- 3)} = 0.6\cdot10^2= 6.0\cdot10$
Example $2$ : Exponential Notation
Evaluate the following, giving your answer in correct exponential notation:
1. $(3.89 \cdot 10^5) \cdot (1.09 \cdot 10^{-3})$
2. $(6.41\cdot10^{-5}) \cdot(2.72 \cdot10^{-2})$
3. $\frac{(5.0\cdot10^6)}{(3.98\cdot10^8)}$
4. $\frac{(7.53\cdot10^{-3})}{(8.57\cdot10^{-5})}$
Solution:
1. $(3.89\cdot10^5)\cdot(1.09 \cdot10^{-3}) = 3.89\cdot1.09\cdot10^{5 + (-3)}= 4.24\cdot10^2$
2. $(6.41\cdot10^{-5})\cdot(2.72 \cdot10^{-2})$= $6.41\cdot2.72\cdot10^{-5 + (-2)}$ = $17.43 \cdot10^{-7}$ = $1.743 \cdot10^{-6}$
3. $\frac{(5.0\cdot10^6)}{(3.98\cdot10^8)}= \frac{5.0}{3.98} \cdot10^{6-8}= 1.26\cdot10^{-2}$
4. $\frac{(7.53\cdot10^{-3})}{(8.57\cdot10^{-5})} = \frac{7.53}{8.57}\cdot10^{-3 - (-5)} = 0.879\cdot10^2 = 8.79\cdot10^1$
Addition and subtraction require that all numbers be converted to the same power of 10. (This corresponds to lining up the decimal points.)
Example $3$ : Scientific Notation
Evaluate the following, giving your answer in scientific notation:
a)$(6.32\cdot10^2) – (1.83 \cdot10^\cdot{-1})$
b)$(3.72 \cdot10^4) + (1.63\cdot10^5) – (1.7 10^3)$
Solution:
a) First convert to the same power of 10; then add the ordinary numbers.
\begin{align}&&6.32\cdot10^2&=& 632 \&&–1.83\cdot10^{-1}&=& – 0.183 \632 – 0.183&=& 631.817&=& 6.318 17\cdot10^2 \end{align}
b) Convert all powers of 10 to 104.
\begin{align}3.72\cdot10^4 &=& 3.72 \cdot10^4 &=& 3.72 × 10^4 \1.63\cdot10^5&=& 1.63\cdot10 \cdot10^4&=& 16.3 \cdot10^4 \–1.7 \cdot10^3&=& –1.7 \cdot10^{-1}\cdot10^4&=& – 0.17\cdot10^4 \ (3.72\cdot10^4) + (16.3\cdot10^4) - (0.17 \cdot10^4)&=& 19.85 \cdot10^4&=& 1.985\cdot10^5\end{align}
Scientific notation is becoming more common every day. Many electronic pocket calculators use it to express numbers which otherwise would not fit into their displays. For example, an eight-digit calculator could not display the number 6 800 000 000.The decimal point would remain fixed on the right, and the 6 and the 8 would “overflow” to the left side. Such a number is often displayed as 6.8 09, which means 6.8 × 109. If you use a calculator which does not have scientific notation, we recommend that you express all numbers as powers of 10 before doing any arithmetic. Follow the rules in the last two examples, using your calculator to do arithmetic on the ordinary numbers. You should be able to add or subtract the powers of 10 in your head.
Computers also are prone to print results in scientific notation, and they use yet another minor modification. The printed number 2.3074 E-07 means 2.3074 × 10–7 for example. In this case the E indicates that the number following is an exponent of 10.
Too Many Digits
Again in reference to our air-pollution experiment(opens in new window), we could express the mass of smoke collected as 3.42 × 10–2 g and the volume of the balloon as 1.021 926 4 × 107 cm3. There is something strange about the second quantity, though. It contains a number which was copied directly from the display of an electronic calculator and has too many digits.
The reliability of a quantity derived from a measurement is customarily indicated by the number of significant figures (or significant digits) it contains. For example, the three significant digits in the quantity 3.42 × 10–2 g tell us that a balance was used on which we could distinguish 3.42 × 10–2 g from 3.43 × 10–2 g or 3.41 × 10–2 g. There might be some question about the last digit, but those to the left of it are taken as completely reliable. Another way to indicate the same thing is (3.42±0.01) × 10–2 g. Our measurement is somewhere between 3.41 × 10–2 g and 3.43 × 10–2 g.
As another example of choosing an appropriate number of significant digits, let us read the volume of liquid in a graduated cylinder. The bottom of the meniscus lies between graduations corresponding to 38 and 39 cm3. We can estimate that it is at 38.5 cm3, but the last digit might be off a bit-perhaps it looks like 38.4 or 38.6 cm3 to you. Since the third digit is in question, we should use three significant figures. The volume would be recorded as 38.5 cm3. Laboratory equipment is often calibrated similarly to this graduated cylinder—you should estimate to the nearest tenth of the smallest graduation.
In some ordinary numbers, for example, 0.001 23, zeros serve merely to locate the decimal point. They do not indicate the reliability of the measurement and therefore are not significant. Another advantage of scientific notation is that we can assume that all digits are significant. Thus if 0.001 23 is written as 1.23 × 10–3, only the 1, 2, and 3, which indicate the reliability of the measurement, are written. The decimal point is located by the power of 10.
If the rule expressed in the previous paragraph is applied to the volume of air collected in our pollution experiment, 1.021 926 4 × 107 cm3, we find that the volume has eight significant digits. This implies that it was determined to ±1 cm3 out of about 10 million cm3, a reliability which corresponds to locating a grasshopper exactly at some point along the road from Philadelphia to New York City. For experiments as crude as ours, this is not likely. Let us see just how good the measurement was.
You will recall that we calculated the volume from the diameter of the balloon(opens in new window), 106 in. The three significant figures imply that this might have been as large as 107 in or as small as 105 in. We can repeat the calculation with each of these quantities to see how far off the volume would be:
$$\begin{split} r &= \frac{1}{2} \times \text{107 in} = \text{53.5 in} \times \frac{\text{1 cm}}{\text{0.3937 in}} \ \ &=135.890 27 \text{cm} \end{split}$ \nonumber$
\begin{align} V &= \frac{4}{3} \times 3.141 59 \times (135.890 27)^3 \ \ &= 10 511 225 \text{cm}^3 = 1.051 122 5 \times 10^7 \text{cm}^3 \end{align} \nonumber
or
\begin{align} V &= \frac{4}{3} \times 3.14159 \times \left( \frac{1}{2} \times \text{105 in} \frac{\text{1 cm}}{\text{0.3937 in}}\right)^3 \ \ &= 9 932 759 \text{cm}^3 = 0.993 275 9 \times 10^7 \text{cm}^3 \end{align} \nonumber
That is, the volume is between 0.99 × 107 and 1.05 × 107 cm3 or (1.02 ± 0.03) × 107 cm3. We should round our result to three significant figures, for example, 1.02 cm³, because the last digit, namely 2, is in question.
Rules for Rounding Numbers
1. All digits to be rounded are removed together, not one at a time.
2. If the left-most digit to be removed is less than five, the last digit retained is not altered.
3. If the left-most digit to be removed is greater than five, the last digit retained is increased by one.
4. If the left-most digit to be removed is five and at least one of the other digits to be removed is nonzero. the last digit retained is increased by one.
5. If the left-most digit to he removed is five and all other digits to he removed are zero, the last digit retained is not altered if it is even, but is increased by one if it is odd.
Application of the Rules for Rounding Numbers can be illustrated by an example.
Example $4$ : Significant Figures
Round each of the numbers below to three significant figures.
1. 34.7449
2. 34.864
3. 34.754
4. 34.250
5. 34.35
Solution:
a) Apply rules 1 and 2:
(Note that a different result would be obtained if the digits were incorrectly rounded one at a time from the right.)
b) Apply rules 1 and 3: 34.864 → 34.9
c) Apply rules 1 and 4: 34.754 → 34.8
d) Apply rules 1 and 5: 34.250 → 34.2
e) Apply rule 5: 34.35 → 34.4
To how many significant figures should we round our air-pollution results? We have already done a calculation involving multiplication and division to obtain the volume of our gas-collection balloon. It involved the following numbers:
calculation involving multiplication and division to obtain the volume of our gas-collection balloon
106 Three significant figures
0.3937 Four significant figures
3.141 59 Six significant figures (we could obtain more if we wanted)
$\frac{4}{3}$ and $\frac{1}{2}$
An infinite number of significant figures since the integers in these fractions are exact by definition. Exact, or 'defined' numbers are not considered when calculating significant figures.
The result of the calculation contained three significant figures — the same as the least-reliable number. This illustrates the general rule that for multiplication and division the number of significant figures in the result is the same as in the least-reliable measurement. Defined numbers such as π, ½ or 100 cm/1m are assumed to have an infinite number of significant figures.
In the case of addition and subtraction, a different rule applies. Suppose, for example, that we weighed a smoke-collection filter on a relatively inaccurate balance that could only be read to the nearest 0.01 g. After collecting a sample, the filter was reweighed on a single-pan balance to determine the mass of smoke particles.
Final mass: 2.3745 g (colored digits are in question)
Initial mass: –2.32 g
Mass of smoke: 0.0545 g
Since the initial weighing could have been anywhere from 2.31 to 2.33 g, all three figures in the final result are in question. (It must be between 0.0445 and 0.0645 g). Thus there is but one significant digit, and the result is 0.05g. The rule here is that the result of addition or subtraction cannot contain more digits to the right than there are in any of the numbers added or subtracted. Note that subtraction can drastically reduce the number of significant digits when this rule is applied.
Rounding numbers is especially important if you use an electronic calculator, since these machines usually display a large number of digits, most of which are meaningless. The best procedure is to carry all digits to the end of the calculation (your calculator will not mind the extra work!) and then round appropriately. Answers to subsequent calculations in this book will be rounded according to the rules given. You may wish to go back to previous examples and round their answers correctly as well.
Example $5$ : Rounding
Evaluate the following expressions, rounding the answer to the appropriate number of significant figures.
1. $\text{32.61 g} + \text{8.446 g} + \text{7.0 g}$
2. $\text{0.136 cm}^3 \times \text{10.685 g cm}^{-3}$
Solution
1. $\text{32.61 g} + \text{8.446 g} + \text{7.0 g} = \text{48.056 g} = \text{48.1g (7.0 has only one figure to the right of the decimal point.)}$
2. $0.136 \text{cm}^3 \times 10.685 \text{g cm}^{-3} = \text{1.453 g} = \text{1.45g (0.136 has only three significant figures.)}$
When we suggested filling a surplus weather balloon to measure how much gas was pumped through our air-pollution collector, we mentioned that this would be a rather crude way to determine volume. For one thing, it would not be all that simple to measure the diameter of an 8- or 9-ft sphere reliably. Using a yardstick, we would be lucky to have successive measurements agree within half an inch or so. It was for this reason that the result was reported to the nearest inch. The degree to which repeated measurements of the same quantity yield the same result is called precision. Repetition of a highly precise measurement would yield almost identical results, whereas low precision implies numbers would differ by a significant percentage from each other.
A highly precise measurement of the diameter of our balloon could be achieved, but it would probably not be worthwhile. We have assumed a spherical shape, but this is almost certainly not exactly correct. No matter how precisely we determine the diameter, our measurement of gas volume will be influenced by deviations from the assumed shape. When one or more of our assumptions about a measuring instrument are wrong, the accuracy of a result will be affected. An obvious example would be a foot rule divided into 11 equal inches. Measurements employing this instrument might agree very precisely, but they would not be very accurate.
An important point of a different kind is illustrated in the last two paragraphs. A great many common words have been adopted into the language of science. Usually such an adoption is accompanied by an unambiguous scientific definition which does not appear in a normal dictionary. Precision and accuracy are many times treated as synonyms, but in science each has a slightly different meaning. Another example is quantity, which we have defined in terms of “number × unit.” Other English words like bulk, size, amount, and so forth, may be synonymous with quantity in everyday speech, but not in science. As you encounter other words like this, try to learn and use the scientific definition as soon as possible, and avoid confusing it with the other meanings you already know.
Even granting the crudeness of the measurements we have just described, they would be adequate to demonstrate whether or not an air-pollution problem existed. The next step would be to find a chemist or public health official who was an expert in assessing air quality, present your data, and convince that person to lend his or her skill and authority to your contention that something was wrong. Such a person would have available equipment whose precision and accuracy were adequate for highly reliable measurements and would be able to make authoritative public statements about the extent of the air-pollution problem.
Web Sources
Several sites were inspired by Charles Eames' "Powers of Ten" Powers of Ten™ (1977)(opens in new window) [www.powersof10.com(opens in new window)]: Word Wizz(opens in new window) [www.wordwizz.com] Secret Worlds: The Universe Within(opens in new window) [micro.magnet.fsu.edu] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.03%3A_Handling_Large_and_Small_Numbers.txt |
The results of a scientific experiment must be communicated to be of value. This affords an opportunity for other scientists to check them. It also allows the scientific community, and sometimes the general public, to share new knowledge. Communication, however, is not always as straightforward as it might seem. Ambiguous terminology can often turn a seemingly clear statement into a morass of misunderstanding.
As an example, consider someone tells you that the forecast is a high of 25°. Do you put on a winter jacket, or summer wear? If you are thinking winter, then you interpreted the temperature as 25°F. However if 25°C was meant, which is equal to 77°F, a winter jacket would be far too warm. Or consider filling up a car with gasoline. If you are in the US, you will be dealing in dollars per gallon, whereas, if you were in continental Europe, you will be dealing in Euros per liter. Given that the exchange rate from USD to Euros fluctuates and that there are roughly 3.79 liters in 1 gallon, it is difficult to simply compare numbers between gas prices in the USA, and say, France, if you don't know what units you are using. As a final example, consider the speed 24 meters/sec. Do you interpret this as close to highway speed limits in the US? It is the same speed as 53.7 miles per hour, likely a more familiar set of units for speed for people in the US.
While some of these examples may seem tired and overdone, they still showcase that even if you know unit conversions well, the many systems for measuring quantities can make things complex and confusing. Even rocket scientists get it wrong sometimes, as was seen in 1999 when a orbiter sent to Mars was unable propel itself correctly into orbit because one team had used metric system measurements, whereas another had used English system measurements.[1] If in the midst of this hodgepodge, you asked, “Would it not be easier to have a single unit for mass, a single unit for volume, and express all masses or volumes in these units,” you would not be the first person to have such an idea. The main difficulty is that it is hard to get everyone to agree on a single consistent set of units. Some units are especially convenient for some tasks. For example, a yard was originally defined as the distance from a man’s nose to the end of his thumb when his arm was held horizontally to one side. This made it easy to measure cloth or ribbon by holding one end to the nose and stretching an arm’s length with the other hand. Now that yardsticks, meter sticks, and other devices are readily available, the original utility of the yard is gone, but we still measure ribbon and cloth in that same unit. Many people would probably be distressed if a change were made.
Scientists are not all that different from other people—they too have favorite units which are especially suited to certain areas of research. Nevertheless, scientists have constantly pressed for improvement and uniformity in systems of measurement. The first such action occurred nearly 200 years ago when, in the aftermath of the French Revolution, the metric system spread over most of continental Europe and was adopted by scientists everywhere. The United States nearly followed suit, but in 1799 Thomas Jefferson was unsuccessful in persuading Congress that a system based on powers of 10 was far more convenient and would eventually become the standard of the world.
The metric system has undergone continuous evolution and improvement since its original adoption by France. Beginning in 1899, a series of international conferences have been held for the purpose of redefining and regularizing the system of units. In 1960 the Eleventh Conference on Weights and Measures proposed major changes in the metric system and suggested a new name — the International System of Units — for the revised metric system. (The abbreviation SI, from the French Système International, is commonly used.) Scientific bodies such as the U.S. National Bureau of Standards and the International Union of Pure and Applied Chemistry have endorsed the SI.
At the heart of the SI are the seven units, listed here. All other units are derived from these seven so-called base units. For example, units for area and volume may be derived by squaring or cubing the unit for length. Some of the base units are probably familiar to you, while others, such as the mole, candela, and kelvin, may be less so. Rather than defining each of them now, we shall wait until later chapters when the less familiar units, as well as the quantities they are used to measure, can be described in detail. The candela, which measures the intensity of light, is not used often by chemists, and so we shall pay no further attention to it.
Table \(1\): The Seven Base Units of the SI.
Quantity (parameter)
Measured
abbreviation Name of Unit Symbol for Unit
Length L meter m
Mass m kilogram kg
Time t second s
Electric current I ampere A
Temperature T kelvin K
Amount of substance n mole mol
Luminous intensity Iν candela cd
Web Sites
• NIST: National Institutes of Standards and Technology. Physics NIST [physics.nist.gov]
• Dictionary of Units of Measurement: www.unc.edu/~rowlett/units/index.html
• Robin Lloyd. Metric mishap caused loss of NASA orbiter. CNN. September 30, 1999. CNN [www.cnn.com]
1.05: SI Prefixes
The SI base units are not always of convenient size for a particular measurement. For example, the meter would be too big for reporting the thickness of this page, but rather small for the distance from Chicago to Detroit. To overcome this obstacle the SI includes a series of prefixes, each of which represents a power of 10. These allow us to reduce or enlarge the SI base units to convenient sizes. The figures below show how these prefixes can be applied to the meter to cover almost the entire range of lengths we might wish to measure.
Table \(1\): Prefixes Used for Decimal Fractions and Multiples of SI Units.
One non-SI unit of length, the angstrom (Å), is convenient for chemists and will continue to be used for a limited time. Since 1Å = 10–10 m, the angstrom corresponds roughly to the diameters of atoms and small molecules. Such dimensions are also conveniently expressed in picometers, 1 pm = 10–12 m = 0.01Å, but the angstrom is widely used and very familiar. Therefore we will usually write atomic and molecular dimensions in both angstroms and picometers.
The SI base unit of mass, the kilogram, is unusual because it already contains a prefix. The standard kilogram is a cylinder of corrosion-resistant platinum-iridium alloy which is kept at the International Bureau of Weights and Measures near Paris. The kilogram was chosen instead of a gram because the latter would have made an inconveniently small piece of platinum-iridium and would have been difficult to handle. Also, units of force, pressure, energy, and power have been derived using the kilogram instead of the gram.
Despite the fact that the kilogram is the SI unit of mass, the standard prefixes are applied to the gram when larger or smaller mass units are needed. For example, the quantity 106 kg (1 million kilograms) can be written as 1 Gg (gigagram) but not as 1 Mkg (megakilogram). The operative rule here is that one and only one prefix should be attached to the name for a unit. Figure 1.6 illustrates the use of this rule in expressing the wide range of masses available in the universe. Note that the masses of atoms and molecules are usually so small that scientific notation must be used instead of prefixes. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.04%3A_The_International_System_of_Units_%28SI%29.txt |
Measurements
Let us assume that you are faced with a specific problem. Then we can see how scientific thinking might help solve it. Suppose that you live near a large plant which manufactures cement. Smoke from the plant settles on your car and house, causing small pits in the paint. You would like to stop this air-pollution problem—but how?
As an individual you will probably have little influence, and even as part of a group of concerned citizens you may be ineffective, unless you can prove that there is a problem. Scientists have had a hand in writing most air-pollution regulations, and so you will have to employ some scientific techniques (or a scientist) to help solve your problem.
It will probably be necessary to determine how much air pollution the plant is producing. This might be done by comparing the smoke with a scale which ranges from white to gray to black, the assumption being that the darker the smoke, the more there is. For white cement dust, however, this is much less satisfactory than for black coal smoke. A better way to determine how much pollution there is would be to measure the mass of smoke particles which could be collected near your house or car. This could be done by using a pump (such as a vacuum cleaner) to suck the polluted gas through a filter. Weighing the filter before and after such an experiment would determine what mass of smoke had been collected.
Mass and Weight
Because weight is the force of gravity on an object, which varies from place to place by about 0.5% as shown in the Table, we must use mass for accurate measurements of how much matter we have. We still call the process of obtaining an accurate mass "weighing".
Values of g on Earth, m s-2
Latitude, o Altitude = 0 Altitude = 10 km
0 9.78036 9.74950
30 9.79324 9.76238
60 9.81911 9.78825
90 9.83208 9.80122
The weight of an object is actually the force of gravity, and is calculated as
$F=\text{“W”}=m \times g \label{1}$
Weight is measured in newtons (kg m s-2) or pounds (lb), where 1 lb = 4.44822162 newtons. The base unit for mass is kilograms (kg), but the pound may also be used as a unit of mass (1 pound = 0.45359237 kilograms). The average value of g is usually taken to be 9.80665 m s-2, so the weight of a 1 kg mass would be
$F = \text{ “W” } = m \times g = 1 \text{ kg} \times 9.80665 \text{ m s}^{-2} = 9.80665 \text{ N or } 2.2046 \text{ lb} \nonumber$
If the weight of the 1 kg mass were measured on an arctic exploration camp with a load cell balance (see below) still calibrated for the average g, its weight would be:
$F = \text{ “W” } = m \times g = 10 \text{ kg} \times 9.83208 \text{ m s}^{-2} = 9.83208 \text{ N or } 2.2103 \text{ lb} \nonumber$
Accurate weighing is usually done with a single pan balance. The empty pan is balanced by a counterweight. When an object is placed on the pan, gravitational attraction forces the pan downward. To restore balance, a series of weights (objects of known mass) are removed from holders above the pan. The force of gravitational attraction is proportional to mass, and when the pan is balanced, the force on it must always be the same. Therefore the mass of the object being weighed must equal that of the weights that were removed. A balance gives the same mass readout regardless of the force of gravity.
Modern laboratory "balances" are based on load cells that convert the force exerted by an object on the balance pan to an electrical signal. The load cell generally is coupled with a dedicated converter and digital display. Because the force exerted by the object depends on gravity, these are really scales that measure weight, and must be calibrated frequently (against standard masses) to read a mass.
Mass Measurements
If you kept a notebook or other record of your measurements, you would probably write down something like 0.0342 g to represent how much smoke had been collected. Such a result, which describes the magnitude of some parameter (in this case the magnitude of the parameter, mass), is called a quantity. Notice that it consists of two parts, a number and a unit. It would be ambiguous to write just 0.0342—you might not remember later whether that was measured in units of grams, ounces, pounds, or something else. A quantity always behaves as though the number and the units are multiplied together. For example, we could write the quantity already obtained as 0.0342 × g. Using this simple rule of number × units, we can apply arithmetic and algebra to any quantity:
\begin{align} & 5 \text{g} + 2 \text{g} = (5 + 2) \text{g} = 7 \text{g} \ & 5 \text{g} \div 2 \text{g} = \dfrac{5 \text{g}}{2 \text{g}} = 2.5 \text{ (the units cancel, and so we get a pure number)} \ & 5 \text{ in} \times 2 \text{ in} = 10 \text{ in}^{2} \text{ (10 square inches)}\end{align} \nonumber
This works perfectly well as long as we do not write equations with different parameters (i.e., those having units which measure different properties, like mass and length, temperature and energy, or volume and amount) on opposite sides of the equal sign. For example, applying algebra to the equation
$5 \text{g} = 2 \text{ in}^{2} \nonumber$
can lead to trouble in much the same way that dividing by zero does and should be avoided, because grams (g) is a unit of the parameter mass, and the inch (in) is a unit of the parameter length.
Conversions with Unity Factors
Mass Unity Factors
Notice also that whether a quantity is large or small depends on the size of the units as well as the size of the number. For example, the mass of smoke has been measured as 0.0342 g, but the balance might have been set to read in milligrams (or grains in the English system). Then the reading would be 34.2 mg (or 0.527 grains). The results (0.0342 g, 34.2 mg, or 0.527 gr) are the same quantity, the mass of smoke. One involves a smaller number and larger unit (0.0342 g), while the others have a larger number and smaller unit. So long as we are talking about the same quantity, it is a simple matter to adjust the number to go with any units we want.
We can convert among the different ways of expressing the mass with unity factors as follows:
Since 1 mg and 0.001 g are the same parameter (mass), we can write the equation
$1 \text{ mg} = 0.001 \text{ g} \nonumber$
Dividing both sides by 1 mg, we have
$1 = \dfrac{1\text{ mg}}{1\text{ mg}} = \dfrac{0.001\text{ g}}{1\text{ mg}} \nonumber$
Since the last term of this equation equals one, it is called a unity factor. It can be multiplied by any quantity, leaving the quantity unchanged.
We can generate another unity factor by dividing both sides by 0.001 g:
$1 = \dfrac{1\text{ mg}}{0.001\text{ g}} = \dfrac{0.001\text{ g}}{0.001\text{ g}} \nonumber$
Example $1$ : Mass Units
What is the mass in grams of a 5.0 grain (5 gr) aspirin tablet, given that 1 gram = 15.4323584 grains?
Solution:
$5.0 \text{ gr} = 5.0 \text{ gr} \times 1 = 5.0 \text{ gr}\times \dfrac{1.0\text{ g}}{\text{15.4323 gr}} \nonumber$
The units gr cancel, yielding the result
$5.0 \text{ gr} = 5.0 \div 15.4323 \text{ g} = 0.324 \text{ g} \nonumber$
Length Unity Factors
The parameter length may be measured in inches (in) in the English system, but scientific measurements (all measurements in the world exclusive of the U.S.) are reported in the metric units of meters (m) or some more convenient derived unit like centimeters (cm).
Example $2$ : Unit Conversions
Express the length 8.89 cm in inches, given that 1 cm = 0.3937 in.
Solution
Since 1 cm and 0.3937 in are the same quantity, we can write the equation
$1 \text{ cm} = 0.3937 \text{ in} \nonumber$
Dividing both sides by 1 cm, we have
$1 = \dfrac{0.3937\text{ in}}{1\text{ cm}} \nonumber$
Since the right side of this equation equals one, it is called a unity factor. It can be multiplied by any quantity, leaving the quantity unchanged.
$8.89 \text{ cm} = 8.89 \text{ cm} \times 1 = 8.89 \text{ cm} \times \dfrac{0.3937\text{ in}}{\text{1 cm}} \nonumber$
The units centimeter cancel, yielding the result 8.89 cm = 8.89 × 0.3937 in = 3.50 in
This agrees with the direct observation made in the figure.
Let us look at our air-pollution problem. It has probably already occurred to you that simply measuring the mass of smoke collected is not enough. Some other variables may affect your experiment and should also be measured if the results are to be reproducible. For example, wind direction and speed would almost certainly be important. The time of day and date when a measurement was made should be noted too. In addition you should probably specify what kind of filter you are using. Some are not fine enough to catch all the smoke particles.
Temperature
Another variable which is almost always recorded is the temperature. A thermometer is easy to use, and temperature can vary a good deal outdoors, where your experiments would have to be done.
In scientific work, temperatures are usually reported in degrees Celsius (°C), a scale in which the freezing point of pure water is 0°C and the normal boiling point 100°C. In the United States, however, you would be more likely to have available a thermometer calibrated in degrees Fahrenheit (°F). The relationship between these two scales of temperature is
$\dfrac{\text{T}_{(^{o}\text{F)}} - 32}{\text{T}_{(^{o}\text{C)}}}=\dfrac{9}{5} \nonumber$
Note that the temperature scales cannot be interconverted with simple unity factors, because they do not have a common zero point (0°C = 32°F). Rather, the mathematical function above must be used. The equation above is written in terms of the parameter temperature (T) with the units or dimensions subscripted in parentheses.
Volume Measurements
More important than any of the above variables is the fact that the more air you pump through the filter, the more smoke you will collect. Since air is a gas, it is easier to measure how much you collect in terms of volume than in terms of mass, and so you might decide to do it that way. Running your pump until it had filled a plastic weather balloon would provide a crude, inexpensive volume measurement. Assuming the balloon to be approximately spherical, you could measure its diameter and calculate its volume from the formula
$V=\dfrac{4}{3}\pi r^{3} \nonumber$
Example $3$ : Volume Calculation
Calculate the volume of gas in a sphere whose diameter is 106 in. Express your result in cubic centimeters (cm3).
Solution
Since the radius of a sphere is half its diameter,
$r = \dfrac{1}{2} \times 106 \text{ in} = 53 \text{ in} \nonumber$
We can use the same equality of quantities as in Example 1 to convert the radius to centimeters. When we cube the number and units, our result will be in cubic centimeters.
$1 \text{ cm} = 0.3937 \text{ in} \nonumber$
$\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = 1 \nonumber$
$R = 53 \text{ in} \times \dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} = \dfrac{\text{53}}{\text{0}\text{.3937}} \text{ cm} \nonumber$
Using the formula
\begin{align} & V =\dfrac{\text{4}}{\text{3}}\pi r^{\text{3}}=\dfrac{\text{4}}{\text{3}}\times \text{3}\text{.14159}\times ( \dfrac{\text{53}}{\text{0}\text{.3937}}\text{cm} )^{3} \ & \text{ }=\text{10219264 cm}^{\text{3}} \ & \end{align} \nonumber
You can see from Examples 1 and 2 that two unity factors may be obtained from the equality
$1 \text{ cm} = 0.3937 \text{ in} \nonumber$
We can use one of them to convert inches to centimeters and the other to convert centimeters to inches. The correct factor is always the one which results in cancellation of the units we do not want.
The result in Example 2 also shows that cubic centimeters are rather small units for expressing the volume of the balloon. If we used larger units, as shown in the following example, we would not need more than 10 million of them to report our answer.
Example $4$: Volume Unit Conversion
Express the result of Example 3 in cubic meters, given that 1 m = 100 cm.
Solution
Again we wish to use a unity factor, and since we are trying to get rid of cubic centimeters, centimeters must be in the denominator:
$\text{1 = }\dfrac{\text{1 m}}{\text{100 cm}} \nonumber$
But this will not allow cancellation of cubic centimeters. However, note that 13 = 1 That is, we can raise a unity factor to any power, and it remains unity. Thus
\begin{align} & \text{1 =}\left( \text{ }\dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{100}^{\text{3}}\text{ cm}^{\text{3}}}\text{ = }\dfrac{\text{1 m}^{\text{3}}}{\text{1 000 000 cm}^{\text{3}}} \ \text{ and } \ & \text{10 219 264 cm}^{\text{3}}\text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 m}}{\text{100 cm}} \right)^{\text{3}} \ & \text{ = 10 219 264 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 m}^{\text{3}}}{\text{100 000 cm}^{\text{3}}} \ & \text{ = 10.219 264 m}^{\text{3}} \ & \end{align} \nonumber
1.06: Measurements Quantities and Unity Factors
Convert Mass
Convert 12 lb weight of bowling ball to g, showing unity factors
12.00 lb x 453.59237 g/lb = 5443 g
Calculate and convert volumes
V=4/3 π r3
V = 4/3 * 3.1416 * (1/2 * 8.59 in)3 =
4/3 * 3.1416 * 79.23 in3
=331.9 in
Unity Factor: 2.54 cm = 1 in
331.9 in3 x (2.54 cm / 1 in)3 Note!!!
331.9 in3 x 16.39cm3/in3
= 5439 cm
Densities
Will the bowling ball float in water? Demo [1]
D = 5443 g / 5439 cm3 Too close to call. See Errors in Measurement Lecture Demonstrations
Mass vs. Weight
What is the mass of hydrogen?
Density of hydrogen at room temperature and 1 Atm = 0.082 g/L
What is the volume in L, assuming same size as bowling ball? Unity Factors?
1 cm3 = 1 mL = 10-3L (Note: 1 mL = 1 cm3 = “1 cc”)
V (L) = 5500 cm3 x (1 L / 1000 cm3)
m (g) = V (L) * D (g/L) = 5.500 L x 0.082 g/cm3
= 0.451 g
Why does the Hydrogen balloon float? F = W = m g
F = W = (0.451 g x 1 kg /1000g) * 9.8 m*s-2 =
= 0. 0044 N
Force Upward: (Archimedes)
D of air = 1.2 g /L; Archimedes Principle: buoyancy = mass of air displaced (6.6 g)
F = m g
F = ( 6.6 g x 1 kg /1000g) * 9.8 m*s-2 =
= 0. 065 N
Net force = 0.065 N - 0.044 N upwards. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.06%3A_Measurements_Quantities_and_Unity_Factors/1.6.01%3A_Measurements_Quantities_and_Unity_Factors_Lecture_Demonstrations.txt |
Scientific measurements are of no value (or at least, they're not really scientific) unless they are given with some statement of the errors they contain. If a poll reports that one candidate leads another by 5%, that may be politically useful for the winning candidate to point out. But all respectable polls are scientific, and report errors. If the error in measurement is plus or minus 10%, which indicates anything from the candidate leading by 15% to trailing by 5%, the poll really does not reliably tell who is in the lead. If the poll had an error of 1%, the leading candidate could make a more scientific case that for being in the lead (by a 4% to 6% margin, or 5% +/-1%).
Luckily, we have an easy method for indicating the error in a measurement, and it is suggested in the results we gave for an air-pollution experiment, in which the mass of smoke collected was given as 3.42 × 10–2 g and the volume of the balloon was given as 1.021 926 4 × 107 cm3.
"Significant Figures"
The volume was calculated from a balloon diameter of 106 inches. There is something strange about the second quantity, though. It contains a number which was copied directly from the display of an electronic calculator and has Too Many Digits to represent the error of measurement properly.
The reliability (more specifically, the random error or precision) of a quantity derived from a measurement is customarily indicated by the number of significant figures (or significant digits) it contains. For example, the three significant digits in the quantity 3.42 × 10–2 g tell us that a balance was used on which we could distinguish 3.42 × 10–2 g from 3.43 × 10–2 or 3.41 × 10–2 g. There might be some question about the last digit, but those to the left of it are taken as completely reliable. Another way to indicate the same thing is (3.42±0.01) × 10–2 g. Our measurement is somewhere between 3.41 × 10–2 and 3.43 × 10–2 g.
To illustrate how the number of significant digits indicates the error, suppose we had a measurement reported as 3.42 g. From the measurement, we would assume an error of 0.01 g, and the percent error is
$\text{Percent Error } = \dfrac{\text{Error}}{\text{Value}} \times 100\%\label{1}$
$\text{Percent Error} = \dfrac{0.01 \text{ g}}{3.42 \text{ g}} \times 100\% = 0.29\% \nonumber$
If the correctly recorded measurement had more digits, like 3.4275 g, the measurement itself would indicate that a more expensive balance was used to give better precision (a smaller percent error). In this case, the error is 0.0001 g, and the percent error is:
$\text{Percent Error} = \dfrac{0.0001 \text{ g}}{3.4265 \text{ g}} \times 100\% = 0.0029\% \nonumber$
So we record measurements to the proper number of digits as a rudimentary method[1] for indicating their approximate error.
Example $1$: Graduated Cylinder
As another example of choosing an appropriate number of significant digits, let us read the volume of liquid in a graduated cylinder.
Solution:
The bottom of the meniscus lies between graduations corresponding to 38 and 39 cm3. We can estimate that it is at 38.5 cm3, but the last digit might be off a bit-perhaps it looks like 38.4 or 38.6 cm3 to you. Since the third digit is in question, we should use three significant figures. The volume would be recorded as 38.5 cm3, indicating 0.1 cm3 error. Laboratory equipment is often calibrated similarly to this graduated cylinder—you should estimate to the nearest tenth of the smallest graduation.
Counting Significant Digits
In some ordinary numbers, for example, 0.001 23, zeros serve merely to locate the decimal point. They do not indicate the reliability of the measurement and therefore are not significant. Another advantage of scientific notation is that we can assume that all digits are significant. Thus if 0.001 23 is written as 1.23 × 10–3, only the 1, 2, and 3, which indicate the reliability of the measurement, are written. The decimal point is located by the power of 10.
If the rule expressed in the previous paragraph is applied to the volume of air collected in our pollution experiment, 1.021 926 4 × 107 cm3, we find that the volume has eight significant digits. This implies that it was determined to ±1 cm3 out of about 10 million cm3, a reliability which corresponds to locating a grasshopper exactly at some point along the road from Philadelphia to New York City. For experiments as crude as ours, this is not likely. Let us see just how good the measurement was.
You will recall that we calculated the volume from the diameter of the balloon, 106 in. The three significant figures imply that this might have been as large as 107 in or as small as 105 in. We can repeat the calculation with each of these quantities to see how far off the volume would be:
\begin{align} & \text{ }r\text{ = }\dfrac{\text{1}}{\text{2}}\text{ }\times \text{ 107 in = 53}\text{.5 in }\times \text{ }\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}}\text{ } \ & \text{ = 135}\text{.890 27 cm} \ & \text{ }V\text{ = }\dfrac{\text{4}}{\text{3}}\text{ }\times \text{ 3}\text{.141 59 }\times \text{ (135}\text{.890 27 cm)}^{\text{3}}\text{ } \ & \text{ = 10 511 225 cm}^{\text{3}}\text{ = 1}\text{.051 122 5 }\times \text{ 10}^{\text{7}}\text{ cm}^{\text{3}} \ & \text{or }V\text{ = }\dfrac{\text{4}}{\text{3}}\text{ }\times \text{ 3}\text{.141 59 }\times \text{ }\left( \dfrac{\text{1}}{\text{2}}\text{ }\times \text{ 105 in }\times \text{ }\dfrac{\text{1 cm}}{\text{0}\text{.3937 in}} \right)^{\text{3}}\text{ } \ & \text{ = 9 932 759 cm}^{\text{3}}\text{ = 0}\text{.993 275 9 }\times \text{ 10}^{\text{7}}\text{ cm}^{\text{3}} \ \end{align}
That is, the volume is between 0.99 × 107 and 1.05 × 107 cm3 or (1.02 ± 0.03) × 107 cm3. We should round our result to three significant figures, for example, 1.02 cm³, because the last digit, namely 2, is in question.
So the conclusion is simple: The calculated result cannot be more precise than the measurement from which it was made. To show this in an alternative way, the error in 106 in would be about 1 in, so the percent error is
$\text{Percent Error} = \dfrac{1 \text{ in}}{106 \text{ in}} \times 100\% = 0.9\% \nonumber$
If the result is reported as 10 219 264 cm3, the implied error is 1 part in 10 219 264 and the percent error is:
$\text{Percent Error} = \dfrac{1 \text{ cm}^{\text{3}}}{10219264 \text{ cm}^\text{3}} \times 100\% = 0.000009\% \nonumber$
Clearly, we should not be able to increase the precision of our data by merely manipulating the data mathematically, so we need simple rules for rounding numbers in calculated quantities.
If the result is reported properly as 1.02 × 107 cm3, the percent error is
$\text{Percent Error} = \dfrac{0.01 \text{ cm}^{\text{3}}}{1.02 \text{ cm}^\text{3}} \times 100\% = 1\% \nonumber$,
approximately the same as the original measurement.
Rules for Rounding Numbers
1. All digits to be rounded are removed together, not one at a time.
2. If the left-most digit to be removed is less than five, the last digit retained is not altered.
3. If the left-most digit to be removed is greater than five, the last digit retained is increased by one.
4. If the left-most digit to be removed is five and at least one of the other digits to be removed is nonzero. the last digit retained is increased by one.
5. If the left-most digit to he removed is five and all other digits to he removed are zero, the last digit retained is not altered if it is even, but is increased by one if it is odd.
Application of the Rules for Rounding Numbers can be illustrated by an example.
Example $2$ : Rounding
Round each of the numbers below to three significant figures.
1. 34.7449
2. 34.864
3. 34.754
4. 34.250
5. 34.35
Solution
a) Apply rules 1 and 2:
(Note that a different result would be obtained if the digits were incorrectly rounded one at a time from the right.)
b) Apply rules 1 and 3: 34.864 → 34.9
c) Apply rules 1 and 4: 34.754 → 34.8
d) Apply rules 1 and 5: 34.250 → 34.2
e) Apply rule 5: 34.35 → 34.4
To how many significant figures should we round our air-pollution results? We have already done a calculation involving multiplication and division to obtain the volume of our gas-collection balloon. It involved the following quantities and numbers:
a calculation involving multiplication and division to obtain the volume of our gas-collection balloon
$106 \text{ in}$ A measured quantity with three significant figures
$0.3937 \text{ in/cm}$ A conversion factor given with four significant figures(1)
$3.141 59$ π used with six significant figures (we could obtain more if we wanted)
$\tfrac{\text{4}}{\text{3}} \text{ and } \tfrac{\text{1}}{\text{2}}$ Numbers(2) with an infinite number of significant figures since the integers in these fractions are exact by definition
1. some conversion factors like 1 mg = 0.001 g are exact, and have an infinite number of significant figures).
2. Note that pure numbers are not measurements or quantities, and so they are unlimited in the number of significant figures (numbers have zero error).
The result of the calculation contained three significant figures — the same as the least-reliable number. This illustrates the general rule that for multiplication and division the number of significant figures in the result is the same as in the least-reliable measurement. Defined numbers such as π, ½ or 100 cm/1m are assumed to have an infinite number of significant figures.
In the case of addition and subtraction, a different rule applies. Suppose, for example, that we weighed a smoke-collection filter on a relatively inaccurate balance that could only be read to the nearest 0.01 g. After collecting a sample, the filter was reweighed on a single-pan balance to determine the mass of smoke particles.
• Final mass: 2.3745 g (colored digits are in question)
• Initial mass: –2.32 g
• Mass of smoke: 0.0545 g
Since the initial weighing could have been anywhere from 2.31 to 2.33 g, all three figures in the final result are in question. (It must be between 0.0445 and 0.0645 g). Thus there is but one significant digit, and the result is 0.05g. The rule here is that the result of addition or subtraction cannot contain more digits to the right than there are in any of the numbers added or subtracted. Note that subtraction can drastically reduce the number of significant digits when this rule is applied.
Rounding numbers is especially important if you use an electronic calculator, since these machines usually display a large number of digits, most of which are meaningless. The best procedure is to carry all digits to the end of the calculation (your calculator will not mind the extra work!) and then round appropriately. Answers to subsequent calculations in this book will be rounded according to the rules given. You may wish to go back to previous examples and round their answers correctly as well.
Example $3$: Significant Figures
Evaluate the following expressions, rounding the answer to the appropriate number of significant figures.
1. $32.61 \text{g} + 8.446 \text{g} + 7.0 \text{g}$
2. $0.136 \text{cm}^3 \times 10.685 \text{g cm}^{-3}$
Solution
1. $32.61 \text{g} + 8.446 \text{g} + 7.0 \text{g} = 48.056 \text{g} = 48.1 \text{g (7.0 has only one figure to the right of the decimal point.)}$
2. $0.136 \text{cm}^3 \times 10.685 \text{g cm}^{-3} = 1.453 \text{g} = 1.45 \text{g (0.136 has only three significant figures.)}$
Precision vs. Accuracy
When we suggested filling a surplus weather balloon to measure how much gas was pumped through our air-pollution collector, we mentioned that this would be a rather crude way to determine volume. For one thing, it would not be all that simple to measure the diameter of an 8- or 9-ft sphere reliably. Using a yardstick, we would be lucky to have successive measurements agree within half an inch or so. It was for this reason that the result was reported to the nearest inch. The degree to which repeated measurements of the same quantity yield the same result is called precision. Repetition of a highly precise measurement would yield almost identical results, whereas low precision implies numbers would differ by a significant percentage from each other.
A highly precise measurement of the diameter of our balloon could be achieved, but it would probably not be worthwhile. We have assumed a spherical shape, but this is almost certainly not exactly correct. No matter how precisely we determine the diameter, our measurement of gas volume will be influenced by deviations from the assumed shape. When one or more of our assumptions about a measuring instrument are wrong, the accuracy of a result will be affected. An obvious example would be a foot rule divided into 11 equal inches. Measurements employing this instrument might agree very precisely, but they would not be very accurate. The following data on the mass of the smoke, measured repeatedly with two balances and a scale, show the difference between accuracy and precision. If he mass has been determined to be exactly 0.03420 g by an independent measurement, the first balance is both accurate (the correct value is obtained) and precise (the range of measurements is small); the second balance is precise, but not accurate; the scale is neither accurate nor precise. Note that we are comparing the three to one another; relative to a balance that had a range of 0.00001 g, all three would be imprecise.
Table $1$: Accurate vs. Precise Measurements
Mass, Balance A Mass, Balance B Mass, Scale
0.0342 0.0362 0.0380
0.0341 0.0361 0.0370
0.0342 0.0363 0.0390
0.0343 0.0362 0.0400
average 0.0342 0.0362 0.0385
range 0.001 0.001 0.013
The following figures help to understand the difference between precision (small expected difference between multiple measurements) and accuracy (difference between the result and a known value).
An important point of a different kind is illustrated in the last two paragraphs. A great many common words have been adopted into the language of science. Usually such an adoption is accompanied by an unambiguous scientific definition which does not appear in a normal dictionary. Precision and accuracy are many times treated as synonyms, but in science each has a slightly different meaning. Another example is quantity, which we have defined in terms of “number × unit.” Other English words like bulk, size, amount, and so forth, may be synonymous with quantity in everyday speech, but not in science. As you encounter other words like this, try to learn and use the scientific definition as soon as possible, and avoid confusing it with the other meanings you already know.
Even granting the crudeness of the measurements we have just described, they would be adequate to demonstrate whether or not an air-pollution problem existed. The next step would be to find a chemist or public health official who was an expert in assessing air quality, present your data, and convince that person to lend his or her skill and authority to your contention that something was wrong. Such a person would have available equipment whose precision and accuracy were adequate for highly reliable measurements and would be able to make authoritative public statements about the extent of the air-pollution problem.
1.07: Errors in Measurement
Re-do calculation of density of bowling ball done earlier paying more attention to accurate measurement of diameter, and correct precision of measurements and calculated results.
Measure diameter precisely with large caliper (or bar on ring stand lowered to top of ball).
Care with propagation of errors gives:
D = 5.44 x 103 g / 5.58 x 103 cm3 = 0.974910394 g/cm | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.07%3A_Errors_in_Measurement/1.7.01%3A_Errors_in_Measurement_Lecture_Demonstrations.txt |
Volume is the amount of 3D space a substance or object occupies. In the photo above, the same volume of water (50 mL) is shown in each of the beakers. As you noticed, the 50 mL looks radically different from beaker to beaker. What causes this variation? Volume is a derived unit, depending on 3 quantities: length, width, and height. Therefore, though the beakers each contain the same volume, their differing lengths, widths, and heights makes that volume look deceptively different.
The most commonly used derived units are those of volume. As we have already seen, calculation of the volume of an object requires that all 3 dimensions are multiplied together (length, width, and height). Thus the SI unit of volume is the cubic meter (m3). This is rather large for use in the chemical laboratory, and so the cubic decimeter (dm3) or cubic centimeter (cm3, formerly cc) are more commonly used. The relationship between these units and the cubic meter is easily shown:
$\text{1 dm} = \text{0.1 m} \nonumber$
$\text{1 cm} = \text{0.01 m} \nonumber$
Cubing both sides of each equation, we have
$\text{1 dm}^\text{3} = \text{0.1}^\text{3} \text{m}^\text{3} = \text{0.001 m}^\text{3} = \text{10}^{-\text{3}} \text{m}^\text{3} \nonumber$
$\text{1 cm}^\text{3} = \text{0.01}^\text{3} \text{m}^\text{3} = \text{0.000 001 m}^\text{3} = \text{10}^{-\text{6}} \text{m}^\text{3} \nonumber$
Note that in the expression dm3 the exponent includes the prefix(opens in new window) as well as the base unit. A cubic decimeter is one-thousandth of a cubic meter, not one-tenth of a cubic meter.
Two other units of volume are commonly encountered in the chemical laboratory — the liter (l) and the milliliter (ml — one-thousandth of a liter). The liter was originally defined as the volume of one kilogram of pure water at the temperature of its maximum density (3.98°C) but in 1964 the definition was changed. The liter is now exactly one-thousandth of a cubic meter, that is, 1 dm3. A milliliter is therefore exactly 1 cm3. Because the new definition of liter altered its volume slightly, it is recommended that the results of highly accurate measurements be reported in the SI units cubic decimeters or cubic centimeters, rather than in liters or milliliters. For most situations discussed in this online textbook, however, the units cubic decimeter and liter, and cubic centimeter and milliliter may be used interchangeably. Thus when recording a volume obtained from laboratory glassware calibrated in milliliters, you can just as well write 24.7 cm3 as 24.7 ml.
1.9.01: Density Lecture Demonstrations
The terms heavy and light are commonly used in two different ways. We refer to weight when we say that an adult is heavier than a child. On the other hand, something else is alluded to when we say that oak is heavier than balsa wood. A small shaving of oak would obviously weigh less than a roomful of balsa wood, but oak is heavier in the sense that a piece of given size weighs more than the same-size piece of balsa.
What we are actually comparing is the mass per unit volume, that is, the density. In order to determine these densities, we might weigh a cubic centimeter of each type of wood. If the oak sample weighed 0.71 g and the balsa 0.15 g, we could describe the density of oak as 0.71 g cm–3 and that of balsa as 0.15 g cm–3. (Note that the negative exponent in the units cubic centimeters indicates a reciprocal. Thus 1 cm–3 = 1/cm3 and the units for our densities could be written as $\frac{\text{g}}{\text{cm}^\text{3}}$, g/cm3, or g cm–3. In each case the units are read as grams per cubic centimeter, the per indicating division.) We often abbreviate "cm3" as "cc", and 1 cm3 = 1 mL exactly by definition.
In general it is not necessary to weigh exactly 1 cm3 of a material in order to determine its density. We simply measure mass and volume and divide volume into mass:
$\text{Density} = \dfrac{\text{mass}}{\text{volume}} \nonumber$
or
$\rho = \dfrac{m}{V} \quad \label{1}$
where $ρ$ is the density, $m$ is the mass, and $V$ volume.
Example $1$: Density of Aluminum
Calculate the density of
1. a piece of aluminum whose mass is 37.42 g and which, when submerged, increases the water level in a graduated cylinder by 13.9 ml;
2. an aluminum cylinder of mass 25.07 g, radius 0.750 cm, and height 5.25 cm.
a
Since the submerged metal displaces its own volume,
\begin{align*} \text{Density} &= \rho =\dfrac{m}{V} \[4pt] &= \dfrac{\text{37.42 g}}{\text{13.9 ml}} \[4pt] &= \text{2.69 g}/\text{ml or 2.69 g ml}^{-\text{1}} \end{align*} \nonumber
b
The volume of the cylinder must be calculated first, using the formula
\begin{align*} \text{V} &= \pi r^\text{2} h\[4pt] &= \text{3.142}\times\text{(0.750 cm)}^\text{2}\times\text{5.25 cm} \[4pt] &= \text{9.278 718 8 cm}^\text{3} \end{align*} \nonumber
Then
$\rho = \dfrac{m}{V} = \dfrac{\text{25.07 g}}{\text{9.278 718 8 cm}^\text{3}} = \begin{cases} 2.70 \dfrac{\text{g}}{\text{cm}^\text{3}} \ \text{2.70 g cm}^{-\text{3}} \ \text{2.70 g}/ \text{cm}^\text{3} \end{cases} \nonumber$
which are all acceptable alternatives.
Note that unlike mass or volume (extensive properties), the density of a substance is independent of the size of the sample (intensive property). Thus density is a property by which one substance can be distinguished from another. A sample of pure aluminum can be trimmed to any desired volume or adjusted to have any mass we choose, but its density will always be 2.70 g/cm3 at 20°C. The densities of some common pure substances are listed below.
Tables and graphs are designed to provide a maximum of information in a minimum of space. When a physical quantity (number × units) is involved, it is wasteful to keep repeating the same units. Therefore it is conventional to use pure numbers in a table or along the axes of a graph. A pure number can be obtained from a quantity if we divide by appropriate units. For example, when divided by the units gram per cubic centimeter, the density of aluminum becomes a pure number 2.70:
$\dfrac{\text{Density of aluminum}}{\text{1 g cm}^{-\text{3}}} = \dfrac{\text{2.70 g cm}^{-\text{3}}}{\text{1 g cm}^{-\text{3}}} = \text{2.70} \nonumber$
Therefore, a column in a table or the axis of a graph is conveniently labeled in the following form:
Quantity/units
This indicates the units that must be divided into the quantity to yield the pure number in the table or on the axis. This has been done in the second column of the table.
Table $1$: Density of Several Substances at 20°C.Anchor
Substance Density / g cm-3
Helium gas 0.000 16
Dry air 0.001 185
Gasoline 0.66-0.69 (varies)
Kerosene 0.82
Benzene 0.880
Water 1.000
Carbon tetrachloride 1.595
Magnesium 1.74
Salt 2.16
Aluminum 2.70
Iron 7.87
Copper 8.96
Silver 10.5
Lead 11.34
Uranium 19.05
Gold 19.32
Converting Densities
In our exploration of density, notice that chemists may express densities differently depending on the subject. The density of pure substances may be expressed in kg/m3 in some journals which insist on strict compliance with SI units; densities of soils may be expressed in lb/ft3 in some agricultural or geological tables; the density of a cell may be expressed in mg/µL; and other units are in common use. It is easy to transform densities from one set of units to another, by multiplying the original quantity by one or more unity factors:
Example $2$: Density of Water
Convert the density of water, 1 g/cm3 to (a) lb/cm3 and (b) lb/ft3
a
The equality $\text{454 g} = \text{1 lb}$ can be used to write two unity factors,
$\dfrac{\text{454 g}}{\text{1 lb}}\nonumber$
or
$\dfrac{\text{1 lb}}{\text{454 g}} \nonumber$
The given density can be multiplied by one of the unity factors to get the desired result. The correct conversion factor is chosen so that the units cancel:
$\text{1} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{1 lb}}{\text{454 g}} = \text{0.002203} \dfrac{\text{lb}}{\text{cm}^\text{3}}\nonumber$
b
Similarly, the equalities $\text{2.54 cm} = \text{1 inch}$, and $\text{12 inches} = \text{1 ft}$ can be use to write the unity factors:
$\dfrac{\text{2.54 cm}}{\text{1 in}} \text{, } \dfrac{\text{1 in}}{\text{2.54 cm}} \text{, } \dfrac{\text{12 in}}{\text{1 ft}} \text{ and } \dfrac{\text{1 ft}}{\text{12 in}} \nonumber$
In order to convert the cm3 in the denominator of 0.002203 to in3, we need to multiply by the appropriate unity factor three times, or by the cube of the unity factor:
$\text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}} \times \dfrac{\text{2.54 cm}}{\text{1 in}}\nonumber$
or
$\text{0.002203} \dfrac{\text{g}}{\text{cm}^\text{3}} \times \left(\dfrac{\text{2.54 cm}}{\text{1 in}}\right)^\text{3} = \text{0.0361 lb}/ \text{in}^\text{3}\nonumber$
This can then be converted to lb/ft3:
$\text{0.0361 lb}/ \text{in}^\text{3}\times \left(\dfrac{\text{12 in}}{\text{1 ft}}\right)^\text{3} = \text{62.4 lb}/\text{ft}^\text{3}\nonumber$
It is important to notice that we have used conversion factors to convert from one unit to another unit of the same parameter
1.09: Density
Create Density Layers with 16%, 12%, 8%, 4% and 0% sucrose in water, colored with food dyes, by adding each successively less dense layer in a graduated cylinder. Use a pipet, and prevent mixing of layers by delivering solution into a deflagrating spoon held on the surface of the layer below to stop downward inertia of solution.[1][2]
Layered, colored sucrose solutions of different concentrations
Density Demonstrations Using Diet Soft Drinks [3][4][5]
Mass and Density-A Surprising Classroom Demonstration [6] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.08%3A_Volume.txt |
Earlier we showed how unity factors can be used to express quantities in different units of the same parameter. For example, a density can be expressed in g/cm3 or lb/ft3. Now we will see how conversion factors representing mathematical functions, like D = m/v, can be used to transform quantities into different parameters. For example, what is the volume of a given mass of gold? Unity factors and conversion factors are conceptually different, and we'll see that the "dimensional analysis" we develop for unit conversion problems must be used with care in the case of functions.
When we are referring to the same object or sample of material, it is often useful to be able to convert one parameter into another. For example, in our discussion of fossil-fuel reserves we find that 318 Pg (3.18 × 1017 g) of coal, 28.6 km3 (2.68 × 1010 m3) of petroleum, and 2.83 × 103 km3 (2.83 × 1013 m3) of natural gas (measured at normal atmospheric pressure and 15°C) are available. But none of these quantities tells us what we really want to know ― how much heat energy could be released by burning each of these reserves? Only by converting the mass of coal and the volumes of petroleum and natural gas into their equivalent energies can we make a valid comparison. When this is done, we find that the coal could release 7.2 × 1021 J, , the petroleum 1.1 × 1021 J, and the gas 1.1 × 1021 J of heat energy. Thus the reserves of coal are more than three times those of the other two fuels combined. It is for this reason that more attention is being paid to the development of new ways for using coal resources than to oil or gas. Conversion of one kind of quantity into another is usually done with what can be called a conversion factor, but the conversion factor is based on a mathematical function (D = m / V) or mathematical equation that relates parameters. Since we have not yet discussed energy or the units (joules) in which it is measured, an example involving the more familiar quantities mass and volume will be used to illustrate the way conversion factors are employed. The same principles apply to finding how much energy would be released by burning a fuel, and that problem will be encountered later.
For helpful context about the above discussion, check out the following Crash Course Chemistry video:
Suppose we have a rectangular solid sample of gold which measures 3.04 cm × 8.14 cm × 17.3 cm. We can easily calculate that its volume is 428 cm3 but how much is it worth? The price of gold is about 5 dollars per gram, and so we need to know the mass rather than the volume. It is unlikely that we would have available a scale or balance which could weigh accurately such a large, heavy sample, and so we would have to determine the mass of gold equivalent to a volume of 428 cm3. This can be done by manipulating the equation which defines density, ρ = m / V. If we multiply both sides by V, we obtain
$V \times \rho =\dfrac{m}{V}\times V = m\label{1}$
$m = V \times \rho \nonumber$
or
$mass = \text{volume} \times \text{ density } \nonumber$
Taking the density of gold from a reference table(opens in new window), we can now calculate
$\text{Mass}= m =V \rho =\text{428 cm}^{3}\times \dfrac{\text{10}\text{0.32 g}}{\text{1 cm}^{3}}=8.27\times \text{10}^{3}\text{g}=\text{8}\text{.27 kg} \nonumber$
This is more than 18 lb of gold. At the price quoted above, it would be worth over 40 000 dollars!
The formula which defines density can also be used to convert the mass of a sample to the corresponding volume. If both sides of Eq. $\ref{1}$ are multiplied by 1/ρ, we have
$\dfrac{\text{1}}{\rho }\times m=V \rho \times \dfrac{\text{1}}{\rho }=V \nonumber$
$V=m \times \dfrac{\text{1}}{\rho }\label{2}$
Notice that we used the mathematical function D = m/V to convert parameters from mass to volume or vice versa in these examples. How does this differ from the use of unity factors to change units of one parameter?
An Important Caveat
A mistake sometimes made by beginning students is to confuse density with concentration, which also may have units of g/cm3. By dimensional analysis, this looks perfectly fine. To see the error, we must understand the meaning of the function
$C = \dfrac{m}{V} \nonumber$
In this case, $V$ refers to the volume of a solution, which contains both a solute and solvent.
Given a concentration of an alloy is 10 g gold in 100 cm3 of alloy, we see that it is wrong (although dimensionally correct as far as conversion factors go) to incorrectly calculate the volume of gold in 20 g of the alloy as follows:
$20 \text{g} \times \dfrac{\text{100 cm^3}}{\text{10 g}} = 200 \text{ cm}^{3} \nonumber$
It is only possible to calculate the volume of gold if the density of the alloy is known, so that the volume of alloy represented by the 20 g could be calculated. This volume multiplied by the concentration gives the mass of gold, which then can be converted to a volume with the density function.
The bottom line is that using a simple unit cancellation method does not always lead to the expected results, unless the mathematical function on which the conversion factor is based is fully understood.
Example $1$: Volume of Ethanol
A solution of ethanol with a concentration of 0.1754 g / cm3 has a density of 0.96923 g / cm3 and a freezing point of -9 ° F [1]. What is the volume of ethanol (D = 0.78522 g / cm3 at 25 °C) in 100 g of the solution?
Solution
The volume of 100 g of solution is
$V = m \div D = 100 \text{ g} \div 0.96923 \text{ g} \text{ cm}^{3} = 103.17 \text{ cm}^{3} \nonumber$
The mass of ethanol in this volume is
$m = V \times C = 103.17 \text{ cm}^{3} \times 0.1754 \text{ g /} \text{ cm}^{3} = 18.097 \text{ g} \nonumber$
$\text{The volume of ethanol } = m \div D = 18.097 \text{ g} \div 0.78522 \text{ g / } \text{cm}^{3} = 23.05 \text{cm}^{3} \nonumber$
Note that we cannot calculate the volume of ethanol by
$\dfrac {\dfrac{0.96923 g}{cm^3} \times 100 cm^3}{\dfrac {0.78522 g}{cm^3}} \normalsize = 123.4 \text{cm}^{3} \nonumber$
even though this equation is dimensionally correct.
Note: Note that this result required when to use the function C = m/V, and when to use the function D=m/V as conversion factors. Pure dimensional analysis could not reliably give the answer, since both functions have the same dimensions.
Example $2$: Volume of Benzene
Find the volume occupied by a 4.73-g sample of benzene.
Solution
The density of benzene is 0.880 g cm–3. Using Eq. (2),
$\text{Volume = }V\text{ = }m\text{ }\times \text{ }\dfrac{\text{1}}{\rho }\text{ = 4}\text{.73 g }\times \text{ }\dfrac{\text{1 cm}^{\text{3}}}{\text{0}\text{.880 g}}\text{ = 5}\text{.38 cm}^{\text{3}} \nonumber$
Note: Note that taking the reciprocal of $\Large\tfrac{\text{0}\text{.880 g}}{\text{1 cm}^{3}}$ simply inverts the fraction ― 1 cm3 goes on top, and 0.880 g goes on the bottom.
The two calculations just done show that density is a conversion factor which changes volume to mass, and the reciprocal of density is a conversion factor changing mass into volume. This can be done because the mathematical formula defining density relates it to mass and volume. Algebraic manipulation of this formula gave us expressions for mass and for volume [Eq. $\ref{1}$ and $\ref{2}$], and we used them to solve our problems. If we understand the function D = m/V and heed the caveat above, we can devise appropriate conversion factors by unit cancellation, as the following example shows:
Example $3$: Volume of Mercury
A student weighs 98.0 g of mercury. If the density of mercury is 13.6 g/cm3, what volume does the sample occupy?
Solution
We know that volume is related to mass through density.
Therefore
$V = m \times \text{ conversion factor} \nonumber$
Since the mass is in grams, we need to get rid of these units and replace them with volume units. This can be done if the reciprocal of the density is used as a conversion factor. This puts grams in the denominator so that these units cancel:
$V=m\times \dfrac{\text{1}}{\rho }=\text{98}\text{.0 g}\times \dfrac{\text{1 cm}^{3}}{\text{13}\text{.6 g}}=\text{7}\text{.21 cm}^{3} \nonumber$
If we had multiplied by the density instead of its reciprocal, the units of the result would immediately show our error:
$V=\text{98}\text{.0 g}\times \dfrac{\text{13.6 }g}{\text{1 cm}^{3}}=\text{1}\text{.333}{\text{g}^{2}}/{\text{cm}^{3}}\;$ (no cancellation!)
It is clear that square grams per cubic centimeter are not the units we want.
Using a conversion factor is very similar to using a unity factor — we know the conversion factor is correct when units cancel appropriately. A conversion factor is not unity, however. Rather it is a physical quantity (or the reciprocal of a physical quantity) which is related to the two other quantities we are interconverting. The conversion factor works because of the relationship [ie. the definition of density as defined by Eqs. $\ref{1}$ and $\ref{2}$ includes the relationships between density, mass, and volume], not because it is has a value of one. Once we have established that a relationship exists, it is no longer necessary to memorize a mathematical formula. The units tell us whether to use the conversion factor or its reciprocal. Without such a relationship, however, mere cancellation of units does not guarantee that we are doing the right thing.
A simple way to remember relationships among quantities and conversion factors is a “road map“of the type shown below:
$\text{Mass }\overset{density}{\longleftrightarrow}\text{ volume or }m\overset{\rho }{\longleftrightarrow}V\text{ } \nonumber$
This indicates that the mass of a particular sample of matter is related to its volume (and the volume to its mass) through the conversion factor, density. The double arrow indicates that a conversion may be made in either direction, provided the units of the conversion factor cancel those of the quantity which was known initially. In general the road map can be written
$\text{First quantity }\overset{\text{conversion factor}}{\longleftrightarrow}\text{ second quantity} \nonumber$
As we come to more complicated problems, where several steps are required to obtain a final result, such road maps will become more useful in charting a path to the solution.
Example $1$: Volume to Mass Conversion
Black ironwood has a density of 67.24 lb/ft3. If you had a sample whose volume was 47.3 ml, how many grams would it weigh? (1 lb = 454 g; 1 ft = 30.5 cm).
Solution
The road map
$V\xrightarrow{\rho }m\text{ } \nonumber$
tells us that the mass of the sample may be obtained from its volume using the conversion factor, density. Since milliliters and cubic centimeters are the same, we use the SI units for our calculation:
$\text{Mass} = m = 47.3 \text{cm}^{3} \times \dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{3}} \nonumber$
Since the volume units are different, we need a unity factor to get them to cancel:
$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\left( \dfrac{\text{1 ft}}{\text{30}\text{.5 cm}} \right)^{\text{3}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}} \nonumber$
We now have the mass in pounds, but we want it in grams, so another unity factor is needed:
$m\text{ = 47}\text{.3 cm}^{\text{3}}\text{ }\times \text{ }\dfrac{\text{1 ft}^{\text{3}}}{\text{30}\text{.5}^{\text{3}}\text{ cm}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{67}\text{.24 lb}}{\text{1 ft}^{\text{3}}}\text{ }\times \text{ }\dfrac{\text{454 g}}{\text{ 1 lb}}\text{ = 50}\text{0.9 g} \nonumber$
In subsequent chapters we will establish a number of relationships among physical quantities. Formulas will be given which define these relationships, but we do not advocate slavish memorization and manipulation of those formulas. Instead we recommend that you remember that a relationship exists, perhaps in terms of a road map, and then adjust the quantities involved so that the units cancel appropriately. Such an approach has the advantage that you can solve a wide variety of problems by using the same technique. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/01%3A_Introduction_-_The_Ambit_of_Chemistry/1.10%3A_Conversion_Factors_and_Functions.txt |
The atomic theory, first proposed in modern form by John Dalton, is one of the most important and useful ideas in chemistry. It interprets observations of the every-day world in terms of particles called atoms and molecules. Macroscopic events—those which humans can observe or experience with their unaided senses—are interpreted by means of microscopic objects—those so small that a special instrument or apparatus must be used to detect them. (Perhaps the term submicroscopic really ought to be used, because most atoms and molecules are much too small to be seen even under a microscope.) In any event, chemists continually try to explain the macroscopic world in microscopic terms.
To get a sense for just how small the atoms we will be working with in the next chapter are, check out this Ted-Ed video called 'Just How Small is an Atom'.
• 2.1: Prelude to Atoms and Reactions
The atomic theory, first proposed in modern form by John Dalton, is one of the most important and useful ideas in chemistry. It interprets observations of the every-day world in terms of particles called atoms and molecules. Macroscopic events—those which humans can observe or experience with their unaided senses—are interpreted by means of microscopic objects—those so small that a special instrument or apparatus must be used to detect them.
• 2.2: Macroscopic Properties and Microscopic Models
As a simple example of how the macroscopic properties of a substance can be explained on a microscopic level, consider the liquid mercury. Macroscopically, mercury at ordinary temperatures is a silvery liquid which can be poured much like water—rather unusual for a metal. Mercury is also the heaviest known liquid. Its density is 13.6 -fold greater than water. When cooled below –38.9°C mercury solidifies and behaves very much like more familiar solid metals such as copper and iron.
• 2.3: The Atomic Theory
• 2.4: Macroscopic and Microscopic Views of a Chemical Reaction
• 2.5: Testing the Atomic Theory
To test a theory, we first use it to make a prediction about the macroscopic world. If the prediction agrees with existing data, the theory passes the test. If it does not, the theory must be discarded or modified. If data are not available, then more research must be done. Eventually the results of new experiments can be compared with the predictions of the theory.
• 2.6: Atomic Weights
The relative masses of the atoms are usually referred to as atomic weights. The atomic-weight scale was originally based on a relative mass of 1 for hydrogen. As more accurate methods for determining atomic weight were devised, it proved convenient to shift to oxygen and then carbon, but the scale was adjusted so that hydrogen’s relative mass remained close to 1. Thus nitrogen’s atomic weight of 14.0067 tells us that a nitrogen atom has about 14 times the mass of a hydrogen atom.
• 2.7: The Amount of Substance- Moles
"How much?" in the above sense of the quantity of atoms or molecules present is not the same thing as "how much" in terms of volume or mass. The International System of Measurements (IUPAC) has a measure of amount that reflects the number of atoms present, and it is called the mole.
• 2.8: The Mole
The very large numbers involved in counting microscopic particles are inconvenient to think about or to write down. Therefore chemists have chosen to count atoms and molecules using a unit called the mole. One mole (abbreviated mol) is $6.022 \times 10^{23}$ of the microscopic particles which make up the substance in question.
• 2.9: The Amount of Substance
In the International System this quantity is called the amount of substance and is given the symbol n.
• 2.10: The Avogadro Constant
To obtain such a pure number, we need a conversion factor which involves the number of particles per unit amount of substance. The appropriate factor is given the symbol $N_A$ and is called the Avogadro constant.
• 2.11: The Molar Mass
It is often convenient to express physical quantities per unit amount of substance (per mole), because in this way equal numbers of atoms or molecules are being compared. Such molar quantities often tell us something about the atoms or molecules themselves.
• 2.12: Formulas and Composition
When a reaction is carried out for the first time, little is known about the microscopic nature of the products. It is therefore necessary to determine experimentally the composition and formula of a newly synthesized substance. One way to approach this involves quantitative analysis—the determination of the percentage by mass of each element in the compound. Such data are usually reported as the percent composition.
• 2.13: Balancing Chemical Equations
Thumbnail: Spinning Buckminsterfullerene ($\ce{C60}$). (CC BY-SA 3.0; unported; Sponk).
02: Atoms Molecules and Chemical Reactions
Two very important things that chemists (and scientists in general) do include making quantitative measurements, and communicating the results of experiments as clearly and unambiguously as possible. We will now deal with another important activity of chemists—the use of their imaginations to devise theories or models to interpret their observations and measurements. Such theories or models are useful in suggesting new observations or experiments that yield additional data. They also serve to summarize existing information and aid in its recall.
The atomic theory, first proposed in modern form by John Dalton, is one of the most important and useful ideas in chemistry. It interprets observations of the every-day world in terms of particles called atoms and molecules. Macroscopic events—those which humans can observe or experience with their unaided senses—are interpreted by means of microscopic objects—those so small that a special instrument or apparatus must be used to detect them. (Perhaps the term submicroscopic really ought to be used, because most atoms and molecules are much too small to be seen even under a microscope.) In any event, chemists continually try to explain the macroscopic world in microscopic terms. The contrasting properties of solids, liquids, and gases, for example, may be ascribed to differences in spacing between and speed of motion of the constituent atoms or molecules. In the form originally proposed by John Dalton(opens in new window), the atomic theory distinguished elements from compounds and was used to explain the law of constant composition and predicted the law of multiple proportions. The theory also agreed with Lavoisier's law of conservation of mass. An important aspect of the atomic theory is the assignment of relative masses (atomic weights) to the elements. Atoms and molecules are extremely small. Therefore, when calculating how much of one substance is required to react with another, chemists use a unit called the mole. One mole contains 6.022 × 1023 of whatever kind of microscopic particles one wishes to consider. Referring to 2 mol Br2 specifies a certain number of Br2 molecules in the same way that referring to 10 gross of pencils specifies a certain number of pencils. The quantity which is measured in the units called moles is known as the amount of substance. The somewhat unusual number 6.022 × 1023, also referred to as the Avogadro Constant, which specifies how many particles are in a mole, has been chosen so that the mass of 1 mol of atoms of any element is the atomic weight of that element expressed in grams. Similarly, the mass of a mole of molecules is the molecular weight expressed in grams. The molecular weight is obtained by summing atomic weights of all atoms in the molecule. This choice for the mole makes it very convenient to obtain molar masses–simply add the units grams per mole to the atomic or molecular weight. Using molar mass and the Avogadro constant, it is possible to determine the masses of individual atoms or molecules and to find how many atoms or molecules are present in a macroscopic sample of matter. A table of atomic weights and the molar masses which can be obtained from it can also be used to obtain the empirical formula of a substance if we know the percentage by weight of each element present. The opposite calculation, determination of weight percent from a chemical formula, is also possible. Once formulas for reactants and products are known, a balanced chemical equation can be written to describe any chemical change. Balancing an equation by adjusting the coefficients applied to each formula depends on the postulate of the atomic theory which states that atoms are neither created, destroyed, nor changed into atoms of another kind during a chemical reaction. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.01%3A_Prelude_to_Atoms_and_Reactions.txt |
As a simple example of how the macroscopic properties of a substance can be explained on a microscopic level, consider the liquid mercury. Macroscopically, mercury at ordinary temperatures is a silvery liquid which can be poured much like water—rather unusual for a metal. Mercury is also the heaviest known liquid. Its density is 13.6 g cm–3, as compared with only 1.0 g cm–3 for water. When cooled below –38.9°C mercury solidifies and behaves very much like more familiar solid metals such as copper and iron. Mercury frozen around the end of a wooden stick can be used to hammer nails, as long as it is kept sufficiently cold. Solid mercury has a density of 14.1 g cm–3 slightly greater than that of the liquid.
When mercury is heated, it remains a liquid until quite a high temperature, finally boiling at 356.6°C to give an invisible vapor. Even at low concentrations gaseous mercury is extremely toxic if breathed into the lungs. It has been responsible for many cases of human poisoning. In other respects mercury vapor behaves much like any other gas. It is easily compressible. Even when quite modest pressures are applied, the volume decreases noticeably. Mercury vapor is also much less dense than the liquid or the solid. At 400°C and ordinary pressures, its density is 3.6 × 10–3 g cm–3 about one four-thousandth that of solid or liquid mercury.
Three states of Matter
The gaseous state The liquid state The solid state
A modern chemist would interpret these macroscopic properties in terms of a microscopic model involving atoms of mercury. As shown in the following figure, the atoms may be thought of as small, hard spheres. Like billiard balls they can move around and bounce off one another. In solid mercury the centers of adjacent atoms are separated by only 300 pm (300 × 10–12 m or 3.00Å). Although each atom can move around a little, the others surround it so closely that it cannot escape its allotted position. Hence the solid is rigid. Very few atoms move out of position even when it strikes a nail. As temperature increases, the atoms vibrate more violently, and eventually the solid melts. In liquid mercury, the regular, geometrically rigid structure is gone and the atoms are free to move about, but they are still rather close together and difficult to separate. This ability of the atoms to move past each other accounts for the fact that liquid mercury can flow and take the shape of its container. Note that the structure of the liquid is not as compact as that of the solid; a few gaps are present. These gaps explain why liquid mercury is less dense than the solid.
In gaseous mercury, also called mercury vapor, the atoms are very much farther apart than in the liquid and they move around quite freely and rapidly. Since there are very few atoms per unit volume, the density is considerably lower than for the liquid and solid. By moving rapidly in all directions, the atoms of mercury (or any other gas for that matter) are able to fill any container in which they are placed. When the atoms hit a wall of the container, they bounce off. This constant bombardment by atoms on the sub-microscopic level accounts for the pressure exerted by the gas on the macroscopic level. The gas can be easily compressed because there is plenty of open space between the atoms. Reducing the volume merely reduces that empty space. The liquid and the solid are not nearly so easy to compress because there is little or no empty space between the atoms.
You may have noticed that although our sub-microscopic model can explain many of the properties of solid, liquid, and gaseous mercury, it cannot explain all of them. Mercury’s silvery color and why the vapor is poisonous remain a mystery, for example. There are two approaches to such a situation. We might discard the idea of atoms in favor of a different theory that can explain more macroscopic properties. On the other hand it may be reasonable to extend the atomic theory so that it can account for more facts. The second approach has been followed by chemists. In the current section on Atoms, Molecules and Chemical Reactions as well as Using Chemical Equations in Calculations we shall discuss in more detail those facts that require only a simple atomic theory for their interpretation. Many of the subsequent sections will describe extensions of the atomic theory that allow interpretations of far more observations. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.02%3A_Macroscopic_Properties_and_Microscopic_Models.txt |
The development of the atomic theory owes much to the work of two men: Antoine Lavoisier, who did not himself think of matter in terms of atoms but whose work laid organization groundwork for thinking about elements, and John Dalton, to whom the atomic theory is attributed. Much of Lavoisier’s work as a chemist was devoted to the study of combustion. He became convinced that when a substance is burned in air, it combines with some component of the air. Eventually he realized that this component was the dephlogisticated air which had been discovered by Joseph Priestly (1733 to 1804) a few years earlier. Lavoisier renamed this substance oxygen. In an important series of experiments he showed that when mercury is heated in oxygen at a moderate temperature, a red substance, calx of mercury, is obtained. (A calx is the ash left when a substance burns in air.) At a higher temperature this calx decomposes into mercury and oxygen. Lavoisier’s careful experiments also revealed that the combined masses of mercury and oxygen were exactly equal to the mass of calx of mercury. That is, there was no change in mass upon formation or decomposition of the calx. Lavoisier hypothesized that this should be true of all chemical changes, and further experiments showed that he was right. This principle is now called the law of conservation of mass.
As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an element—an ultimately simple substance which could not be decomposed by chemical changes.
Lavoisier did not originate the idea that certain substances (elements) were fundamental and all others could be derived from them. This had first been proposed in Greece during the fifth century B.C. by Empedocles, who speculated that all matter consisted of combinations of earth, air, fire, and water. These ideas were further developed and taught by Aristotle and remained influential for 2000 years.
Lavoisier did produce the first table of the elements which contained a large number of substances that modern chemists would agree should be classifies as elements. He published it with the knowledge that further research might succeed decomposing some of the substances listed, thus showing them not to be elements. One of his objectives was to prod his contemporaries into just that kind of research. Sure enough the “earth substances” listed at the bottom were eventually shown to be combinations of certain metals with oxygen. It is also interesting to note that not even Lavoisier could entirely escape from Aristotle’s influence. The second element in his list is Aristotle’s “fire,” which Lavoisier called “caloric,” and which we now call “heat.” Both heat and light, the first two items in the table, are now regarded as forms of energy rather than of matter.
Although his table of elements was incomplete, and even incorrect in some instances, Lavoisier’s work represented a major step forward. By classifying certain substances as elements, he stimulated much additional chemical research and brought order and structure to the subject where none had existed before. His contemporaries accepted his ideas very readily, and he became known as the father of chemistry.
John Dalton (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it!
The postulates of the atomic theory are given below. The first is no advance on the ancient Greek philosopher Democritus who had theorized almost 2000 years earlier that matter consists of very small particles.
The Postulates of Dalton's Atomic Theory
1. All matter is composed of a very large number of very small particles called atoms.
2. For a given element, all atoms are identical in all respects. In particular all atoms of the same element have the same constant mass, while atoms of different elements have different masses.
3. The atoms are the units of chemical changes. Chemical reactions involve the combination, separation, or rearrangement of atoms, but atoms are neither created, destroyed, divided into parts, or converted into atoms of any other kind.
4. Atoms combine to form molecules in fixed ratios of small whole numbers.
The second postulate, however, shows the mark of an original genius; here Dalton links the idea of atom to the idea of element. Lavoisier’s criterion for an element had been essentially a macroscopic, experimental one. If a substance could not be decomposed chemically, then it was probably an element. By contrast, Dalton defines an element in theoretical, sub-microscopic terms. An element is an element because all its atoms are the same. Different elements have different atoms. There are just as many different kinds of elements as there are different kinds of atoms.
Now look back a moment to the physical states of mercury, where sub-microscopic pictures of solid, liquid, and gaseous mercury were given. Applying Dalton’s second postulate to this figure, you can immediately conclude that mercury is an element, because only one kind of atom appears. Although mercury atoms are drawn as spheres in the figure, it would be more common today to represent them using chemical symbols. The chemical symbol for an element (or an atom of that element) is a one- or two-letter abbreviation of its name. Usually, but not always, the first two letters are used. To complicate matters further, chemical symbols are sometimes derived from a language other than English. For example the symbol for Hg for mercury comes from the first and seventh letters of the element’s Latin name, hydrargyrum.
Table \(1\): Names, Chemical Symbols, and Atomic Weights of the Element
Name Symbol Atomic Number Atomic Weight Name Symbol Atomic Number Atomic Weight
Actinium2 Ac 89 (227) Molybdenum Mo 42 95.96(2)
Aluminum Al 13 26.981 5386(8) Neodymium Nd 60 144.242(3)
Americium2 Am 95 (243) Neon Ne 10 20.1797(6)
Antimony Sb 51 121.760(1) Neptunium2 Np 93 (237)
Argon Ar 18 39.948(1) Nickel Ni 28 58.6934(4)
Arsenic As 33 74.92160(2) Niobium Nb 41 92.90638(2)
Astatine2 At 85 (210) Nitrogen N 7 14.0067(2)
Barium Ba 56 137.327(7) Nobelium2 No 102 (259)
Berkelium2 Bk 97 (247) Osmium Os 76 190.23(3)
Beryllium Be 4 9.012182(3) Oxygen O 8 15.9994(3)
Bismuth Bi 83 208.98040(1) Palladium Pd 46 106.42(1)
Bohrium2 Bh 107 (272) Phosphorus P 15 30.973762(2)
Boron B 5 10.811(7) Platinum Pt 78 195.084(9)
Bromine Br 35 79.904(1) Plutonium2 Pu 94 (244)
Cadmium Cd 48 112.411(8) Polonium2 Po 84 (209)
Calcium Ca 20 40.078(4) Potassium K 19 39.0983(1)
Californium2 Cf 98 (251) Praseodymium Pr 59 140.90765(2)
Carbon C 6 12.0107(8) Promethium2 Pm 61 (145)
Cerium Ce 58 140.116(1) Protactinium2 Pa 91 231.03588(2)
Cesium Cs 55 132.9054519(2) Radium2 Ra 88 (226)
Chlorine Cl 17 35.453(2) Radon2 Rn 86 (222)
Chromium Cr 24 51.9961(6) Rhenium Re 75 186.207(1)
Cobalt Co 27 58.933195(5) Rhodium Rh 45 102.90550(2)
Copper Cu 29 63.546(3) Roentgenium2 Rg 111 (280)
Curium2 Cm 96 (247) Rubidium Rb 37 85.4678(3)
Darmstadtium2 Ds 110 (281) Ruthenium Ru 44 101.07(2)
Dubnium2 Db 105 (268) Rutherfordium2 Rf 104 (267)
Dysprosium Dy 66 162.500(1) Samarium Sm 62 150.36(2)
Einsteinium2 Es 99 (252) Scandium Sc 21 44.955912(6)
Erbium Er 68 167.259(3) Seaborgium2 Sg 106 (271)
Europium Eu 63 151.964(1) Selenium Se 34 78.96(3)
Fermium2 Fm 100 (257) Silicon Si 14 28.0855(3)
Fluorine F 9 18.9984032(5) Silver Ag 47 107.8682(2)
Francium2 Fr 87 (223) Sodium Na 11 22.98976928(2)
Gadolinium Gd 64 157.25(3) Strontium Sr 38 87.62(1)
Gallium Ga 31 69.723(1) Sulfur S 16 32.065(5)
Germanium Ge 32 72.64(1) Tantalum Ta 73 180.94788(2)
Gold Au 79 196.966569(4) Technetium2 Tc 43 (98)
Hafnium Hf 72 178.49(2) Tellurium Te 52 127.60(3)
Hassium2 Hs 108 (277) Terbium Tb 65 158.92535(2)
Helium He 2 4.002602(2) Thallium Tl 81 204.3833(2)
Holmium Ho 67 164.93032(2) Thorium2 Th 90 232.03806(2)
Hydrogen H 1 1.00794(7) Thulium Tm 69 168.93421(2)
Indium In 49 114.818(3) Tin Sn 50 118.710(7)
Iodine I 53 126.90447(3) Titanium Ti 22 47.867(1)
Iridium Ir 49 192.217(3) Tungsten W 74 183.84(1)
Iron Fe 26 55.845(2) Uranium2 U 92 238.02891(3)
Krypton Kr 36 83.798(2) Vanadium V 23 50.9415(1)
Lanthanum La 57 138.90547(7) Xenon Xe 54 131.293(6)
Lawrencium2 Lr 103 (262) Ytterbium Yb 70 173.054(5)
Lead Pb 82 207.2(1) Yttrium Y 39 88.90585(2)
Lithium Li 3 [6.941(2)]1 Zinc Zn 30 65.38(2)
Lutetium Lu 71 174.9668(1) Zirconium Zr 40 91.224(2)
Magnesium Mg 12 24.3050(6) -2,3,4 112 (285)
Manganese Mn 25 54.938045(5) -2,3 113 (284)
Meitnerium2 Mt 109 (276) - 2,3 114 (287)
Mendelevium2 Md 101 (258) -2,3 115 (288)
Mercury Hg 80 200.59(2) -2,3 116 (293)
-2,3 118 (294)
The chemical symbols for all the currently known elements are listed above in the table, which also includes atomic weights. These symbols are the basic vocabulary of chemistry because the atoms they represent make up all matter. You will see symbols for the more important elements over and over again, and the sooner you know what element they stand for, the easier it will be for you to learn chemistry. These more important element have been indicated in the above table by colored shading around their names.
Dalton’s fourth postulate states that atoms may combine to form molecules. An example of this is provided by bromine, the only element other than mercury which is a liquid at ordinary room temperature (20°C). Macroscopically, bromine consists of dark-colored crystals below –7.2°C and a reddish brown gas above 58.8°C. The liquid is dark red-brown and has a pungent odor similar to the chlorine used in swimming pools. It can cause severe burns on human skin and should not be handled without the protection of rubber gloves.
Three states of Matter
(a) in the gaseous state (b) as a liquid (c) in solid form
Figure \(1\): Sub-microscopic view of the diatomic molecules of the element bromine (a) in the gaseous state (above 58°C); (b) in liquid form (between -7.2 and 58.8°C); and (c) in solid form (below -7.2°C).
The sub-microscopic view of bromine in the following figure is in agreement with its designation as an element—only one kind of atom is present. Except at very high temperatures, though, bromine atoms always double up. Whether in solid, liquid, or gas, they go around in pairs. Such a tightly held combination of two or more atoms is called a molecule
The composition of a molecule is indicated by a chemical formula. A subscript to the right of the symbol for each element tells how many atoms of that element are in the molecule. For example, the atomic weights table gives the chemical symbol Br for bromine, but each molecule contains two bromine atoms, and so the chemical formula is Br2. According to Dalton’s fourth postulate, atoms combine in the ratio of small whole numbers, and so the subscripts in a formula should be small whole numbers.
2.03: The Atomic Theory
The Atomic Theory in Foods
Introduction: Minerals, Elements, and Compounds
Nutrition books are often confusing when they discuss human nutrient requirements. For example, one, in discussing essential minerals in sports nutrition[1] says "Minerals are elements". But only one common essential mineral, iron (symbolized Fe for the Latin "Ferrum") is commonly ingested as an element. Iron is an element because it is made up of a single kind of atom, designated by the symbol Fe. Other symbols are given in the Table below. Iron metal particles are actually found in breakfast cereals[2], and the easiest way to prove it is to float a cereal flake on water and pull it around with a magnet. You can also see it extracted in several YouTube videos like this one. Sometimes iron is added as a compound, like "ferrous sulfate", FeSO4. A compound has atoms of several elements bonded together is specific ratios. The ratios are given by the subscripts in the formula; in this case, ferrous sulfate is made from 1 atom of iron, 1 of the element sulfur, and 4 atoms of the element oxygen.
Another essential mineral that is commony misidentified is "iodine". The nutrient we actually ingest is not the element iodine, but a compound of iodine like potassium iodide, KI (K is the symbol for potassium, after the German Kallium).
While the mineral KI looks and tastes like table salt, the element iodine is a dark purple (almost black) solid, sometimes dissolved in alcohol to give a brown solution used to disinfect minor wounds. It's a toxic element with the formula I2<, showing that an element may be made up of molecules which contain two or more atoms bonded together, as long as the atoms are the same.
When iodine crystals are heated, they "sublime" (turn directly to a gas) to give purple vapors composed of diatomic molecules. The liquid phase, which also contains diatomic molecules, only forms under higher than atmospheric pressure. All the phases are the element iodine, because they contain only I atoms. (See Bromine as an example)
Other essential minerals are referred to as "elements", but are actually either toxic or useless to our body as elements. Foods must contain nutrients in chemical compounds that are easily digested and absorbed in our bodies in order for them to be nutritious. For example, the element phosphorus may have several forms. One, P4, is very toxic and actually burns when exposed to air, making a glow that's responsible for its name.
The nutrient form of phosporus is illustrated by mineral phosphates, like Ca3(PO4)2, which contains 3 calcium atoms (we will see later, they're in the form of "ions" each with a 2+ charge, Ca2+) and 2 "phosphate ions", each made of 1 phosphorus atom bonded to 4 oxygen atoms, and having an overall 3- charge. Phosphorus is also supplied in biomolecules found in meats and vegetables.
"Elemental Diets"
An "Elemental diet" is a solution of nutrients that can be administered intravenously (or with a gastric feeding tube) for people with digestive disorders, like Crohn's disease or colitis. There are no elements in an elemental diet, but the simplest chemical compounds that can provide complete (or near complete) nutrition. A typical elemental diet shows us what is required to maintain our bodies.[3] We require ounces to pounds of macronutrients daily, but typically less than 5 g (.2 oz) of micronutrients.
• Macronutrients (needed for energy and construction of body parts)
• Carbohydrates: Corn Syrup Solids, Mono and Diglycerides, Esters of Mono and Diglycerides
• Essential lipids:Fractionated Coconut Oil, Canola Oil, High Oleic Safflower Oil,
• Amino Acids: (L-Glutamine), L-Isoleucine, L-Leucine, L-Lysine-L-Aspartate, N-Acetyl-L-Methionine, L-Phenylalanine, L-Threonine, (L-Proline), L-Tryptophan, (Glycine), L-Histidine, L-Arginine, L-Valine, (L-Alanine), (L-Serine), (L-Tyrosine), (L-Cystine), (Taurine, a sulfonic acid).
• Micronutrients (used mostly in regulators of body processes)
• Minerals: Tripotassium Citrate, Tricalcium Phosphate, Dicalcium Phosphate, Sodium Chloride, Magnesium Acetate, Ferrous Sulfate, Zinc Sulfate, Manganese Sulfate, Cupric Sulfate, Potassium Iodide, Chromium Chloride, Sodium Molybdate, Sodium Selenite, (missing: F, V, B, Sn, Ni).
• Vitamins: Choline Bitartrate (B), (M-Inositol), Niacinamide (B3), Calcium-D-Pantothenate (B6), Thiamine Chloride Hydrochloride (B1), Pyridoxine Hydrochloride (B6), Riboflavin (B2), Folic Acid (B9), Cyanocobalamin (B12), D-Biotin (D7), Vitamin D3, L-Ascorbic Acid (C), DL-Alpha Tocopherol Acetate (E), Vitamin A Acetate, Phylloquinone (K1).
• Other
• Nutritional Supplement: L-Carnitine (normally synthesized from amino acids).
• Emulsifier, thickeners: Diacetyl Tartaric Acid, Propylene Glycol Alginate.
The ratio of chemical elements in our bodies is sometimes presented as a "formula for a human being", which does not represent any actual chemical compound, but just tells us the relative numbers of atoms:
$\ce{H_{375,000,000}O_{132,000,000}C_{85,700,000}N_{6,430,000}Ca_{1,500,000}P_{1,020,000}S_{206,000}Na_{183,000}K_{177,000}Cl_{127,000}Mg_{40,000}Si_{38, 600}Fe_{2,680}Zn_{2,110}Cu_{76}I_{14}Mn_{13}F_{13}Cr_{7}Se_{4}Mo_{3}Co_{1}} \nonumber$
So for just 1 cobalt (Co) atom, we need 2,680 iron atoms, 1,500,000 calcium (Ca) atoms, etc. Clearly, there's a lot that chemistry can tell us about nutrition. But how do we know what is an "element" and what is a "compound", and what compounds provide the elements we need to maintain our bodies?
How Do We Know What Substances Contain Essential Elements?
The early greek philosophers (Empedocles, Lucretius and Democritus) proposed that everything was made of atoms, but had no way of providing evidence for the claim. Evidence that each chemical element is composed of one kind of atom (that is unchanged during chemical reactions) finally developed between 1750 and 1850. Looking at the pictures of potassium iodide (KI) and iodine (I2) above, will help to imagine how implausible it must have seemed that they share a common iodine atom.
Evidence for unchanging atoms as the components of elements came from weight measurements. Joseph Priestley and Antoine Lavoisier finally correctly interpreted the loss of weight when some substances burn (wood or sugar) and the gain in weight of others, by recognizing the fact that the mass of atoms in gases had been neglected. Lavoisier showed that mercury (Hg) gains weight because it combines with oxygen molecules from air to make solid mercuric oxide:
$\ce{2 Hg + O2 → 2 HgO.} \nonumber$
But wood loses weight because it is converted to gaseous molecules of carbon dioxide and water of equal mass. The combustion reaction is similar to the combusion of the sugar, glucose, which the basis for the metabolism of carbohydrates in our bodies to provide energy:
$\ce{C6H12O6 + 6 O2 → 6 CO2 + 6 H2O} \nonumber$
Earlier, van Helmont[5] had failed to recognize that incorporation of carbon dioxide gas contributed the most to the weight gain of trees; he thought it was the water, because the soil showed virtually no change in mass. Nonetheless, he showed committment to the idea of concervation of mass, that no mass should be lost or gained during a chemical change. The equation for photosynthesis is just the reverse of the combustion equation above,
\begin{align*} \ce{6 CO2} + \ce{6 H2O} &\rightarrow \ce{C6H12O6} + \ce{6 O2} \[4pt] \text{carbon dioxide} + \text{water} + \text{light energy} &\rightarrow \text{carbohydrate} + \text{oxygen} \end{align*} \nonumber
so we could say that trees are mostly carbohydrate, and that when they burn, they release the energy that they had absorbed from the sun while growing:
As Lavoisier continued his experiments with oxygen, he noticed something else. Although oxygen combined with many other substances, it never behaved as though it were itself a combination of other substances. Lavoisier was able to decompose the red calx into mercury and oxygen, but he could find no way to break down oxygen into two or more new substances. Because of this he suggested that oxygen must be an element—an ultimately simple substance which could not be decomposed by chemical changes.
This was the fundamental discovery that allows us to identify tricalcium phosphate as a good source of "phosphorus" or potassium iodide as a good source of "iodine".
John Dalton[1] (1766 to 1844) was a generation younger than Lavoisier and different from him in almost every respect. Dalton came from a working class family and only attended elementary school. Apart from this, he was entirely self-taught. Even after he became famous, he never aspired beyond a modest bachelor’s existence in which he supported himself by teaching mathematics to private pupils. Dalton made many contributions to science, and he seems not to have realized that his atomic theory was the most important of them. In his “New System of Chemical Philosophy” published in 1808, only the last seven pages out of a total of 168 are devoted to it! | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.03%3A_The_Atomic_Theory/2.3.01%3A_Foods-_Elemental_Diets.txt |
Dalton’s third postulate(opens in new window) states that atoms are the units of chemical changes. What this means can be seen in the macroscopic and microscopic views of a chemical change in the video and figure farther down this page. The video shows that when macroscopic quantities of mercury and bromine are mixed at room temperature, a chemical reaction occurs, and a new substance, Mercury(II) bromide (mercuric bromide), is produced. Mercury(II) bromide is a white solid, quite different in appearance from the two elements from which it was formed.
A chemist’s sub-microscopic (nanoscale or atomic scale) theory about what is going on is shown below the video of the macroscopic reaction. Soon after the two liquids are mixed together, a rearrangement of atoms begins. The two bromine atoms of each Br2 molecule become separated and combine instead with mercury atoms. When the chemical reaction is complete, all that remains is a collection of mercury(II) bromide molecules, each of which contains one mercury atom and two bromine atoms. Notice that there are just as many mercury atoms after the reaction as there were before the reaction. The same applies to bromine atoms. Atoms were neither created, destroyed, divided into parts, or changed into other kinds of atoms during the chemical reaction.
The view of solid mercury(II) bromide shown in part f is our first sub-microscopic example of a compound. Each molecule of a compound is made up of two (or more) different kinds of atoms. Since these atoms may be rearranged during a chemical reaction, the molecules may be broken apart and the compound can be decomposed into two (or more) different elements.
The formula for a compound involves at least two chemical symbols—one for each element, present. In the case of mercury(II) bromide each molecule contains one mercury atom and two bromine atoms, and so the formula is HgBr2. Both part f of the figure and the formula tell you that any sample of pure mercury(II) bromide contains twice as many bromine atoms as mercury atoms. This 2:1 ratio agrees with Dalton’s fourth postulate that atoms combine in the ratio of small whole numbers.
Although John Dalton originally used circular symbols like those in the figure to represent atoms in chemical reactions, a modern chemist would use chemical symbols and a chemical equation like:
$\underset{\text{Reactants}}{\ce{Hg + Br2}} \rightarrow \underset{\text{Products}}{\ce{HgBr2}} \label{1}$
This equation may be interpreted at the sub-microscopic scale to mean that 1 mercury atom and 1 bromine molecule react to form 1 mercury(II) bromide molecule. It should also call to mind the macroscopic change shown in the video, in which a silvery liquid and a red-brown liquid change into a white solid. This macroscopic interpretation is often strengthened by specifying physical states of the reactants and products:
$\text{Hg}(l) + \text{Br}_2 (l) \rightarrow \text{HgBr}_2 (s) \label{2}$
Thus liquid mercury and liquid bromine react to form solid mercuric bromide. [If the bromine had been in gaseous form, Br2(g) might have been written as a product of the reaction. Occasionally (c) may be used instead of (s) to indicate a crystalline solid.] Chemical equations such as $\ref{1}$ and $\ref{2}$ summarize a great deal of information on both macroscopic and sub-microscopic levels, for those who know how to interpret them.
Macroscopic and sub-microscopic descriptions of a chemical reaction.
Figure $1$ Macroscopic and sub-microscopic descriptions of a chemical reaction. On the macroscopic level a silvery liquid, mercury, is mixed with a red-brown liquid, bromine, and white crystals are produced, as shown in the video above. On the sub-microscopic (nanoscale) level Hg atoms combine with Br2 molecules to form HgBr2 molecules packed together in a regular array. These images are below in sequential order.
2.04: Macroscopic and Microscopic Views of a Chemical Reaction
In a previous section on Elemental diet we looked at essential minerals that are sometimes called "elements", but are really only usable to our bodies when they are supplied as "compounds". Both potassium and iodine are essential nutrients, but each must be supplied in a compound if it is to be of any use (let alone nontoxic) to the body. Iodine deficiency leads to thyroid dysfunction and goiter, but very little iodine is required, so goiter is rare.[1] Potassium deficiency (hypokalemia, after kallium, the German name for potassium and source of the symbol, K) leads to muscle weakness, cramps, and constipation [2]. The Recommended Daily Allowances (RDA) for iodine and potassium are 150 µg and 4700 mg respectively, so we need about 30,000 times as much potassium as iodine. Luckily, most foods (especially oranges, potatos, and bananas) supply potassium, and KI is only necessary as therapy.
Iodine deficiency ranges from <50 per 100,000 individuals (yellow) to over 800 (red)[3]
Goiter
As we saw previously, the element iodine I2 is a toxic, purple crystalline solid, while the nutrient must be supplied as part of a compound, like potassium iodide (KI). Iodine has a melting point of just 113 oC, where it also vaporizes significantly, and it boils at just 184.3 oC. Iodine has a high density of 4.93 g/cc. We'll see below that KI has entirely different properties.
Similarly, the element potassium (K) is a metal with the lowest density (0.89 g/cc) of any metal except lithium. It cuts like butter, has a melting point of only 63.38 oC and a boiling point of 759 oC. In the solid, K atoms are arranged in the lattice shown below. Potassium reacts explosively with water as shown in the video below or on YouTube
Body centered cubic jmol: SID10534500
Because of the toxicity and reactivity of the element potassium, nutrient sources must contain the potassium in a compound like potassium iodide, KI, where it has entirely different properties. Potassium iodide is a white crystalline solid that looks and tastes like table salt (NaCl), dissolves in water, has a melting point of 631 oC, and a boiling point of 1330 oC. It's used, along with sodium iodide (NaI) to "iodize" salt.
Potassium metal, K (the element)
Iodine, I2 (the element))
Potassium iodide, KI (the mineral nutrient)
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Violet, solid iodine (I2)
White crystalline potassium iodide
2 K (s) + I2 (s) → 2 KI (c)
Reactants → Products
Here we have three views of the same reaction: At the top, the macroscopic appearance of the reactants and products; below them, the microscopic or atomic level representation of the atoms or molecules; and finally, a symbolic represenation in the form of a chemical equation. The solid state of the reactants is indicated by the "s" in parentheses, and the crystalline state of the product is indicated by "(c)". Liquids are designated by (l), gases by (g), and aqueous (water) solutions as (aq).
This equation may be interpreted microscopically to mean that 1 potassium atom and 1 iodine molecule react to form 2 potassium iodide units, but there is no such thing as an isolated KI unit. Rather, there's an extended lattice of alternating K1+ ions and I1- ions.
When the elements potassium and iodine are heated together, they react even more energetically than potassium and water, to give the compound KI according to the equation above.
This illustrates Dalton’s third postulate, which states that atoms are the units of chemical changes. Notice that there are just as many potassium atoms after the reaction as there were before the reaction. The same applies to iodine atoms. Atoms were neither created, destroyed, divided into parts, or changed into other kinds of atoms during the chemical reaction. The balanced chemical equation reinforces this idea.
The view of solid KI shown above is our first microscopic example of a compound. A compound is made up of two (or more) different kinds of atoms. Since these atoms may be rearranged during a chemical reaction, the compound can be decomposed into two (or more) different elements. The 1:1 ratio of K atoms to I atoms implied by the formula KI (subscript "1"s are assumed, so KI = K1I1) agrees with Dalton’s fourth postulate that atoms combine in the ratio of small whole numbers. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.04%3A_Macroscopic_and_Microscopic_Views_of_a_Chemical_Reaction/2.4.01%3A_Foods-_The_Mineral_Nutrients_Potassium_and_Iodine.txt |
Two criteria are usually applied to any theory. First, does it agree with facts which are already known? Second, does it predict new relationships and stimulate additional observation and experimentation? Dalton’s atomic theory was able to do both of these things. It was especially useful in dealing with data regarding the masses of different elements which were involved in chemical compounds or chemical reactions.
To test a theory, we first use it to make a prediction about the macroscopic world. If the prediction agrees with existing data, the theory passes the test. If it does not, the theory must be discarded or modified. If data are not available, then more research must be done. Eventually the results of new experiments can be compared with the predictions of the theory.
Several examples of this process of testing a theory against the facts are afforded by Dalton’s work. For example, postulate 3 in Dalton’s Atomic Theory(opens in new window) states that atoms are not created, destroyed, or changed in a chemical reaction. Postulate 2 says that atoms of a given element have a characteristic mass: By logical deduction, then, equal numbers of each type of atom must appear on left and right sides of chemical equations such as
$\text{Hg}(l) + \text{Br}_2(l) \rightarrow \text{HgBr}_2(s) \label{1}$
and the total mass of reactants must equal the total mass of products. Dalton’s atomic theory predicts Lavoisier’s experimental law of conservation of mass.
A second prediction of the atomic theory is a bit more complex. A compound is a definite number of two or more types of atom. No matter how, when, or where a compound is made, it must always have the same ratios of different atoms. Thus mercuric bromide molecules always have the formula HgBr2 no matter how much we have or where the compound came from, there will always be twice as many bromine atoms as mercury atoms. Since each type of atom has a characteristic mass, the mass of one element which combines with a fixed mass of the other should always be the same. In mercuric bromide, for example, if each mercury atom is 2.510 times as heavy as a bromine atom, the ratio of masses would be
$\dfrac{\text{Mass of 1 mercury atom}}{\text{Mass of 2 bromine atoms}} = \dfrac{2.510 \times \color{red}\cancel{\color{black}\text{mass of 1 bromine atom}}}{2 \times \color{red}\cancel{\color{black}\text{mass of 1 bromine atom}}} = 1.255 \nonumber$
No matter how many mercury (II) bromide molecules we have, each has the same proportion of mercury, and so any sample of mercury (II) bromide must have that same proportion of mercury. We have just derived the law of constant composition, sometimes called the law of definite proportions. When elements combine to form a compound, they always do so in exactly the same ratio of masses. This law had been postulated in 1799 by the French chemist Proust (1754 to 1826) 4 years before Dalton proposed the atomic theory, and its logical derivation from the theory contributed to the latter’s acceptance. The law of constant composition makes the important point that the composition and other properties of a pure compound are independent of who prepared it or where it came from. The carbon dioxide found on Mars, for instance, can be expected to have the same composition as that on Earth, while the natural vitamin C extracted and purified from rose hips has exactly the same composition as the synthetic vitamin C prepared by a drug company. Absolute purity is, however, an ideal limit which we can only approach, and the properties of many substances may be affected by the presence of very small quantities of impurities.
A third law of chemical composition may be deduced from the atomic theory. It involves the situation where two elements can combine in more than one way, forming more than one compound. For example, if mercury (II) bromide is ground and thoroughly mixed with liquid mercury, a new compound, mercury (I) bromide (mercurous bromide), is formed. Mercury (I) bromide is a white solid which is distinguishable from mercury (II) bromide because of its insolubility in hot or cold water. Mercury (I) bromide also changes directly from a solid to a gas at 345°C. From the microscopic view of solid mercury (I) bromide in the above figure , you can readily determine that its chemical formula is $Hg_2Br_2$. (Since there are two atoms of each kind in the molecule, it would be incorrect to write the formula as HgBr.) The chemical equation for synthesis of mercurous bromide is
$\text{Hg} (l) + \text{HgBr}_2 (s) \rightarrow \text{Hg}_2\text{Br}_2 (s) \label{2}$
From the formulas HgBr2 and Hg2Br2 (and the image above) we can see that mercury (II) bromide has only 1 mercury atom for every 2 bromines, while mercury (I) bromide has 2 mercury atoms for every 2 bromines. Thus, for a given number of bromine atoms, mercury (I) bromide will always have twice as many mercury atoms as mercury (II) bromide. Again using postulate 2 from Dalton’s Atomic Theory, the atoms have characteristic masses, and so a given number of bromine atoms corresponds to a fixed mass of bromine. Twice as many mercury atoms correspond to twice the mass of mercury.
Therefore we can say that for a given mass of bromine, mercury (I) bromide will contain half the mass of mercury that mercury (II) bromide will (the doubled mass of mercury was provided by adding liquid mercury to mercuric bromide in Equation $\ref{2}$).
Example $1$
Given that the mass of a mercury atom is 2.510 times the mass of a bromine atom, calculate the mass ratio of mercury to bromine in mercury (I) bromide.
Solution
The formula $Hg_2Br_2$ tells us that there are 2 mercury atoms and 2 bromine atoms in each molecule. Thus the mass ratio is
$\dfrac{\text{Mass of 2 mercury atoms}}{\text{Mass of 2 bromide atoms}} = \dfrac{\text{2}\times(\text{2.510}\times\text{Mass of 1 bromide atom})}{\text{2}\times\text{Mass of 1 bromide atom}}=\text{2.510}\nonumber$
Note that the mass of mercury per unit mass of bromine is double that calculated earlier for mercury (II) bromide.
The reasoning and calculations above illustrate the law of multiple proportions. When two elements form several compounds, the mass ratio in one compound will be a small whole-number multiple of the mass ratio in another. In the case of mercury (II) bromide and mercury (I) bromide the mass ratios of mercury to bromine are 1.255 and 2.510, respectively. The second value is a small whole-number multiple of (2 times) the first.
Until the atomic theory was proposed, no one had expected any relationship to exist between mass ratios in two or more compounds containing the same elements. Because the theory predicted such relationships, Dalton and other chemists began to look for them. Before long, a great deal of experimental evidence was accumulated to show that the law of multiple proportions was valid. Thus the atomic theory was able to account for previously known facts and laws, and it also predicted a new law. In the process of verifying that prediction, Dalton and his contemporaries did many additional quantitative experiments. These led onward to more facts, more laws, and, eventually, new or modified theories. This characteristic of stimulating more research and thought put Dalton’s postulates in the distinguished company of other good scientific theories.
2.05: Testing the Atomic Theory
Two criteria are usually applied to any theory. First, does it agree with facts which are already known? Second, does it predict new relationships and stimulate additional observation and experimentation? Dalton’s atomic theory was able to do both of these things.It was especially useful in dealing with data regarding the masses of different elements which were involved in chemical compounds or chemical reactions.
To test a theory, we first use it to make a prediction about the macroscopic world. If the prediction agrees with existing data, the theory passes the test. If it does not, the theory must be discarded or modified. If data are not available, then more research must be done. Eventually the results of new experiments can be compared with the predictions of the theory. One hallmark of a scientific theory is that it suggests tests that may falsify the theory. Non-scientific theories do not. For example, Intelligent Design (ID) theory proposes that the world was created by an intelligent designer, but proposes no tests of that claim. It's a theory, but not a scientific theory.
Several examples of this process of testing a theory against the facts are afforded by Dalton’s work. For example, postulate 3 in Dalton’s Atomic Theory states that atoms are not created, destroyed, or changed in a chemical reaction. Postulate 2 says that atoms of a given element have a characteristic mass: By logical deduction, then, equal numbers of each type of atom must appear on left and right sides of chemical equations such as
O2(g) + 2 H2(g) → 2 H2O(l) (1)
and the total mass of reactants must equal the total mass of products. Dalton’s atomic theory predicts Lavoisier’s experimental law of conservation of mass.
A second prediction of the atomic theory is a bit more complex. A compound is made up of molecules, each of which contains a certain number of each type of atom. No matter how, when, or where a compound is made, its molecules will always be the same. Thus water molecules always have the formula H2O. No matter how much we have or where the compound came from, there will always be twice as many hydrogen atoms as oxygen atoms. Since each type of atom has a characteristic mass, the mass of one element which combines with a fixed mass of the other should always be the same. In water, for example, if each oxygen atom is 15.873 times as heavy as a hydrogen atom, the ratio of masses would be
$\dfrac{\text{mass of 1 O atom}}{\text{mass of 2 H atoms}}= \dfrac{\text{15.873 x mass of 1 H atom}}{\text{2 x mass of 1 H atom}} \nonumber$
and cancelling "mass of 1 H atom" in the numerator and denominator we get
$\dfrac{\text {15.873}}{2} = \dfrac{7.937}{1} \nonumber$
No matter how many water molecules we have, each has the same proportion of oxygen, and so any sample of water must have 7.937 times as much oxygen as hydrogen. We have just derived the law of constant composition, sometimes called the law of definite proportions. When elements combine to form a compound, they always do so in exactly the same ratio of masses. This law had been postulated in 1799 by the French chemist Proust (1754 to 1826) four years before Dalton proposed the atomic theory, and its logical derivation from the theory contributed to the latter’s acceptance. The law of constant composition makes the important point that the composition and other properties of a pure compound are independent of who prepared it or where it came from. The carbon dioxide found on Mars, for instance, can be expected to have the same composition as that on Earth, while the natural vitamin C extracted and purified from rose hips has exactly the same composition as the synthetic vitamin C prepared by a drug company. Absolute purity is, however, an ideal limit which we can only approach, and the properties of many substances may be affected by the presence of very small quantities of impurities.
A third law of chemical composition may be deduced from the atomic theory. It involves the situation where two elements can combine in more than one way, forming more than one compound. For example, hydrogen and oxygen form another compound, hydrogen peroxide, modeled here:
This is a "Jmol" model. If you place the mouse pointer on the molecule, hold down the left mouse key, and move the mouse, you can rotate the model to get a 3D perspective. The oxygen atoms are red, hydrogen gray.
Hydrogen peroxide is a pale blue liquid that freezes just 0.4oC below 0o, boils at 150.2oC, and is slightly more viscous than water. It is used in 3% water solution as an antiseptic (it's used to disinfect biological safety cabinets) and bleaching agent. Pure hydrogen peroxide is a dangerous oxidizer, used in rocket engines, which burns skin even in 10% water solutions. Surprisingly, is naturally produced in organisms as a byproduct of oxygen metabolism. Nearly all living things possess enzymes known as peroxidases, which very rapidly catalytically decompose low concentrations of hydrogen peroxide to water and oxygen.
From the molecular model of hydrogen peroxide, you can readily see that its chemical formula is H2O2. (Since there are two atoms of each kind in the molecule, it would be incorrect to write the formula as HO). Hydrogen peroxide cannot be synthesized directly by the reaction of hydrogen and oxygen, and it decomposes to water and oxygen:
2 H2O2(aq) → 2 H2O(l) + O2(g) (2)
From the formulas H2O and H2O2 we can see that water has only 1 oxygen atom for every 2 hydrogens, while hydrogen peroxide has 2 oxygen atoms for every 2 hydrogens. Thus, for a given number of bromine atoms, hydrogen peroxide will always have twice as many oxygen atoms as water. Again using postulate 2 from Dalton’s Atomic Theory, the atoms have characteristic masses, and so a given number of hydrogen atoms corresponds to a fixed mass of hydrogen. Twice as many oxygen atoms correspond to twice the mass of oxygen.
Therefore we can say that for a given mass of hydrogen, hydrogen peroxide will contain twice the mass of oxygen that water will.
Example $1$: Mass Ratio
Given that the mass of an oxygen atom is 7.937 times the mass of a hydrogen atom, calculate the mass ratio of oxygen to hydrogen in hydrogen peroxide.
Solution The formula H2O2 tells us that there are 2 oxygen atoms and 2 hydrogen atoms in each molecule. Thus the mass ratio is
$\dfrac{\text{mass of 2 O atoms}}{\text{mass of 2 H atoms}}= \dfrac{\text{2 x 15.873 x mass of 1 H atom}}{\text{2 x mass of 1 H atom}} \nonumber$
and again cancelling "mass of 1 H atom" in the numerator and denominator we get
$\dfrac{\text {2 x 15.873}}{2} = \dfrac{15.873}{1} \nonumber$
Note that the mass of oxygen per unit mass of hydrogen is double that calculated earlier for water.
The reasoning and calculations above illustrate the law of multiple proportions. When two elements form several compounds, the mass ratio in one compound will be a small whole-number multiple of the mass ratio in another. In the case of water and hydrogen peroxide, the mass ratios of mercury to bromine are 7.937 and 15.873, respectively. The second value is a small whole-number multiple of (2 times) the first.
Until the atomic theory was proposed, no one had expected any relationship to exist between mass ratios in two or more compounds containing the same elements. Because the theory predicted such relationships, Dalton and other chemists began to look for them. Before long, a great deal of experimental evidence was accumulated to show that the law of multiple proportions was valid. Thus the atomic theory was able to account for previously known facts and laws, and it also predicted a new law. In the process of verifying that prediction, Dalton and his contemporaries did many additional quantitative experiments. These led onward to more facts, more laws, and, eventually, new or modified theories. This characteristic of stimulating more research and thought put Dalton’s postulates in the distinguished company of other good scientific theories.
From ChemPRIME: 2.4: Testing the Atomic Theory | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.05%3A_Testing_the_Atomic_Theory/2.5.01%3A_Biology-_Water.txt |
Our discussion of the atomic theory has indicated that mass is a very important characteristic of atoms — it does not change as chemical reactions occur. Volume, on the other hand, often does change, because atoms or molecules pack together more tightly in liquids and solids or become more widely separated in gases when a reaction takes place. From the time Dalton’s theory was first proposed, chemists realized the importance of the masses of atoms, and they spent much time and effort on experiments to determine how much heavier one kind of atom is than another.
Dalton, for example, studied a compound of carbon and oxygen which he called carbonic oxide. He found that a 100-g sample contained 42.9 g C and 57.1 g O. In Dalton’s day there were no simple ways to determine the microscopic nature of a compound, and so he did not know the composition of the molecules (and hence the formula) of carbonic oxide. Faced with this difficulty, he did what most scientists would do — make the simplest possible assumption. This was that the molecules of carbonic oxide contained the minimum number of atoms: one of carbon and one of oxygen. Carbonic oxide was the compound we now know as carbon monoxide, CO, and so in this case Dalton was right. However, erroneous assumptions about the formulas for other compounds led to half a century of confusion about atomic weights.
Since the formula was CO, Dalton argued that the ratio of the mass of carbon to the mass of oxygen in the compound must be the same as the ratio of the mass of 1 carbon atom to the mass of 1 oxygen atom:
$\dfrac{\text{Mass of 1 C atom}}{\text{Mass of 1 O atom}}=\frac{\text{mass of C in CO}}{\text{mass of O in CO}}=\dfrac{\text{42}\text{.9 g}}{\text{57}\text{.1 g}}=\dfrac{\text{0}\text{.751}}{\text{1}}=\text{0.751}\label{1}$
In other words, the mass of a carbon atom is about three-quarters (0.75) as great as the mass of an oxygen atom.
Notice that this method involves a ratio of masses and that the units grams cancel, yielding a pure number. That number (0.751, or approximately ¾) is the relative mass of a carbon atom compared with an oxygen atom. It tells nothing about the actual masses of carbon or oxygen atoms – only that carbon is three-quarters as heavy as oxygen.
The relative masses of the atoms are usually referred to as atomic weights. Their values are in the Table of Atomic Weights, along with the names and symbols for the elements. The atomic-weight scale was originally based on a relative mass of 1 for the lightest atom, hydrogen. As more accurate methods for determining atomic weight were devised, it proved convenient to shift to oxygen and then carbon, but the scale was adjusted so that hydrogen’s relative mass remained close to 1. Thus nitrogen’s atomic weight of 14.0067 tells us that a nitrogen atom has about 14 times the mass of a hydrogen atom.
The fact that atomic weights are ratios of masses and have no units does not detract at all from their usefulness. It is very easy to determine how much heavier one kind of atom is than another.
Example $1$: Mass of Mercury Atom
Use the Table of Atomic Weights to show that the mass of a mercury atom is 2.510 times the mass of a bromine atom.
Solution
The actual masses of the atoms will be in the same proportion as their relative masses. The atomic weight of mercury is 200.59 and bromine is 79.904. Therefore:
$\frac{\text{Mass of a Hg atom}}{\text{Mass of a Br atom}}=\frac{\text{relative mass of a Hg atom}}{\text{relative mass of a Br atom}}=\frac{\text{200}\text{.59}}{\text{79}\text{0.904}}=\text{2.5104} \nonumber$
or: Mass of a Hg atom = 2.5104 $\cdot$ Mass of a Br atom
The atomic-weight table also permits us to obtain the relative masses of molecules. These are called molecular weights and are calculated by summing the atomic weights of all atoms in the molecule.
Example $2$: Mass Comparison
How heavy would a mercurous bromide molecule be in comparison to a single bromine atom?
Solution
First, obtain the relative mass of an Hg2Br2 molecule (the molecular weight):
\begin{align*} \text{2 Hg atoms: relative mass}= 2 \cdot200.59 &=& 401.18 \ \text{2 Br atoms: relative mass}= 2 \cdot79.904 &=& 159.808 \ \text{1 Hg2Br2 molecule: relative mass} &=& 560.99 \end{align*} \nonumber
Therefore
$\frac{\text{Mass of a Hg}_{\text{2}}\text{Br}_{\text{2}}\text{ molecule}}{\text{Mass of a Br atom}}=\dfrac{\text{560}\text{0.99}}{\text{79}\text{0.904}}=\text{7.0208} \nonumber$
The Hg2Br2 molecule is about 7 times heavier than a bromine atom. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.06%3A_Atomic_Weights.txt |
According to the atomic theory, atoms are the units of chemical reactions. The formula HgBr2 indicates that each molecule of this substance contains one mercury and two bromine atoms. Therefore, if we ask how much bromine is required to make a given quantity of mercury (II) bromide, the answer is two bromine atoms for each mercury atom or two bromine atoms per molecule. In other words, how much substance we have depends in a very important way on how many atoms or molecules are present.
So far, we've dealt with mass ratios. Is there a way to change masses of atoms into numbers of atoms, so it is easy to see how much of one element will react with another, just by looking at the number of atoms that are needed?
As we see below, there seems to be no fundamental connection between the number of atoms or molecules in the chemical equations, and typical measures of "how much":
the number of atoms or molecules in the chemical equations
$1 \text{ Hg } (l)$ + $1 \text{Br}_{2} (l)$ $\rightarrow$ $1 \text{ HgBr}_{2} (s)$
1 atom
1 molecule
1 molecule
1.00 g 0.797 g 1.797 g
1.00 ml 3.47 ml 0.30 ml
"How much?" in the above sense of the quantity of atoms or molecules present is not the same thing as "how much" in terms of volume or mass. It takes 3.47 cm3 Br2(l) to react with a 1 cm3 sample of Hg(l). That same 1 cm3 Hg(l) would weigh 13.59 g, but only 10.83 g Br2(l) would be needed to react with it. In terms of volume, more bromine than mercury is needed, while in terms of mass, less bromine than mercury is required. In the atomic sense, however, there are exactly twice as many bromine atoms as mercury atoms and twice as much bromine as mercury.
Luckily, the International System of Measurements (IUPAC) has a measure of amount that reflects the number of atoms present, and it is called the mole. For perspective, 1 mole of salt cubes (like those seen below) would form a cube 27 miles square. For additional perspective, it would take the fastest marathoner in the world just about 2 hours to run the length of one side. A mole is a huge number... Find out more about it on the next page!
Image credits go to Tony L. Wong on Flickr.
2.08: The Mole
Because atoms and molecules are extremely small, there are a great many of them in any macroscopic sample. The 1 cm3 of mercury referred to in the introduction to moles(opens in new window) would contain 4.080 x 1022 mercury atoms, for example, and the 3.47 cm3 of bromine would contain twice as many (8.160 x 1022) bromine atoms. The very large numbers involved in counting microscopic particles are inconvenient to think about or to write down. Therefore chemists have chosen to count atoms and molecules using a unit called the mole. One mole (abbreviated mol) is 6.022 x 1023 of the microscopic particles which make up the substance in question. Thus 6.022 x 1023 Br atoms is referred to as 1 mol Br. The 8.160 x 1022 atoms in the sample we have been discussing would be:
$\dfrac {8.160\cdot10^{22}} {6.022\cdot10^{23}\text{ mol Br}} = \text {0.1355 mol Br} \nonumber$
The idea of using a large number as a unit with which to measure how many objects we have is not unique to chemists. Eggs, doughnuts, and many other things are sold by the dozen—a unit of twelve items. Smaller objects, such as pencils, may be ordered in units of 144, that is, by the gross, and paper is packaged in reams, each of which contains 500 sheets. A chemist who refers to 0.1355 mol Br is very much like a bookstore manager who orders 2½ dozen sweat shirts, 20 gross of pencils, or 62 reams of paper.
There is a difference in degree, however, because the chemist’s unit, 6.022 x 1023, is so large. A stack of paper containing a mole of sheets would extend more than a million times the distance from the earth to the sun, and 6.022 x 1023 grains of sand would cover all the land in the world to a depth of nearly 2 ft. Obviously there are a great many particles in a mole of anything.
Why have chemists chosen such an unusual number as 6.022 x 1023 as the unit with which to count the number of atoms or molecules? Surely some nice round number would be easier to remember. The answer is that the number of grams in the mass of 1 mol of atoms of any element is the atomic weight of that element. For example, 1 mol of mercury atoms not only contains 6.022 x 1023 atoms, but its mass of 200.59 g is conveniently obtained by adding the unit gram to the Table of Atomic Weights. Some other examples are
\begin{align} &\text{1 mol H contains 6.022} \times 10^{23} \text{H atoms;} & \text{its mass is 1.008 g.} \&\text{1 mol C contains 6.022} \times 10^{23} \text{C atoms;} &\text{its mass is 12.01 g.} \&\text{1 mol O contains 6.022} \times 10^{23} \text{O atoms;} &\text{its mass is 16.00 g.} \&\text{1 mol Br contains 6.022} \times 10^{23} \text{Br atoms;} &\text{its mass is 79.90 g.} \end{align} \nonumber
Here and in subsequent calculations atomic weights are rounded to two decimal places, unless, as in the case of H, fewer than four significant figures would remain.
The mass of a mole of molecules can also be obtained from atomic weights. Just as a dozen eggs will have a dozen whites and a dozen yolks, a mole of CO molecules will contain a mole of C atoms and a mole of O atoms.
The mass of a mole of CO is thus:
$\text{Mass of 1 mol C + mass of 1 mol O = mass of 1 mol CO} \nonumber$
$\text{12.01 g + 16.00 g = 28.01 g} \nonumber$
The molecular weight of CO (28.01) expressed in grams is the mass of a mole of CO. Some other examples are in Table $1$.
Table $1$: Molecular Weight
Molecule Molecular Weight Mass of 1 Mol of Molecules
Br2 2(79.90) = 159.80 159.80 g
O2 2(16.00) = 32.00 32.00 g
H2O 2(1.008) + 16 = 18.02 18.02 g
HgBr2 200.59 + 2(79.90) = 360.39 360.39 g
Hg2Br2 2(200.59) + 2(79.90) = 560.98 560.98 g
It is important to specify to what kind of particle a mole refers. A mole of Br atoms, for example, has only half as many atoms (and half as great a mass) as a mole of Br2 molecules. It is best not to talk about a mole of bromine without specifying whether you mean 1 mol Br or 1 mol Br2
2.09: The Amount of Substance
Chemists use the mole so often to measure how much of a substance is present that it is convenient to have a name for the quantity which this unit measures. In the International System this quantity is called the amount of substance and is given the symbol n. Here again a common English word has been given a very specific scientific meaning. Although amount might refer to volume or mass in everyday speech, in chemistry it does not. When a chemist asks what amount of Br2 was added to a test tube, an answer like “0.0678 mol Br2” is expected. This would indicate that 0.0678 × 6.022 × 1023 or 4.08 × 1022, Br2 molecules had been added to the test tube.
2.10: The Avogadro Constant
Although chemists usually work with moles as units, occasionally it is helpful to refer to the actual number of atoms or molecules involved. When this is done, the symbol N is used. For example, in referring to 1 mol of mercury atoms, we could write
$n_{\text{Hg}} = 1 \text{mol} \nonumber$
and
$N_{\text{Hg}} = 6.022\times 10^{23} \nonumber$
Notice that $N_{Hg }$ is a unitless quantity, which requires the use of a conversion factor to obtain. This conversion factor involves the number of particles per unit amount of substance and is given the symbol $N_A$ and called the Avogadro constant. It is defined by the equation
$N_{\text{A}} = \dfrac{N}{n} \label{1}$
Since for any substance there are 6.022 × 1023 particles per mole,
$\textit{N}_\text{A}=\dfrac{6.022\cdot10^{23}} {1\text{ mol}}=6.022\times 10^{23}\text{ mol}^{\text{–1}} \nonumber$
Example $1$: Moles to Molecules
Calculate the number of O2 molecules in 0.189 mol O2.
Solution
Rearranging Equation $\ref{1}$, we obtain
$N = n \times N_{\text{A}} = 0.189 \text{ mol} \times 6.022 \times 10^{23} \tfrac{1}{\text{ mol}} \ = 1.14 \times 10^{23} \nonumber$
Alternatively, we might include the identity of the particles involved:
\begin{align*}\ce{N} &= \text{0.189 mol O}_{\text{2}}\cdot \left( \dfrac{6.022 \times 10^{23}\text{ O}_2\text{ molecules}} {\text{1 mol O}_2}\right) \[4pt] &= 1.14\cdot10^{23}\text{ O}_{\text{2}}\text{ molecules}\end{align*} \nonumber
Notice that Equation $\ref{1}$, which defines the Avogadro constant, has the same form as the equation which defined density. The preceding example used the Avogadro constant as a conversion factor in the same way that density was used. As in previous examples, all that is necessary is to remember that number of particles and amount of substance are related by a conversion factor, the Avogadro constant.
$\large\text{Number of particles } \large\overset{\text{Avogadro constant}}{\longleftrightarrow} \large\text{amount of substance} \ \quad \ \large N \large\overset{\text{N}_{\text{A}}}{\longleftrightarrow} { \space} \large n\label{2}$
As long as the units mole cancel, NA is being used correctly. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.07%3A_The_Amount_of_Substance-_Moles.txt |
As we saw in The Amount of Substance: Moles, there is no relationship between the mass or volume of a substance and the number of molecules. But we then defined the amount of substance, n, to represent the number of particles. The amount is therefore useful in determining how much of each substance will react. While 1 g of Hg, or 1 cm3 of Hg, reacts with a mass or volume of Br2 that is not related to the coefficients of the chemical equation, 1 mol of Hg always reacts with 1 mol of Br2, since one atom of Hg reacts with one molecule of Br2:
What we need is a convenient way to convert masses to amounts, and the necessary conversion factor is called the molar mass. A molar quantity is one which has been divided by the amount of substance. For example, an extremely useful molar quantity is the molar mass M:
$\text{Molar mass}=\frac{\text{mass}}{\text{amount of substance}}\\textit{M}\text{(g/mol)}=\frac{\textit{m}\text{ (g)}}{\textit{n}\text{ (mol)}} \nonumber$
It is often convenient to express physical quantities per unit amount of substance (per mole), because in this way equal numbers of atoms or molecules are being compared. Such molar quantities often tell us something about the atoms or molecules themselves. For example, if the molar volume of one solid is larger than that of another, it is reasonable to assume that the molecules of the first substance are larger than those of the second. (Comparing the molar volumes of liquids, and especially gases, would not necessarily give the same information since the molecules would not be as tightly packed.)
It is almost trivial to obtain the molar mass, since atomic and molecular weights expressed in grams give us the masses of 1 mol of substance.
Example $1$: Molar Mass
Obtain the molar mass of (a) Hg and (b) Hg2Br2.
Solution
a) The atomic weight of mercury is 200.59, and so 1 mol Hg weighs 200.59 g.
$M_{\text{Hg}}=\frac{m_{\text{Hg}}}{n_{\text{Hg}}} = \frac{\text{200}\text{.59 g}}{\text{1 mol}}= \text{200.59 g mol}^{-1}$
b) Similarly, for Hg2Br2 the molecular weight is 560.98, and so
$\text{M}_ {\text{Hg}_2\text{Br}_2} =\frac {\text{m}_ {\text{Hg}_2\text{Br}_2} } {\text{n}_ {\text{Hg}_2\text{Br}_2} } =\text{560.98 g mol}^{-1} \nonumber$
The molar mass is numerically the same as the atomic or molecular weight, but it has units of grams per mole. The equation, which defines the molar mass, has the same form as those defining density, and the Avogadro constant. As in the case of density or the Avogadro constant, it is not necessary to memorize or manipulate a formula. Simply remember that mass and amount of substance are related via molar mass.
$\text{Mass}\overset{\text{Molar mass}} {\longleftrightarrow} \text{amount of substance}\textit{ m}\overset{\textit{M}}{\longleftrightarrow}\textit{n} \nonumber$
The molar mass is easily obtained from atomic weights and may be used as a conversion factor, provided the units cancel.
Example $2$: Moles
Calculate the amount of octane (C8H18) in 500 g of this liquid.
Solution
Any problem involving interconversion of mass and amount of substance requires molar mass
$\textit{M}=8\cdot12.01\text{ + } 18\cdot1.008 \text{ g mol}^{\text{-1}}= 114.2\text{ g mol}^{\text{-1}} \nonumber$
The amount of substance will be the mass times a conversion factor which permits cancellation of units:
$\textit{n}=\textit{m}\cdot\text{conversion factor}=\text{m}\cdot \tfrac {1} {\textit{M}}= 500 \text{ g}\cdot \tfrac {\text{1 mol}} {\text{114.2 g}}=\text{4.38 mol} \nonumber$
In this case the reciprocal of the molar mass was the appropriate conversion factor.
The Avogadro constant, molar mass, and density may be used in combination to solve more complicated problems.
Example $3$: Molecules
How many molecules would be present in 25.0 ml of pure carbon tetrachloride (CCl4)?
Solution
In previous examples, we showed that the number of molecules may be obtained from the amount of substance by using the Avogadro constant. The amount of substance may be obtained from mass by using the molar mass, and mass from volume by means of density. A road map to the solution of this problem is:
$\text{Volume}\xrightarrow{\text{density}}\text{mass }\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount}\overset{\text{Avogadro constant}}{\longleftrightarrow}\text{number of molecules} \nonumber$
or in shorthand notation:
$\text{V}\xrightarrow{\rho}\textit{m}\xrightarrow{\textit{M}}\textit{n}\xrightarrow{\textit{N}_{\textit{A}}}\textit{N} \nonumber$
The road map tells us that we must look up the density of CCl4(opens in new window):
$\rho=1.595\text{ g}\text{ cm}^{-3} \nonumber$
The molar mass must be calculated from the Table of Atomic Weights(opens in new window)
$\textit{M}=\left(12.01 + 4\cdot35.45\right) \text{ g mol}^{-1}=153.81 \text{ g mol}^{-1} \nonumber$
and we recall that the Avogadro constant is:
$\textit{N}_\textit{A}=6.022\cdot10^{23}\text{ mol}^{-1} \nonumber$
The last quantity (N) in the road map can then be obtained by starting with the first (V) and applying successive conversion factors:
\begin{align}\textit{N}&=& 25\text{ cm}^{3}\cdot\tfrac{\text{1.595 g}} {1 \text{ cm}^{3}}\cdot\tfrac {1 \text{ mol}} {\text{153.81 g}}\cdot\tfrac {6.022\cdot10^{23} \text{ molecules}}{1 \text{ mol}} \&=&1.56\cdot10^{23} \text{ molecules} \end{align}
Notice that in this problem we had to combine techniques from previous examples. To do this you must remember relationships among quantities. For example, a volume was given, and we knew it could be converted to the corresponding mass by means of density, and so we looked up the density in a table. By writing a road map, or at least seeing it in your mind’s eye, you can keep track of such relationships, determine what conversion factors are needed, and then use them to solve the problem.
2.11: The Molar Mass
It should now be clear that knowing the number of water (or other biological) molecules is central to many issues in biology, and it's an issue that chemistry can shed light on.
For example, photosynthesis involves the reaction of carbon dioxide and water to make sugars. Richard Feynman noted chemistry's contribution in a very poetic way:
“The world looks so different after learning science. For example, trees are made of air, primarily. When they are burned, they go back to air, and in the flaming, heat is released, the flaming heat of the sun which was bound in to convert the air into tree. And in the ash is the small remnant of the part which did not come from air, that came from the solid earth, instead.
Sugar (and other "carbohydrates" in plants) is made photosynthetically according to overall equations like
${6 CO_2 + 6 H_2O + Light Energy \longrightarrow\ C_6H_{12}O_6 + 6 O_2} \nonumber$
SID 24715013 CID 24749
http://employees.csbsju.edu/hjakubowski/Jmol/Beta-D-Glucopyranose/betaglucose.pdb
This explains why plants die without water. But how much water is required to make a pound (457 grams) of sugar? The equation tells us 6 water molecules make 6 glucose (sugar) molecules. How do we get useful, macroscopic answers, like how much water (and CO2) is needed for a pound of sugar? We start with molar masses.
It is often convenient to express physical quantities per unit amount of substance (per mole), because in this way equal numbers of atoms or molecules are being compared. Such molar quantities often tell us something about the atoms or molecules themselves. For example, if the molar volume of one solid is larger than that of another, it is reasonable to assume that the molecules of the first substance are larger than those of the second. (Comparing the molar volumes of liquids, and especially gases, would not necessarily give the same information since the molecules would not be as tightly packed.)
A molar quantity is one which has been divided by the amount of substance. For example, an extremely useful molar quantity is the molar mass M:
${{Molar\ mass} =\tfrac {\text {mass} } {\text {amount of substance} } } \nonumber$
It is almost trivial to obtain the molar mass, since atomic and molecular weights expressed in grams give us the masses of 1 mol of substance.
EXAMPLE 1 Obtain the molar mass of (a) C (carbon) and (b) H2O.
Solution
a) The atomic weight of carbon is 12.01, and so 1 mol C weighs 12.01 g.
$M_{\text{C}}=\tfrac{\text{m}_{\text{C}}}{n_{\text{C}}}=\tfrac{\text{12}\text{.01 g}}{\text{1 mol}}=\text{12}\text{.01 g mol}^{-1} \nonumber$
b) Similarly, for H2O, the molecular weight is 18.02 (16.00 for O and 2.02 for 2 H), and so
$M_{\text{H}_{\text{2}}\text{O}}= \tfrac {\text{m}_{\text{H}_{\text{2}}\text{O}}} {n_{\text{H}_{\text{2}}\text{O}}}=\text{18}\text{.02 g mol}^{-1} \nonumber$
The molar mass is numerically the same as the atomic or molecular weight, but it has units of grams per mole. The equation, which defines the molar mass, has the same form as those defining density, and the Avogadro constant. As in the case of density or the Avogadro constant, while it's always good to know a formula and be able to manipulate it, it's often only necessary to remember that mass and amount of substance are related via molar mass.
$\text{Mass}\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount of substance}m\overset{M}{\longleftrightarrow}n \nonumber$
The molar mass is easily obtained from atomic weights and may be used as a conversion factor, provided the units cancel.
EXAMPLE 2 Calculate the amount of glucose (C6H12O6) in 457 g of this solid.
Solution Any problem involving interconversion of mass and amount of substance requires molar mass
M = (6 x 12.01 + 12 x 1.008 + 6 x 16) g mol–1 = 180 g mol–1
The amount of substance will be the mass times a conversion factor which permits cancellation of units:
$\textit{n} = \textit{m}\ \cdot\text{ conversion factor} = m\ \cdot \tfrac {\text{1}} {M} = \text{457 g}\cdot \tfrac {\text{1 mol}} {\text{180 g}} = 2.54\ \text{mol} \nonumber$
In this case the reciprocal of the molar mass was the appropriate conversion factor. The Avogadro constant, molar mass, and density may be used in combination to solve more complicated problems.
EXAMPLE 3 The chemical equation above tells us that 6 glucose molecules require 6 water molecules, so 6 moles of glucose require 6 moles of water, and since they're equal, the 2.54 moles of glucose above would require 2.54 moles of water. What mass of water is that?
The mass of substance will be the amount times a conversion factor which permits cancellation of units:
$\textit{m} = \textit{n} \cdot\text{conversion factor} = \text{n} \cdot \text{M} = \text{2.54 mol} \cdot 18.02 \tfrac{\text{g}}{\text{mol}} = \text{45.7 g} \nonumber$
EXAMPLE 4 How many molecules would be present in 50 mL of pure water?
Solution In previous examples, we showed that the number of molecules may be obtained from the amount of substance by using the Avogadro constant. The amount of substance may be obtained from mass by using the molar mass, and mass from volume by means of density. A road map to the solution of this problem is
$\text{Volume}\xrightarrow{\text{density}}\text{mass }\overset{\text{Molar mass}}{\longleftrightarrow}\text{amount}\overset{\text{Avogadro constant}}{\longleftrightarrow}\text{number of molecules} \nonumber$
or in shorthand notation
$V\xrightarrow{\rho }m\xrightarrow{M}n\xrightarrow{\text{N}_{\text{A}}}N \nonumber$
The road map tells us that we must look up the [[File:Chapter 1 page 23.jpg|density of H2]O]:
ρ = 1.0 g cm–3
The molar mass must be calculated from the Table of Atomic Weights.
M = (2 1.008 + 1 × 16.00) g mol–1 = 18.02 g mol–1
and we recall that the Avogadro constant is
NA = 6.022 × 1023 mol–1
The last quantity (N) in the road map can then be obtained by starting with the first (V) and applying successive conversion factors:
\begin{align} & \text{N}=\text{50}\text{.0 cm}^{\text{3}}\cdot \tfrac{\text{1}\text{.00 g}}{\text{1 cm}^{\text{3}}}\cdot \tfrac{\text{1 mol}}{\text{18}\text{.02 g}}\cdot \tfrac{\text{6}\text{.022}\cdot \text{10}^{\text{23}}\text{ molecules}}{\text{1 mol}} \ & \text{ }=\text{1}\text{.67}\cdot \text{10}^{\text{24}}\text{ molecules} \ \end{align}
Notice that in this problem we had to combine techniques from previous examples. To do this you must remember relationships among quantities. For example, a volume was given, and we knew it could be converted to the corresponding mass by means of density, and so we looked up the density in a table. By writing a road map, or at least seeing it in your mind’s eye, you can keep track of such relationships, determine what conversion factors are needed, and then use them to solve the problem.
Example 5 A student's body may contain 1027 water molecules and would need to accumulate about 1018 (ten million million million) water molecules per second over 18 years.
a. Show that the number of water molecules is about right, assuming a body weight of 150 lb and that water is about 70% of body weight.
b. Show that the rate of accumulation of water molecules per second is approximately right.
Solution
a. $150 lb \cdot \dfrac{\text{0.453 kg}}{\text{1 lb}} \cdot .70 = 47.6 kg water \nonumber$
$\dfrac{\text{47,600 g water}}{18\tfrac{\text{g}}{\text{mol}}} = 2.6 \cdot 10^3 \text{mol} \nonumber$
$2.6 \cdot 10^3 \text {mol} \cdot 6.02 \cdot 10^{23} \tfrac{\text{molecules}}{\text{mol}} = 1.6 \cdot 10^{27} \text{molecules} \nonumber$
b.$\dfrac{10^{27} \text {water molecules}} { \text{18y} \cdot \tfrac{\text{365d}}{\text{y}} \cdot \tfrac{\text {24h}}{\text{d}} \cdot \tfrac{\text{60 min}}{\text{h}} \cdot \tfrac{\text{60 sec}}{\text{min}} } = 4 \cdot 10^{18} \tfrac{\text {molecules}}{\text{sec}} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.11%3A_The_Molar_Mass/2.11.01%3A_Biology-_Water.txt |
We have presented a microscopic view of the chemical reaction between mercury and bromine. The equation
chemical reaction between mercury and bromine
$\ce{Hg(l)^{+}}$ $\ce{Br2(l)} \rightarrow$ $\ce{HgBr2(s)}$
represents the same event in terms of chemical symbols and formulas, while the pictures below represent the macroscopic view. But how does a practicing chemist find out what is occurring on the microscopic scale? When a reaction is carried out for the first time, little is known about the microscopic nature of the products. It is therefore necessary to determine experimentally the composition and formula of a newly synthesized substance.
The first step in such a procedure is usually to separate and purify the products of a reaction. For example, although the combination of mercury with bromine produces mainly mercuric bromide, a little mercurous bromide is often formed as well. A mixture of mercurous bromide with mercuric bromide has properties which are different from a pure sample of HgBr2, and so the Hg2Br2 must be removed. The low solubility of Hg2Br2 in water would permit purification by recrystallization. The product could be dissolved in a small quantity of hot water and filtered to remove undissolved Hg2Br2. Upon cooling and partial evaporation of the water, crystals of relatively pure HgBr2 would form.
Once a pure product has been obtained, it may be possible to identify the substance by means of its physical and chemical properties. The reaction of mercury with bromine yields white crystals which melt at 236°C. The liquid which is formed boils at 322°C. Since it is made by combining two elements, the product is a compound. Comparing its properties with a handbook or table of data leads to the conclusion that it is mercuric bromide.
But suppose you were the first person who ever prepared mercuric bromide. There were no tables which listed its properties then, and so how could you determine that the formula should be HgBr2? One answer involves quantitative analysis—the determination of the percentage by mass of each element in the compound. Such data are usually reported as the percent composition
Example $1$: Percent Composition
When 10.0 g mercury reacts with sufficient bromine, 18.0 g of a pure compound is formed. Calculate the percent composition from these experimental data.
Solution:
The percentage of mercury is the mass of mercury divided by the total mass of compound times 100 percent:
$\%\text{ Hg} = \dfrac{m_{\text{Hg}}}{m_{\text{compound}}} \cdot 100\%\ = \dfrac{\text{10.0 g}}{\text{18.0 g}} \cdot 100\%\text{ = 55.6 }\% \nonumber$
The remainder of the compound (18.0 g – 10 g = 8.0 g) is bromine:
$\%\text{ Br = } \dfrac{m_{\text{Br}}}{m_{\text{compound}}} \cdot \text{ 100 }\%\text{ = } \dfrac{\text{8.0 g}}{\text{18.0 g}} \cdot \text{ 100 } \%\text{ = 44.4 } \% \nonumber$
As a check, verify that the percentages add to 100:
$55.6\% + 44.4\% = 100\% \nonumber$
To obtain the formula from percent-composition data, we must find how many bromine atoms there are per mercury atom. On a macroscopic scale this corresponds to the ratio of the amount of bromine to the amount of mercury. If the formula is HgBr2, it not only indicates that there are two bromine atoms per mercury atom, it also says that there are 2 mol of bromine atoms for each 1 mol of mercury atoms. That is, the amount of bromine is twice the amount of mercury. The numbers in the ratio of the amount of bromine to the amount of mercury (2:1) are the subscripts of bromine and mercury in the formula.
Example $2$ : Formula
Determine the formula for the compound whose percent composition was calculated in the previous example.
Solution:
For convenience, assume that we have 100 g of the compound. Of this, 55.6 g (55.6%) is mercury and 44.4 g is bromine. Each mass can be converted to an amount of substance
\begin{align*} & n_{\text{Hg}}=\text{55.6 g}\cdot \dfrac{\text{1 mol Hg}}{\text{200.59 g}} =\text{0.277 mol Hg} \ { } \ & n_{\text{Hg}}=\text{44.4 g}\cdot \dfrac{\text{1 mol Br}}{\text{79.90 g}} =\text{0.556 mol Br} \end{align*} \nonumber
Dividing the larger amount by the smaller, we have
$\dfrac{n_{\text{Br}}}{n_{\text{Hg}}} = \dfrac{\text{0}\text{.556 mol Br}}{\text{0}\text{.227 mol Hg}} = \dfrac{\text{2}\text{.01 mol Br}}{\text{1 mol Hg}} \nonumber$
The ratio 2.01 mol Br to 1 mol Hg also implies that there are 2.01 Br atoms per 1 Hg atom. If the atomic theory is correct, there is no such thing as 0.01 Br atom; furthermore, our numbers are only good to three significant figures. Therefore we round 2.01 to 2 and write the formula as HgBr2.
Example $3$: Formula Calculation
A bromide of mercury has the composition 71.5% Hg, 28.5% Br. Find its formula.
Solution:
Again assume a 100-g sample and calculate the amount of each element:
\begin{align*} & n_{\text{Hg}}=\text{71}\text{.5 g}\cdot \dfrac{\text{1 mol Hg}}{\text{200.59 g}} = \text{0.356 mol Hg} \ { } \ & n_{\text{Hg}}=\text{28.5 g}\cdot \dfrac{\text{1 mol Br}}{\text{79.90 g}} =\text{0.357 mol Br} \end{align*} \nonumber
The ratio is
$\dfrac{n_{\text{Br}}}{n_{\text{Hg}}} = \dfrac{\text{0.357 mol Br}}{\text{0.356 mol Hg}} = \dfrac{\text{1.00 mol Br}}{\text{1 mol Hg}} \nonumber$
We would therefore assign the formula HgBr.
The formula obtained in Example $3$ does not correspond to either of the two mercury bromides we have already discussed. Is it a third one? The answer is no because our method can only determine the ratio of Br to Hg. The ratio 1:1 is the same as 2:2, and so our method will give the same result for HgBr or Hg2Br2 (or Hg7Br7, for that matter, should it exist). The formula determined by this method is called the empirical formula or simplest formula. Occasionally, as in the case of mercurous bromide, the empirical formula differs from the actual molecular composition, or the molecular formula. Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula.
Example $4$:
A compound whose molecular weight is 28 contains 85.6% C and 14.4% H. Determine its empirical and molecular formulas.
Solution:
\begin{align*} & n_{\text{C}}=\text{85.6 g}\cdot \dfrac{\text{1 mol C}}{\text{12.01 g}} =\text{7.13 mol C} \ { } \ & n_{\text{H}}=\text{14.4 g}\cdot \dfrac{\text{1 mol H}}{\text{1.008 g}} =\text{14.3 mol H} \end{align*} \nonumber
$\dfrac{n_{\text{H}}}{n_{\text{C}}} = \dfrac{\text{14.3 mol H}}{\text{7.13 mol C}} = \dfrac{\text{2.01 mol H}}{\text{1 mol C}} \nonumber$
The empirical formula is therefore CH2. The molecular weight corresponding to the empirical formula is
$12.01 + 2 \cdot 1.008 = 14.03 \nonumber$
Since the experimental molecular weight is twice as great, all subscripts must be doubled and the molecular formula is C2H4.
Occasionally the ratio of amounts is not a whole number.
Example $5$: Empirical Formula
Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical formula?
Solution:
\begin{align*} & n_{\text{H}}=\text{14.4 g}\cdot \dfrac{\text{1 mol H}}{\text{1.008 g}} =\text{14.3 mol H} \ { } \& n_{\text{C}}=\text{85.6 g}\cdot \dfrac{\text{1 mol C}}{\text{12.01 g}} =\text{7.13 mol C} \ { } \ & n_{\text{O}}=\text{35.5 g}\cdot \dfrac{\text{1 mol O}}{\text{16.00 g}} =\text{2.22 mol O} \end{align*} \nonumber
Divide all three by the smallest amount of substance
\begin{align*} & \dfrac{n_{\text{C}}}{n_{\text{O}}} = \dfrac{\text{5.00 mol C}}{\text{2.22 mol O}} =\dfrac{\text{2.25 mol H}}{\text{1 mol O}} \ { } \ & \dfrac{n_{\text{H}}}{n_{\text{O}}}=\dfrac{\text{4.44 mol H}}{\text{2.22 mol O}}= \dfrac{\text{2.00 mol H}}{\text{1 mol O}} \end{align*} \nonumber
Clearly there are twice as many H atoms as O atoms, but the ratio of C to O is not so obvious. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes $\small \dfrac{1}{4}$. Thus $2.25 = 2 \small\dfrac{1}{4} \normalsize = \tfrac{\text{9}}{\text{4}}$, and
$\dfrac{n_{\text{C}}}{n_{\text{O}}} = \dfrac{\text{2.25 mol C}}{\text{1 mol O}} = \dfrac{\text{9 mol C}}{\text{4 mol O}} \nonumber$
We can also write
$\dfrac{n_{\text{H}}}{n_{\text{O}}} = \dfrac{\text{2 mol H}}{\text{1 mol O}} = \dfrac{\text{8 mol H}}{\text{4 mol O}} \nonumber$
Thus the empirical formula is C9H8O4.
Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows.
Example $6$: Percent Nitrogen
In order to replenish nitrogen removed from the soil when plants are harvested, the compounds NaNO3 (sodium nitrate), NH4NO3 (ammonium nitrate), and NH3 (ammonia) are used as fertilizers. If a farmer could buy each at the same cost per gram, which would be the best bargain? In other words, which compound contains the largest percentage of nitrogen?
Solution
We will show the detailed calculation only for the case of NH4NO3.
1 mol NH4NO3 contains 2 mol N, 4 mol H, and 3 mol O. The molar mass is thus:
$M = (2 \cdot 14.01 + 4 \cdot 1.008 + 3 \cdot 16.00) \text{ g} \text{ mol}^{-1} = 80.05 \text{ g} \text{ mol}^{-1} \nonumber$
A 1-mol sample would weigh 80.05 g. The mass of 2 mol N it contains is:
$m_{\text{N}}\text{ = 2 mol N }\cdot \text{ } \dfrac{\text{14.01 g}}{\text{1 mol N}} \text{ = 28.02 g} \nonumber$
Therefore the percentage of N is:
$\text{ } \%\text{ N = } \dfrac{m_{\normalsize\text{N}}}{m_{\normalsize\text{NH}_{\text{4}}\text{NO}_{\text{3}}}}\text{ } \cdot \text{ 100 }\%\text{ = } \dfrac{\text{28.02 g}}{\text{80.05 g}}\text{ } \cdot \text{ 100 }\%\text{ = 35}\text{.00 }\%\text{ } \nonumber$
The percentages of H and O are easily calculated as:
\begin{align*} m_{\text{H}}& = \text{4 mol H }\cdot\dfrac{\text{1.008 g}}{\text{1 mol H}}\text{ = 4.032 g} \ { } \ \text{ }\%\text{ H } & = \dfrac{\text{4.032 g}}{\text{80.05 g}} \cdot \text{ 100 }\%\text{ = 5.04 }\%\ \ { } \ m_{\text{O}}& = \text{3 mol O }\cdot \dfrac{\text{16.00 g}}{\text{1 mol O}} \text{ = 48.00 g} \ { } \ \%\text{ O } & = \dfrac{\text{48.00 g}}{\text{80.05 g}}\text{ }\cdot \text{ 100 }\%\text{ = 59.96 }\%\text{ } \end{align*} \nonumber
Though not strictly needed to answer the problem, the latter two percentages provide a check of the results. The total $35.00 + 5.04\% + 59.96\% = 100.00\%$ as it should. Similar calculations for NaNO3 and NH3 yield 16.48% and 82.24% nitrogen, respectively. The farmer who knows chemistry chooses ammonia!
2.12: Formulas and Composition
We have presented a microscopic view of the chemical reaction between oxygen and hydrogen. The equation
$\ce{O2 (g) + 2 H2 (g) -> 2H2O (l)} \nonumber$
represents the same event in terms of chemical symbols and formulas.
We also saw that plants form glucose from CO2 and H2O:
$\ce{6CO2 (g) + 6 H2O (l) + Light Energy -> C6H12O6 (s) + 6O2(g)} \nonumber$
But how does a practicing chemist find out what is occurring on the microscopic scale? When a reaction is observed for the first time, little is known about the microscopic nature of the products. It is therefore necessary to determine experimentally the composition and formula of a newly synthesized substance.
The first step in such a procedure is usually to separate and purify the products of a reaction. The products above are easy to separate, because they are liquids or solids, while the reactants are gaseous. But how could you determine that the formula should be H2O, and not H2O2? And plants produce mixtures of carbohydrates during photosynthesis, which must be separated by chromatography or other techniques before they can be identified. Glucose is perhaps the most common simple sugar, and it exists in two molecular forms, with the chemical formula C6H12O6:
the open chain form of glucose
the cyclic form of glucose
Plants may produce other simple sugars with similar structures, like D-ribulose (C5H10O5), D-fructose (C6H12O6) and di- or polysaccharides (with two or more simple sugars linked together) like sucrose (C12H22O11) or Maltose (C12H22O11).
Maltose
How can the actual product be identified?
One answer involves quantitative analysis—the determination of the percentage by mass of each element in the compound. Such data are usually reported as the percent composition.
Example $1$: Percent Composition
When 10.0 g oxygen reacts with sufficient hydrogen, 11.26 g of a pure compound is formed. Calculate the percent composition from these experimental data.
Solution
The percentage of oxygen is the mass of oxygen divided by the total mass of compound times 100 percent:
$\text{%O} = \tfrac {\text{m}_{\text{O}}} {\text{m}_{\text{compound}}} \cdot \text{100%} = \tfrac{\text{10.0 g}} {\text{11.26 g}} \cdot \text{100%} = \text{88.81%} \nonumber$
The remainder of the compound (11.26 g – 10 g = 1.26 g) is hydrogen:
$\text{%H} = \tfrac{\text{m}_{\text{H}}} {\text{m}_{\text{compound}}} \cdot \text{100%} = \tfrac{\text{1.26 g}} {\text{11.26 g}} \cdot \text{100%} = \text{11.19%} \nonumber$
As a check, verify that the percentages add to 100:
88.81% + 11.19% = 100%
To obtain the formula from percent-composition data, we must find how many hydrogen atoms there are per oxygen atom. On a macroscopic scale this corresponds to the ratio of the amount of hydrogen to the amount of oxygen. If the formula is H2O, it not only indicates that there are two hydrogen atoms per oxygen atom, it also says that there are 2 mol of hydrogen atoms for each 1 mol of oxygen atoms. That is, the amount of hydrogen is twice the amount of oxygen. The numbers in the ratio of the amount of bromine to the amount of mercury (2:1) are the subscripts of hydrogen and oxygen in the formula.
Example $2$: Formula Determination
Determine the formula for the compound whose percent composition was calculated in the previous example.
Solution
For convenience, assume that we have 100 g of the compound. Of this, 88.81 g (88.81%) is oxygen and 11.19 g is hydrogen. Each mass can be converted to an amount of substance
\begin{align} & n_{\text{O}}=\text{88}\text{.81 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{5}\text{.550 mol O} \ & n_{\text{H}}=\text{11}\text{.19 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{11}\text{.101 mol H} \ \end{align} \nonumber
So the formula is H11.101O5.550, but we know there are no fractions of atoms, so we have to put this in a standard form. To do this, we divide the larger amount by the smaller:
$\frac{n_{\text{H} }}{n_ { \text{O} }}=\frac{\text{11.19 mol H}}{\text{5.550 mol O}}=\frac{\text{1.998 mol H}}{\text{1 mol O}} \nonumber$
To the formula is H1.998O1, but again, there seems to be a problem. The ratio 1.998 mol H to 1 mol O also implies that there are 1.998 H atoms per 1 O atom. If the atomic theory is correct, there is no such thing as 0.998 atom; but remembering that our measurements are only good to three significant figures, we round 1.998 to 2.00 and write the formula as H2O.
Example $3$: Formula of Compound
An oxide of hydrogen has the composition 94.07% O and 5.93% H. Find its formula.
Solution
Again assume a 100-g sample and calculate the amount of each element:
\begin{align} & n_{\text{O}}=\text{94}\text{.07 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{5}\text{.88 mol O} \ & n_{\text{H}}=\text{5}\text{.93 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{5}\text{.88 mol H} \ \end{align}
The ratio is
$\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{5.88 mol H}}{\text{5.88 mol O}}=\frac{\text{1 mol H}}{\text{1 mol O}} \nonumber$
So the formula is H5.88O5.88, which is a 1:1 ratio within experimental precision. We would therefore assign the formula HO.
The formula obtained in Example 3 does not correspond to either of the two hydrogen oxides we have already discussed. Is it a third one? The answer is no because our method can only determine the ratio of H to O. The ratio 1:1 is the same as 2:2, and so our method will give the same result for HO or H2O2 (or H7O7, for that matter, should it exist). The formula determined by this method is called the empirical formula or simplest formula. Occasionally, as in the case of hydrogen peroxde, the empirical formula differs from the actual molecular composition, or the molecular formula. Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula.
Example $4$: Empirical and Molecular Formulas
A compound whose molecular weight is 28 contains 85.6% C and 14.4% H. Determine its empirical and molecular formulas.
Solution
\begin{align} & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \ & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \ \end{align} \nonumber
So the formula is C7.13H14.3, but to get integral subscripts we divide each by the smaller:
$\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{14.3 mol H}}{\text{7.13 mol C}}=\frac{\text{2.01 mol H}}{\text{1 mol C}} \nonumber$
The empirical formula is therefore CH2. The molecular weight corresponding to the empirical formula is
12.01 + 2 × 1.008 = 14.03
Since the experimental molecular weight is twice as great, all subscripts must be doubled and the molecular formula is C2H4.
Example $5$: Empirical Formula
D-Xylose contains 40.0% C, 6.71% H, and 53.29% O. What is its empirical formula?
Solution
\begin{align} & n_{\text{H}}=\text{6}\text{.71 g}\times \tfrac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{6}\text{.66 mol H} \ & n_{\text{C}}=\text{40}\text{.00 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{3}\text{.33 mol C} \ & n_{\text{O}}=\text{53}\text{.29 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{3}\text{.33 mol O} \ \end{align} \nonumber
So the formula is C3.33H6.66O3.33, and dividing all three by the smallest amount of substance we obtain CH2O.
\begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{3}\text{.33 mol C}}{\text{3}\text{.33 mol O}}=\frac{\text{1}\text{. mol H}}{\text{1 mol O}} \ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \ \end{align} \nonumber
Now we see where carbohydrates get their name. They all have formulas like Cn(H2O)m, so they all look like hydrates (water containing compounds) of carbon. D-Xylose is actually C5(H2O)5 or C5H10O5, Glucose is C6(H12O)6 or C6H12O6, and Sucrose is C12(H2O)11 or C12H22O11.
We have seen that experimental error can give mole ratios that are not exact units. Occasionally the ratio of amounts is farther from a whole number than can be explained by experimental error, as in the following example.
Example $6$: Empirical Formula of Aspirin
Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical formula?
Solution
\begin{align} & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \ & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \ & n_{\text{O}}=\text{35}\text{.5 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.22 mol O} \ \end{align}
Divide all three by the smallest amount of substance
\begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{5}\text{.00 mol C}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.25 mol H}}{\text{1 mol O}} \ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \ \end{align}
Clearly there are twice as many H atoms as O atoms, but the ratio of C to O is not so obvious. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus 2.25 = 2¼ = 9/4, and
$\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}} \nonumber$
We can also write
$\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{2 mol H}}{\text{1 mol O}}=\frac{\text{8 mol H}}{\text{4 mol O}} \nonumber$
Thus the empirical formula is C9H8O4.
Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows.
Example $7$: Percent Composition of a Sugar
As we saw above, all carbohydrates have the general formula Cn(H2O)m. All simple sugars have the general formula Cn(H2O)n, so they all have the same percentage of C, H, and O.
1. Calculate the percent C in glucose.
2. Show that the compound C8(H2O)4 cannot be a simple sugar by calculating its percent composition.
Solution
a. C6(H2O)6 contains 6 mol C, 12 mol H, and 6 mol O. The molar mass is thus
M = (6 x 12.011) + (12 x 1.008) + (6 x 16) = 180 g mol–1
A 1-mol sample would weigh 180.56 g. The mass of 6 mol C it contains is
$\text{mC}=\text{6 mol C}\cdot\frac{\text{12.011 g}}{\text{1 mol C}}=\text{72.066 g} \nonumber$
Therefore the percentage of C is
$\text{%C}=\frac{\text{mC}}{\text{mC6H_{12}O_6}}\cdot\text{ 100%}=\frac{\text{72.066 g}}{\text{180.56 g}}\cdot\text{100%}=\text{40.00%} \nonumber$
The percentages of H and O are easily calculated as
\begin{align} & m_{\text{H}}=\text{12 mol H}\cdot=\frac{\text{1.008 g}}{\text{1 mol H}}=\text{12.096 g}\&\text{%H}= \frac{\text{12.096 g}}{\text{180.56 g}}\cdot\text{100%}=\text{6.70%}\& m_{\text{O}}=\text{6 mol O}\cdot=\frac{\text{16.00 g}}{\text{1 mol O}}=\text{96.00 g}\& \text{%O}= \frac{\text{96.00 g}}{\text{180.56 g}}\cdot\text{100%}=\text{53.17%}\\end{align}
Though not strictly needed to answer the problem, the latter two percentages provide a check of the results. The total 40.00 + 6.70% + 53.17% = 100.00% as it should.
Note that for any simple sugar,
$\text{%C}=\tfrac{\textit{n}\text{(12.011)}}{\textit{n}\text{(12.011)} +\textit{n}\text{(18)}} = \text{40%} \nonumber$
regardless of how many carbons it contains (simple sugars are trioses, C3H6O3, tetroses, C4H8O4, pentoses, and hexoses. The same goes for the percent H and O.
b. The molar mass of C8(H2O)4 is 168.15;
%C = [(8 x 12.011)/168.15] x 100% = 57.14%, not the 40% characteristic of simple sugars.
Similarly,
%H = [(8 x 1.008) / 168.15] x 100% = 4.80% and the percent O is 38.06%. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.12%3A_Formulas_and_Composition/2.12.01%3A_Biology-_Formula_and_Composition_of_Water_and_Glucose.txt |
Fertilizers, Formulas and Ecological Stoichiometry
Fertilizer Composition
When you shop for fertilizers, you will find them labeled according to the "NPK" (nitrogen/phosphorus/potassium) analysis. For example, one fertilizer might be labeled "18-51-20".
Deciphering this label requires understanding of chemical formulas.
The nitrogen value (18) is the actual percent nitrogen in the fertilizer, but the phosphorus and potassium values are not.
Fertilizer application to corn field
Phosphorus Content
The phosphorus content is actually expressed as if the phosphorus compound were P2O5. So the "51" in the label means 51% P2O5.
But what percent of this is actually phosphorus itself?
We can determine that from the formula.
EXAMPLE 1 Determine the percent phosphorus and the percent oxygen in P2O5.
Solution The formula indicates the relative amounts (in mol) of phosphorus and oxygen, so 1 mol of the compound contains 2 mol of P and 5 mol of O.
The percent phosphorus is
$\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!$
1 mol of P2O5 has a mass of (2 mol P * 31 g/mol P + 5 mol O * (16.0 g/mol O) = 142 g, and it contains 2 mol of P, weighing 2 mol P * 31.0 g/mol = 62.0 g, so...
$\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{P}_{\text{2}}\text{O}_{\text{5}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{62}\text{.0 g}}{\text{142}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 43}\text{.7 }\!\!%\!\!\text{ }$
Similarly,
%O = 100% x (5 mol O * 16 g/mol) / (2 mol P * 31 g/mol P + 5 mol O * (16.0 g/mol O)
% O = 56.3%
If the fertilizer contains 51% P2O5, it contains only 43.7% of that as pure phosphorus, or 0.437 * 51 = 22.3% P.
How does an analytical laboratory determine which phosphorus compound is present? First, the compound must be separated from the others and purified, possibly by dissolving it in a solvent that does not dissolve the others, then evaporating the solvent until the compound "recrystallizes". Then techniques of quantitative analysis are used to determine the percentage by mass of each element in the compound. Such data are usually reported as the percent composition.
EXAMPLE 2 When 10.0 g of phosphorus burns in oxygen, 22.9 g of a pure compound is formed. Calculate the percent composition from these experimental data.
Solution The percentage of phosphorus is the mass of phosphorus divided by the total mass of compound times 100 percent:
$\text{ }\!\!%\!\!\text{ P = }\frac{m_{\text{P}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{10}\text{.0 g}}{\text{22}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 43}\text{.7 }\!\!%\!\!\text{ }$ The remainder of the compound (22.9 g – 10 g = 12.9 g) is oxygen: $\text{ }\!\!%\!\!\text{ O = }\frac{m_{\text{O}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{12}\text{.9 g}}{\text{22}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 56}\text{.3 }\!\!%\!\!\text{ }$ As a check, verify that the percentages add to 100: 56.3% + 43.7% = 100%
Potassium Content
The potassium (K) content of a fertilizer is actually expressed as if the potassium compound were K2O. So the "20" in the label (above) means 20% K2O, which is 17% oxygen and 83% elemental potassium:
% K = 100% x m (K, g) / m (Total, g) = (2 mol K * 39.1 g/mol) / (2 mol K * 39.1 g/mol + 1 mol O * 16.0 g/mol)
So the true percentage of potassium in the fertilizer is 0.83 x % K2O = 0.83 x 20% = 16.6% K
So 18−51−20 fertilizer actually contains (by weight) 18% elemental (N), 22% elemental (P), and 16% elemental (K).
Ecological Stoichiometry and the Growth Rate Hypothesis
Ecological stoichiometry is an approach to ecology which studies how the chemical elements in a population of plants or animals matches the elements in the food supply for the population. Most plants and animals (including us) show a high degree of chemical hormesis, that is, they have a characteristic chemical composition regardless of their environment, or food supply. They need food which supplies the elements in the proper ratios.
If we want plants to grow well, we need to supply elements in the proper ratios.
One claim of the practitioners of ecological stoichiometry is that fast growth requires high concentrations of RNA, which in turn requires a lot of phosphorus. This is the "Growth Rate Hypothesis" [1] [2]
Nutrient Ratios
The nitrogen to phosphorus ratio in RNA is similar to that in ATP, which is one of its constituents (along with UTP, CTP and GTP). Suppose that this N:P ratio establishes the requirements for the plant. We can then calculate the optimal N:P food ratio by calculating the N:P ratio in ATP, whose structure is:
Adenosine Triphosphate (ATP, C10H16N5O13P3)
From the formula, C10H16N5O13P3,we see that % N = 100* (5*14.0 / 507.2 ) = 13.8%
%P = 100* (3*31 / 507.2) = 18.3% P
So the optimal N/P ratio is 13.8/18.3 = 0.753:1
Is the 18−51−20 fertilizer optimal for ATP synthesis in the plant? The N:P ratio is 18/51 = 0.35, much less than that required. This particular fertilizer is designed for soils that are depleted in phosphorus.
Ecological stoichiometry focuses on the chemical requirements of each trophic level, in addition to energy requirements. Proponents say that "Ecological stoichiometry recognizes that organisms themselves are outcomes of chemical reactions and thus their growth and reproduction can be constrained by supplies of key chemical elements [especially carbon (C), nitrogen (N) and phosphorus (P)]".[3]. Organisms in each trophic level of an ecosystem must have evolved to utilize the nutrient ratios provided by the lower trophic levels.
Formulas from Percent Composition
To obtain the formula from percent-composition data for a phosphorus oxide, we must find how many phosphorus atoms there are per oxygen atom. On a macroscopic scale this corresponds to the ratio of the amount of phosphorus to the amount of oxygen. If the formula is P2O5, it not only indicates that there are five oxygen atoms for every two phosphorus atoms, it also says that there are 5 mol of oxygen atoms for each 2 mol of phosphorus atoms. That is, the amount of oxygen is 2.5 times the amount of phosphorus. The numbers in the ratio of the amount of phosphorus to the amount of oxygen (2:5) are the subscripts of phosphorus and oxygen in the formula.
EXAMPLE 3 Phosphorus forms several oxides, including phosphorus pentoxide, P2O5, phosphorus trioxide, P2O3, P4O7, P4O8, P4O9, PO and P2O6. Determine the formula for the compound whose percent phosphorus is 49.2%.
Solution For convenience, assume that we have 100 g of the compound. Of this, 49.2 g (49.2%) is phosphorus and 50.8 g is oxygen (100% - 49.2%). Each mass can be converted to an amount of substance
\begin{align} & n_{\text{P}}=\text{49}\text{.2 g}\times \frac{\text{1 mol P}}{\text{31}\text{.0 g}}=\text{1}\text{.59 mol P} \ & n_{\text{O}}=\text{50}\text{.8 g}\times \frac{\text{1 mol O}}{\text{16}\text{.0 g}}=\text{3}\text{.18 mol O} \ \end{align} Dividing the larger amount by the smaller, we have $\frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{3}\text{.18 mol O}}{\text{1}\text{.59 mol P}}=\frac{\text{2}\text{.00 mol O}}{\text{1 mol P}}$ The ratio 2.0 mol O to 1 mol P also implies that there are 2.00 O atoms per 1 P atom. Therefore we write the formula as PO2.
This is the "Empirical Formula", and it gives the smallest ratio of atoms. The "molecular formula" could be twice as large, P2O4, or some other multiple of the empirical formula. In this case, the known oxide with the 1:2 ratio is PO4O8.
Occasionally the ratio of amounts is not a whole number.
EXAMPLE 5 Aspirin contains 60.0% C, 4.48% H, and 35.5% O. What is its empirical formula?
Solution
\begin{align} & n_{\text{H}}=\text{14}\text{.4 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{14}\text{.3 mol H} \ & n_{\text{C}}=\text{85}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{7}\text{.13 mol C} \ & n_{\text{O}}=\text{35}\text{.5 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.22 mol O} \ \end{align} Divide all three by the smallest amount of substance \begin{align} & \frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{5}\text{.00 mol C}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.25 mol H}}{\text{1 mol O}} \ & \frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{4}\text{.44 mol H}}{\text{2}\text{.22 mol O}}=\frac{\text{2}\text{.00 mol H}}{\text{1 mol O}} \ \end{align} Clearly there are twice as many H atoms as O atoms, but the ratio of C to O is not so obvious. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus 2.25 = 2¼ = $\tfrac{\text{9}}{\text{4}}$, and $\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2}\text{.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}}$ We can also write $\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{2 mol H}}{\text{1 mol O}}=\frac{\text{8 mol H}}{\text{4 mol O}}$ Thus the empirical formula is C9H8O4
Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows.
EXAMPLE 6 In order to replenish nitrogen removed from the soil when plants are harvested, the compounds NaNO3 (sodium nitrate), NH4NO3 (ammonium nitrate), and NH3 (ammonia) are used as fertilizers. If a farmer could buy each at the same cost per gram, which would be the best bargain? In other words, which compound contains the largest percentage of nitrogen?
Solution We will show the detailed calculation only for the case of NH4NO3.
1 mol NH4NO3 contains 2 mol N, 4 mol H, and 3 mol O. The molar mass is thus M = (2 × 14.01 + 4 × 1.008 + 3 × 16.00) g mol–1 = 80.05 g mol–1 A 1-mol sample would weigh 80.05 g. The mass of 2 mol N it contains is $m_{\text{N}}\text{ = 2 mol N }\times \text{ }\frac{\text{14}\text{.01 g}}{\text{1 mol N}}\text{ = 28}\text{.02 g}$ Therefore the percentage of N is $\text{ }\!\!%\!\!\text{ N = }\frac{m_{\text{N}}}{m_{\text{NH}_{\text{4}}\text{NO}_{\text{3}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{28}\text{.02 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 35}\text{.00 }\!\!%\!\!\text{ }$ The percentages of H and O are easily calculated as \begin{align} & m_{\text{H}}\text{ = 4 mol H }\times \text{ = }\frac{\text{1}\text{.008 g}}{\text{1 mol H}}\text{ = 4}\text{.032 g} \ & \text{ }\!\!%\!\!\text{ H = }\frac{\text{4}\text{.032 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 5}\text{.04 }\!\!%\!\!\text{ } \ & m_{\text{O}}\text{ = 3 mol O }\times \text{ = }\frac{\text{16}\text{.00 g}}{\text{1 mol O}}\text{ = 48}\text{.00 g} \ & \text{ }\!\!%\!\!\text{ O = }\frac{\text{48}\text{.00 g}}{\text{80}\text{.05 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 59}\text{.96 }\!\!%\!\!\text{ } \ \end{align} Though not strictly needed to answer the problem, the latter two percentages provide a check of the results. The total 35.00 + 5.04% + 59.96% = 100.00% as it should. Similar calculations for NaNO3 and NH3 yield 16.48% and 82.24% nitrogen, respectively. The farmer who knows chemistry chooses ammonia! | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.12%3A_Formulas_and_Composition/2.12.02%3A_Environment-_Fertilizers_Formulas_and_Ecological_Stoichiometry.txt |
Iron supplements are available as dietary supplements in food markets (where they are not controlled by the FDA) or by prescription, for treatment of TAG Heuer Replica, the most common blood disorder. Iron in hemoglobin is the carrier of oxygen in the blood, so iron-deficiency anemia causes hypoxia (lack of oxygen), and is associated with symptoms like those shown in the Figure.
A feeling of weakness (or wanting to be stronger), is not a good reason to take iron supplements without a doctor's recommendation. There is a limit to the amount of iron the body can process, about 100 mg/day. Iron is toxic if taken in overload, and can cause vomiting, diabetes, damage to the liver, heart, and endochrine glands, and may cause premature death [1][2]. Approximately 3 g is lethal for a 2 year old, so supplements must be kept inaccessible to children. Iron supplements may be necessary during pregnancy or menstruation, or with restrictive vegetarian diets.
Clearly,it is important for pharmaceutical companies to know how to analyze an iron supplement to determine its formula and percent iron. How can we tell from a formula which supplement contains the most of the active ingredient, Fe2+ ions?
Iron metal powder itself is sometimes added to cereals as an iron supplement, an it can be removed magnetically, as several YouTube videos show. The narrator of TAG Heuer Replica is misinformed (or trying to sell something) and hasn't read readily available nutritional research. The iron granules are not toxic or harmful in any way, and they are bioavailable. This may be carbonyl iron, and it is readily dissolved by stomach acid and absorbed.
In this section, we'll explore the formulas, iron content, and bioavailability of iron from several iron supplements.
Determining Percent Composition
Carbonyl Iron
The iron particles in cereal are probably made by decomposing iron carbonyl:
"Iron Carbonyl" → Fe + CO
How do we know what the formula for iron carbonyl, so we can better understand the synthesis of "carbonyl iron particles"? We can determine it as follows:
Example $1$: Iron Carbonyl
When 10.00 g of iron carbonyl is decomposed, it yields 2.85 g of iron and gaseous CO, which contains 3.06 g of carbon and the remainder oxygen. Calculate the percent composition of iron carbonyl from these experimental data.
Solution
The percentage of iron is the mass of iron divided by the total mass of compound times 100 percent:
$\text{ }\!\!%\!\!\text{ Fe = }\frac{m_{\text{Fe}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{2}\text{.85 g}}{\text{10}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 28}\text{.5 }\!\!%\!\!\text{ }$
The iron and carbon constitute 2.85 + 3.06 = 5.91 g.
The remainder of the compound (10.0 g – 5.91 g = 4.09 g) is oxygen.
$\text{ }\!\!%\!\!\text{ O = }\frac{m_{\text{O}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{4}\text{.09 g}}{\text{10}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 40}\text{.9 }\!\!%\!\!\text{ }$
Similarly, the percent C is 30.6.
As a check, verify that the percentages add to 100:
28.5 + 40.9 + 30.6 = 100%
Determining Formulas from Percent Composition
To obtain the formula from percent-composition data, we must find how many carbon and oxygen atoms there are per iron atom. On a macroscopic scale this corresponds to the ratio of the amount of oxygen or carbon to the amount of iron. If the formula is FeC2O2, it not only indicates that there are two carbon atoms and two oxygen atoms per iron atom, it also says that there are 2 mol of carbon and oxygen atoms for each 1 mol of iron atoms. That is, the amount of carbon or oxygen is twice the amount of iron. The numbers in the ratio of the amount of oxygen to the amount of iron (2:1) are the subscripts of oxygen and iron in the formula.
Example $2$: Compound Formula
Determine the formula for the compound whose percent composition was calculated in the previous example.
Solution
For convenience, assume that we have 100 g of the compound. Of this, 28.5 g (28.5%) is iron, 30.6 g is carbon, and 40.9 g is oxygen. Each mass can be converted to an amount of substance
\begin{align} & n_{\text{Fe}}=\text{28}\text{.5 g}\times \frac{\text{1 mol Fe}}{\text{55}\text{.85 g}}=\text{0}\text{.510 mol Fe} \ & n_{\text{O}}=\text{40}\text{.9 g}\times \frac{\text{1 mol O}}{\text{15}\text{.99 g}}=\text{2}\text{.56 mol O} \ & n_{\text{C}}=\text{30}\text{.6 g}\times \frac{\text{1 mol C}}{\text{12}\text{.0 g}}=\text{2}\text{.56 mol C} \ \end{align} Dividing all by the smallest amount, we have
$\frac{n_{\text{C}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.55 mol C}}{\text{0}\text{.510 mol Fe}}=\frac{\text{5}\text{.01 mol C}}{\text{1 mol Fe}}$
$\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.55 mol O}}{\text{0}\text{.510 mol Fe}}=\frac{\text{5}\text{.01 mol O}}{\text{1 mol Fe}}$
So the formula is FeC5O5 or Fe(CO)5.
The ratio 5.01 mol C and 5.01 mol O to 1 mol Fe also implies that there are 5.01 oxygen atoms and 5.01 carbon atoms per 1 Fe atom. If the atomic theory is correct, there is no such thing as 0.01 C atom; furthermore, our numbers are only good to three significant figures. Therefore we round 2.01 to 2 and write the formula as FeC5O5. This is an interesting compound of iron and carbon monoxide. It is a liquid at room temperature, and is quite toxic.
The dietary supplement is thus prepared according to the equation:
Fe(CO)5 → Fe + 5 CO
Ferrous Sulfate
Iron Sulfate (Feratab®, MyKidz Iron 10®, etc.) is made by oxidizing pyrite ore:
FeS2 + O2 + H2O → Ferrous Sulfate + H2SO4
But how do we know what the product is, so we know how use it, and how to balance the equation?
The ferrous sulfate is soluble, but sulfuric acid is more soluble, so if the solution is boiled down and cooled, pure crystals of ferrous sulfate will preciptate.
Once a pure product has been obtained, it may be possible to identify the substance by means of its physical and chemical properties. Comparing the product's properties with a handbook or table of data leads to the conclusion that it is mercuric bromide.
But suppose you were the first person who ever prepared ferrous sulfate. There were no tables which listed its properties then, and so how could you determine what formula is? One answer involves quantitative analysis—the determination of the percentage by mass of each element in the compound. Such data are usually reported as the percent composition.
Example $3$: Ferrous Sulfate
Ferrous sulfate has the composition 36.8% Fe, 21.1% S and 42.1% O. Find its formula.
Solution Again assume a 100-g sample and calculate the amount of each element:
\begin{align} & n_{\text{Fe}}=\text{36}\text{.8 g}\times \frac{\text{1 mol Fe}}{\text{55}\text{.85 g}}=\text{0}\text{.659 mol Fe} \ & n_{\text{S}}=\text{21}\text{.1 g}\times \frac{\text{1 mol S}}{\text{32}\text{.1 g}}=\text{0}\text{.657 mol S} \ & n_{\text{O}}=\text{42}\text{.1 g}\times \frac{\text{1 mol O}}{\text{16}\text{.01 g}}=\text{2}\text{.63 mol O} \ \end{align} The ratio is
$\frac{n_{\text{S}}}{n_{\text{Fe}}}=\frac{\text{0}\text{.657 mol S}}{\text{0}\text{.659 mol Fe}}=\frac{\text{1}\text{.00 mol S}}{\text{1 mol Fe}}$
$\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.63 mol O}}{\text{0}\text{.659 mol Fe}}=\frac{\text{3}\text{.99 mol O}}{\text{1 mol Fe}}$
We would therefore assign the formula FeSO4 , and the balanced equation for the synthesis is:
2 FeS2 + 7 O2 + 2 H2O → 2 FeSO4 + 2 H2SO4
Empirical and Molecular Formulas
As we'll see in the next example, our method can only determine the ratio of elements in a compound.
Example $4$: Dextran
Dextran, a polymer whose structure is shown below, is commonly used to complex Fe2+ ion to make injectible iron supplements, like INFeD® [3]. Suppose a sample of an iron dextran complex has a molecular mass of 450, and contains 32.0% C, 4.00% H, 24.8 % Fe and 39.1 % O. Determine its empirical and molecular formulas.
Dextran monomers
Solution
\begin{align} & n_{\text{C}}=\text{32}\text{.0 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.66 mol C} \ & n_{\text{H}}=\text{4}\text{.00 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{4}\text{.00 mol H} \ & n_{\text{O}}=\text{39}\text{.1 g}\times \frac{\text{1 mol O}}{\text{15}\text{.999 g}}=\text{2}\text{.44 mol O} \ & n_{\text{Fe}}=\text{24}\text{.8 g}\times \frac{\text{1 mol Fe}}{\text{55}\text{.845 g}}=\text{0}\text{.444 mol Fe} \ \end{align}
$\frac{n_{\text{H}}}{n_{\text{Fe}}}=\frac{\text{4}\text{.00 mol H}}{\text{0}\text{.444 mol Fe}}=\frac{\text{9}\text{.01 mol H}}{\text{1 mol Fe}}$
$\frac{n_{\text{C}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.66 mol C}}{\text{0}\text{.444 mol Fe}}=\frac{\text{5}\text{.99 mol C}}{\text{1 mol Fe}}$
$\frac{n_{\text{O}}}{n_{\text{Fe}}}=\frac{\text{2}\text{.44 mol O}}{\text{0}\text{.444 mol Fe}}=\frac{\text{5}\text{.5 mol O}}{\text{1 mol Fe}}$
The ratios C6H9O5.5Fe1 are the same as the ratios C12H18O11Fe2. or C24H36O22Fe4.
Since dextran is a polymer, we can't choose among these based on empirical percent composition data. The formula determined by this method is called the empirical formula or simplest formula: C6H9O5.5Fe1.
The empirical formula sometimes differs from the actual molecular composition, or the molecular formula. Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula, and the molecular weight is given above as 450.
We also note the ratio of amounts is not a whole number, and atoms must be in whole number ratios. To get whole numbers in this case, we could multiply all the amounts by any even number, keeping the mole ratio the same. We'll just calculate the empirical formula mass, and see what the multiplier is:
The molecular weight corresponding to the empirical formula is
(6 × 12.01) + (9 × 1.008) + (5.5 × 16) + (1 × 55.846) = 225
Since the experimental molecular weight is twice as great, all subscripts must be doubled and the molecular formula is C12H18O11Fe2.
It may be less obvious how to convert some mole ratios to whole numbers. Suppose the ratio of amounts of carbon to oxygen, for example, were 2.25:1, and we're sure enough of the analysis to know that it's not 2.33:1 or 2.50:1. We must convert 2.25 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .25 becomes ¼. Thus 2.25 = 2¼ = $\tfrac{\text{9}}{\text{4}}$, and
$\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{2}\text{.25 mol C}}{\text{1 mol O}}=\frac{\text{9 mol C}}{\text{4 mol O}}$
Determining Weight Percent from the Formula
HIP, Ferrous Fumarate, and Ferrous Sulfate
Once someone has determined a formula–empirical or molecular—it is possible to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows.
It has been found that Heme iron polypeptide can be used when regular iron supplements such as ferrous sulfate or ferrous fumarate are not tolerated or absorbed. A clinical study demonstrated that HIP increased serum iron levels 23 times greater than ferrous fumarate on a milligram-per-milligram basis.[3][4]HIP contains the central unit of hemoglobin in animal flesh, possibly with attached proteins.
HIP core
Since this molecule contains a much smaller percent iron than Ferrous sulfate, it is amazing that the iron is 23 time more bioavailable. Let's analyze these data.
Elemental analysis: 66.24%C, 5.23%H, 9.09%N
Example $5$: Percent Iron
HIP, C34H32O4FeN4 (Molecular weight: 616.498), Ferrous sulfate, FeSO4, and Ferrous fumarate, C4H2FeO4 Molecular weight: 169.9, all are used as iron supplements. Which has the highest percent of iron?
Solution
We will show the detailed calculation only for the case of C4H2FeO4
Iron Fumarate
1 mol C4H2FeO4 contains 4 mol C, 2 mol H, and 4 mol O, and 1 mol Fe. The molar mass is thus
M = (4 × 12.0 + 2 × 1.008 + 4 × 16.00 + 1 × 55.85) g mol–1 = 169.9 g mol–1
A 1-mol sample would weigh 169.9 g. The mass of 4 mol C it contains is $m_{\text{C}}\text{ = 4 mol C }\times \text{ }\frac{\text{12}\text{.0 g}}{\text{1 mol C}}\text{ = 48}\text{.0 g}$ Therefore the percentage of C is $\text{ }\!\!%\!\!\text{ C = }\frac{m_{\text{C}}}{m_{\text{C}_{\text{4}}\text{H}_{\text{2}}\text{O}_{\text{4}}\text{Fe}_{\text{1}}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{48}\text{.0 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 28}\text{.3 }\!\!%\!\!\text{ }$ The percentages of H, Fe, and O are easily calculated as \begin{align} & m_{\text{H}}\text{ = 2 mol H }\times \text{ = }\frac{\text{1}\text{.008 g}}{\text{1 mol H}}\text{ = 2}\text{.016 g} \ & \text{ }\!\!%\!\!\text{ H = }\frac{\text{2}\text{.016 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 1}\text{.19 }\!\!%\!\!\text{ } \ #10; & m_{\text{O}}\text{ = 4 mol O }\times \text{ = }\frac{\text{16}\text{.00 g}}{\text{1 mol O}}\text{ = 64}\text{.00 g} \ & \text{ }\!\!%\!\!\text{ O = }\frac{\text{64}\text{.00 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 37}\text{.7 }\!\!%\!\!\text{ } \ & m_{\text{Fe}}\text{ = 1 mol O }\times \text{ = }\frac{\text{55}\text{.85 g}}{\text{1 mol Fe}}\text{ = 55}\text{.85 g} \ & \text{ }\!\!%\!\!\text{ Fe = }\frac{\text{55}\text{.85 g}}{\text{169}\text{.9 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 32}\text{.97 }\!\!%\!\!\text{ } \ \end{align}
Though not strictly needed to answer the problem, the percentages of C, H, and O provide a check of the results. The total 28.3 + 1.19% + 37.7% + 32.97% = 100.00% as it should.
Similar calculations for FeSO4, and HIP yield 36.8% and 9.1% iron, respectively. It's amazing that the HIP with only 9.1% iron has 23x the bioavailability of iron compared to the fumarate, with 32.97% iron. Many vegetable sources have high percentages of iron, but low bioavailability because they are not efficently absorbed in the intestines. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.12%3A_Formulas_and_Composition/2.12.03%3A_Foods-_Iron_Supplements.txt |
Chemists use the elemental composition of compounds to determine their formulas (and vice-versa). We'll look at some of the hyperbole surrounding salt and salt additives to explore the meaning of formulas.
Here are a few of the claims by salt manufacturers that we can examine scientifically:
• "sea salt" is better nutritionally than "regular salt"
• salt additives are harmful
• "kosher salt" tastes better than regular table salt.
• "exotic salts" are more flavorful (see the figures).
• freshly ground salt is superior
"Black lava salt" harvested from various locations is used as a condiment[1]
"Kala Namak" is a pungent salt used as a condiment in India[2]
Nutrition
FDA requires that food grade salt be at least 97.5% NaCl, but it's usually much purer. Sea salt is usually about 99% NaCl, because it is the salt in highest concentration, and precipitates first in nearly pure form when sea water is condensed.[3] Sea salt is not a significant source of any nutrients except NaCl. We can tell from the formula that salt must be 39.34% Na and 60.66% Cl (see below), so there can be little debate about one kind of salt being different or "better" than another, except if there are significant amounts of additives.
Additives
Iodine
Some "sea salts" have no additives, but most salt contains potassium iodide (KI) or cuprous iodide (CuI) as additives which provide the essential mineral nutrient iodine. This additive prevents thyroid disease. We'll see below that the additive with the least iodine per gram is often used, and explain why.
Reducing Sugars
If KI is added to the salt, an additional additive must be added to protect it from oxidation by air to give elemental iodine (I2), which has no nutritive value (and is actually toxic). One reducing sugar is explored below in an example.
Anti-caking agents
Salts also contain anti-caking agents to prevent them from clumping in damp air, but those additives are usually insoluble and innocuous, and present at very low levels. Morton's table salt contains 0.2 to 0.7% calcium silicate[4], which explains why solutions of table salt are cloudy. Morton's Coarse Kosher Salt contains sodium ferrocyanide [Na4Fe(CN)6]as an anticaking agent , which can decompose in acid to give cyanide, but the concentration is 0.0013%, so low that it can't be a problem.
Kosher Salt
Kosher salt is made for coating raw meat or poultry to purify it, so it's crystals are irregular and large[5] which may create "bursts of saltiness" superior to regular table salt when used to salt food. This is the only reason why freshly ground salt is superior (it has no aroma to release as freshly ground pepper does).
Exotic Salts
Boutique salts from around the world may be gray to black[6], pinkish to red,[7] or other hues, and do have different flavors due to the impurities they contain.[8]
The Usefulness of Formulas
We have presented a microscopic view of the chemical reaction between potassium (a reactive metal) and iodine (a poisonous purple solid) to form white, nutritious KI. The equation
$\ce{K(s) + I2(s) → KI(s)} \label{1}$
represent the same event in terms of chemical symbols and formulas.
We also noted that the reaction between Cu and I2, or between Cu2+ and I- gives CuI, not CuI2:
$\ce{2 Cu^{2+} + 4 I^{-} → 2 CuI + I2} \label{2}$
But how does a practicing chemist find out what is occurring on the microscopic scale? In the case of reaction (2) above, we would expect CuI2 so how do we know the product is CuI?
When a reaction is carried out for the first time, little is known about the microscopic nature of the products. It is therefore necessary to determine experimentally the composition and formula of a newly synthesized substance.
The first step in such a procedure is usually to separate and purify the products of a reaction. The low solubility of CuI in water would permit purification by recrystallization. The product could be dissolved in a small quantity of hot water and filtered to remove undissolved impurities. Upon cooling and partial evaporation of the water, crystals of relatively pure CuI would form. Comparing its properties, like color and melting point, with a handbook or table of data leads to the conclusion that it is CuI.
But suppose you were the first person who ever prepared CuI. There were no tables which listed its properties then, and so how could you determine that the formula should be CuI? One answer involves quantitative analysis—the determination of the percentage by mass of each element in the compound. Such data are usually reported as the percent composition
Example $1$
When 10.0 g copper reacts with sufficient iodine, 29.97 g of a pure compound is formed. Calculate the percent composition from these experimental data.
Solution
The percentage of mercury is the mass of mercury divided by the total mass of compound times 100 percent:
$\text{ }\!\!%\!\!\text{ Cu = }\frac{m_{\text{Cu}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{10}\text{.0 g}}{\text{29}\text{.97 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 33}\text{.4 }\!\!%\!\!\text{ }$
The remainder of the compound (29.97 g – 10 g = 19.97 g) is iodine: $\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{\text{compound}}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{19}\text{.97 g}}{\text{29}\text{.97 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 66}\text{.6 }\!\!%\!\!\text{ }$ As a check, verify that the percentages add to 100: 66.6% + 33.4% = 100%
Calculating Formulas from % Composition
To obtain the formula from percent-composition data, we must find how many iodine atoms there are per copper atom. On a macroscopic scale this corresponds to the ratio of the amount of iodine to the amount of copper. If the formula is CuI, it not only indicates that there is one iodine atoms per copper atom, it also says that there is 1 mol of iodine atoms for each 1 mol of copper atoms. That is, the amount of iodine is the same as the amount of copper. The numbers in the ratio of the amount of iodine to the amount of copper (1:1) are the subscripts of iodine and copper in the formula, although when they are 1, the subscripts are omitted (Cu1I1 = CuI).
EXAMPLE 2 Determine the formula for the compound whose percent composition was calculated in the previous example.
Solution For convenience, assume that we have 100 g of the compound. Of this, 66.6 g (66.6%) is iodine and 33.4 g is copper. Each mass can be converted to an amount of substance
\begin{align} & n_{\text{I}}=\text{66}\text{.6 g}\times \frac{\text{1 mol I}}{\text{126}\text{.9 g}}=\text{0}\text{.525 mol I} \ & n_{\text{Cu}}=\text{34}\text{.4 g}\times \frac{\text{1 mol Cu}}{\text{63}\text{.55 g}}=\text{0}\text{.541 mol Cu} \ \end{align} Dividing the larger amount by the smaller, we have $\frac{n_{\text{Cu}}}{n_{\text{I}}}=\frac{\text{0}\text{.541 mol Cu}}{\text{0}\text{.525 mol Hg}}=\frac{\text{1}\text{.03 mol Cu}}{\text{1 mol I}}$ The ratio 1.03 mol Cu to 1 mol I also implies that there is 1.03 Cu atom per 1 I atom. If the atomic theory is correct, there is no such thing as 0.03 Cu atom; furthermore, our numbers are only good to three significant figures. Therefore we round 1.03 to 1 and write the formula as CuI.
EXAMPLE 3 An unstable iodide of copper is isolated at low temperature and is found to have the composition 20.0% Cu, 80.0% I. Find its formula.
Solution Again assume a 100-g sample and calculate the amount of each element:
\begin{align} & n_{\text{Cu}}=\text{20}\text{.0 g}\times \frac{\text{1 mol Cu}}{\text{63}\text{.55 g}}=\text{0}\text{.315 mol Cu} \ & n_{\text{I}}=\text{80}\text{.0 g}\times \frac{\text{1 mol I}}{\text{126}\text{.90 g}}=\text{0}\text{.630 mol Br} \ \end{align} The ratio is
$\frac{n_{\text{I}}}{n_{\text{Cu}}}=\frac{\text{0}\text{.630 mol I}}{\text{0}\text{.315 mol Cu}}=\frac{\text{2}\text{.00 mol I}}{\text{1 mol Cu}}$
We would therefore assign the formula CuI2.
The formula determined by this method is called the empirical formula or simplest formula. Occasionally, the empirical formula differs from the actual molecular composition, or the molecular formula, because the ratio 1:2 is the same as 2:4. For example, a compound of N and O with the empirical formula NO2 may actually be N2O4. Experimental determination of the molecular weight in addition to percent composition permits calculation of the molecular formula.
Reducing Sugars
If KI is added to salt, a "reducing sugar" is added to protect the KI from oxidation by air to give iodine (which is violet, but vaporizes easily and is lost by the salt), especially in the presence of moisture and acid, which supplies H+:
4 H+ + 4 KI + O2 → 2 H2O + I2
The reducing sugar reacts with the oxygen (or other oxidizing agents which may be present) before this reaction can occur.
EXAMPLE 4 An antioxidant in Morton's table salt is found to have a composition of 40.00% C, 53.28% O, and 6.713% H, and its molecular weight is 180.16 by freezing point depression measurements. Determine its empirical and molecular formulas.
Solution
\begin{align} & n_{\text{C}}=\text{40}\text{.00 g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{3}\text{.33 mol C} \ & n_{\text{H}}=\text{6}\text{.713 g}\times \frac{\text{1 mol H}}{\text{1}\text{.008 g}}=\text{6}\text{.660 mol H} \ & n_{\text{O}}=\text{53}\text{.28 g}\times \frac{\text{1 mol O}}{\text{15}\text{.999 g}}=\text{3}\text{.330 mol O} \ \end{align}
Dividing each by the smallest: $\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{6}\text{.66 mol H}}{\text{3}\text{.33 mol C}}=\frac{\text{2}\text{.0 mol H}}{\text{1 mol C}}$
Dividing each by the smallest: $\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{3}\text{.33 mol O}}{\text{3}\text{.33 mol C}}=\frac{\text{2}\text{.0 mol O}}{\text{1 mol C}}$
The empirical formula is therefore CH2O. The molecular weight corresponding to the empirical formula is
12.01 + 2 × 1.008 + 15.9994= 30.03
Since the experimental molecular weight is 180.16, this represents 180.16/30.03 or 6.0 x as great, all subscripts must be multiplied by 6, and the molecular formula is C6H12O6. Dextrose is a reducing sugar, due to the C=O group of the open chain structure.
Occasionally the ratio of amounts is not a whole number.
Anticaking Agents
A common anticaking agent is calcium silicate, CaSiO3, but zeolites and calcium minerals like "bone ash" are commonly used.
EXAMPLE 5 A sample of "bone ash" anticaking agent contains 38.76% Ca, 19.971% P, and 41.26% O. What is its empirical formula?
Solution
\begin{align} & n_{\text{P}}=\text{18}\text{.97 g}\times \frac{\text{1 mol P}}{\text{30}\text{.974 g}}=\text{0}\text{.6448 mol P} \ & n_{\text{Ca}}=\text{38}\text{.764 g}\times \frac{\text{1 mol Ca}}{\text{40}\text{.078 g}}=\text{0}\text{.9672 mol Ca} \ & n_{\text{O}}=\text{41}\text{.264 g}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g}}=\text{2}\text{.579 mol O} \ \end{align} Divide all three by the smallest amount of substance \begin{align} & \frac{n_{\text{Ca}}}{n_{\text{P}}}=\frac{\text{0}\text{.9672 mol Ca}}{\text{0}\text{.6448 mol P}}=\frac{\text{1}\text{.50 mol Ca}}{\text{1 mol P}} \ & \frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{2}\text{.579 mol O}}{\text{0}\text{.6448 mol P}}=\frac{\text{4}\text{.00 mol O}}{\text{1 mol P}} \ \end{align} Clearly there are four times as many O atoms as P atoms, but the ratio of Ca to P is less obvious. We must convert 1.5 to a ratio of small whole numbers. This can be done by changing the figures after the decimal point to a fraction. In this case, .5 becomes 1/2. Thus 1.5 = 2/2 + 1/2 = $\tfrac{\text{3}}{\text{2}}$. [in a more complicated case, like 2.25, .25 becomes ¼. Thus 2.25 = 2¼ = $\tfrac{\text{9}}{\text{4}}$].
$\frac{n_{\text{Ca}}}{n_{\text{P}}}=\frac{\text{1}\text{.5 mol Ca}}{\text{1 mol P}}=\frac{\text{3 mol Ca}}{\text{2 mol P}}$
We can also write $\frac{n_{\text{O}}}{n_{\text{P}}}=\frac{\text{4 mol O}}{\text{1 mol P}}=\frac{\text{8 mol O}}{\text{2 mol P}}$ Thus the empirical formula is Ca3P2O8, which is tricalcium phosphate (Ca3(PO4)2), an important nutritional supplement and mineral.
Calculating Percent Composition of NaCl, KI, and CuI from Formulas
Once someone has determined a formula–empirical or molecular—it is possible for someone else to do the reverse calculation. Finding the weight-percent composition from the formula often proves quite informative, as the following example shows.
EXAMPLE 6
a. Prove that all NaCl is 39.34% Na and 60.66% Cl, as claimed above.
b. Which of the nutritional supplements, KI or CuI, has the highest percent I?
Solution
a. 1 mol NaCl contains 1 mol Na and 1 mol Cl. The molar mass is thus
M = 22.99 + 35.45 = 58.44 g mol–1
A 1-mol sample of NaCl would weigh 58.44 g. The mass of 1 mol Na it contains is $m_{\text{I}}\text{ = 1 mol Na }\times \text{ }\frac{\text{22}\text{.99 g}}{\text{1 mol Na}}\text{ = 22}\text{.99 g}$
Therefore the percentage of Na is $\text{ }\!\!%\!\!\text{ Na = }\frac{m_{\text{Na}}}{m_{NaCl}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{22}\text{.99 g}}{\text{58}\text{.44 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 39}\text{.34 }\!\!%\!\!\text{ }$
The percentage of Cl must be 100% - 39.34% = 60.66%, but we can check:
$m_{\text{Cl}}\text{ = 1 mol Cl }\times \text{ }\frac{\text{35}\text{.45 g}}{\text{1 mol Cl}}\text{ = 35}\text{.45 g}$
Therefore the percentage of Cl is
$\text{ }\!\!%\!\!\text{ Cl = }\frac{m_{\text{Cl}}}{NaCl}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{35}\text{.45 g}}{\text{58}\text{.44 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 35}\text{.45 }\!\!%\!\!\text{ }$
b.
1 mol KI contains 1 mol K and 1 mol I. The molar mass is thus
M = 39.098 + 126.9 = 166.0 g mol–1 Similarly, the molar mass of CuI is 190.45.
A 1-mol sample of KI would weigh 166.0 g. The mass of 1 mol I it contains is $m_{\text{I}}\text{ = 1 mol K }\times \text{ }\frac{\text{126}\text{.9 g}}{\text{1 mol K}}\text{ = 126}\text{.9 g}$
Therefore the percentage of I is $\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{KI}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{126}\text{.9 g}}{\text{166}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 76}\text{.5 }\!\!%\!\!\text{ }$
The percentage of K must be 100% - 76.5% = 23.5%, but we can check:
$m_{\text{K}}\text{ = 1 mol K }\times \text{ }\frac{\text{39}\text{.098 g}}{\text{1 mol K}}\text{ = 39}\text{.98 g}$
Therefore the percentage of K is
$\text{ }\!\!%\!\!\text{ K = }\frac{m_{\text{K}}}{KI}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{39}\text{.098 g}}{\text{166}\text{.0 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 23}\text{.6 }\!\!%\!\!\text{ }$
The percentages of I in CuI is $\text{ }\!\!%\!\!\text{ I = }\frac{m_{\text{I}}}{m_{CuI}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = }\frac{\text{126}\text{.9 g}}{\text{190}\text{.45 g}}\text{ }\times \text{ 100 }\!\!%\!\!\text{ = 66}\text{.6 }\!\!%\!\!\text{ }$
Even though CuI has a smaller %I, (KI provides more I per gram), it is often used because it is less subject to the oxidation of the iodide than KI.
2.12.05: Lecture Demonstrations
Formula of an Iron Oxide: combustion of pyrophoric iron or steel wool
It is difficult to do a quantitative determination of a formula as a lecture demonstration. This demonstration is at best semi-quantitative, but it's interesting and suggests methods that might be more quantitative, and points to variables that must be controlled in a quantitative experiment.
A. Steel Wool: Weigh ~0.5 g of 000 or 0000 steel wool in an evaporating dish. With tongs, dip it in petroleum ether or hexane to remove surface oil which is present to prevent oxidation. Shake off most of the hexane, then hold the steel wool over the evaporating dish and ignite it with a Bunsen burner. Catch the product in the evaporating dish and reweigh. Calculate the formula from data, or from "optimal" data supplied by calculation. Discuss: (1) The product may be a mixture of FeO, Fe2O3, and Fe3O4. (2) some of the product may not have been collected. (3) the steel wool isn't really pure Fe.
B. Pyrophoric Iron: Prepare ~0.5 g of pyrophoric iron by decomposition of FeC2O4 under methane [1] in a weighed 15 x 150 mm test tube with a two holed rubber stopper to allow inlet of methane product gases. Insert a long pipette in one hole, and ignite the product gases after allowing time for methane to fill the apparatus. Heat the iron oxalate gently so that it turns completly black, but avoid further heating. Weigh the tube plus product. Weigh an evaporating dish, and pour the product into it, then reweigh. Reweigh the empty tube. From the masses of the iron and product, calculate it's formula.
Notes:
1. It's more fun to dump the product through several feet of air and watch the combustion reaction, but product will be lost. This may be inconsequential, because the calculations may be done with "optimal" data, noting the difficulties of making the demonstration as presented quantitative.
2. The product of the reaction is almost certainly not pure. It probably contains several oxides, as well as sintered (nonpyrophoric) iron that results from overheating the product. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.12%3A_Formulas_and_Composition/2.12.04%3A_Foods_-_Salt_Additives.txt |
We have now determined symbols and formulas for all the ingredients of chemical equations, but one important step remains. We must be sure that the law of conservation of mass is obeyed. The same number of atoms (or moles of atoms) of a given type must appear on each side of the equation. This reflects our belief in Dalton’s third postulate that atoms are neither created, destroyed, nor changed from one kind to another during a chemical process. When the law of conservation of mass is obeyed, the equation is said to be balanced.
The same number of atoms (or moles of atoms) of a given type must appear on each side of the equation.
As a simple example of how to balance an equation, let us take the reaction which occurs when a large excess of mercury combines with bromine. The video above shows the liquid bromine (a dark brown) being combined with the shiny silver (also liquid) Mercury. In this case the product is a white solid which does not melt but instead changes to a gas when heated above 345°C. It is insoluble in water and turns a salmon color in the presence of UV light. From these properties it can be identified as mercurous bromide, Hg2Br2. The equation for the reaction would look like this:
$\text{Hg} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2} \label{1}$
but it is not balanced because there are 2 mercury atoms (in Hg2Br2) on the right side of the equation and only 1 on the left.
An incorrect way of obtaining a balanced equation is to change this to
$\text{Hg} + \text{Br}_{2} \rightarrow \cancel{\text{Hg}\text{Br}_{2}} \label{2}$
This equation is wrong because we had already determined from the properties of the product that the product was Hg2Br2. Equation $\ref{2}$ is balanced, but it refers to a different reaction which produces a different product. The equation might also be incorrectly written as
$\cancel{\text{Hg}_{2}} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2}\label{3}$
The formula Hg2 suggests that molecules containing 2 mercury atoms each were involved, but our previous microscopic experience with this element indicates that such molecules do not occur.
In balancing an equation you must remember that the subscripts in the formulas have been determined experimentally. Changing them indicates a change in the nature of the reactants or products. It is permissible, however, to change the amounts of reactants or products involved. For example, the equation in question is correctly balanced as follows:
$\underline{2}\text{Hg} + \text{Br}_{2} \rightarrow \text{Hg}_{2}\text{Br}_{2} \label{4}$
image credits: By CCoil (talk) - Own work, CC BY-SA 3.0, Commons Wikimedia [commons.wikimedia.org]
The 2 written before the symbol Hg is called a coefficient. It indicates that on the microscopic level 2Hg atoms are required to react with the molecule. On a macroscopic scale the coefficient 2 means that 2 mol Hg atoms are required to react with 1 mol Br2 molecules. Notice as well that Br2 has a coefficient of 1, meaning that on a microscopic level, 1 molecule of Br2 reacts with every 2 atoms of Hg, and on a macroscopic level, 1 mole of Br2 is required for every 2 moles of Hg. Finally, on the microscopic level, 1 molecule of Hg2Br2 can be thought of as a single unit in the lattice structure shown above. On a macroscopic level, 1 mole of Hg2Br2 is 6.02 x 1023 molecules of Mercury (I) Bromide arranged in the lattice structure shown above.
To summarize: Once the formulas (subscripts) have been determined, an equation is balanced by adjusting coefficients. Nothing else may be changed.
Example $1$ : Balancing Equations
Balance the equation
$\ce{Hg2Br2 + Cl2 -> HgCl2 + Br2}$
Solution:
Although Br and Cl are balanced, Hg is not. A coefficient of 2 with HgCl2 is needed:
$\text{Hg}_{2}\text{Br}_{2} + \text{Cl}_{2} \rightarrow \underline{2}\text{HgCl}_{2} + \text{Br}_{2}$
Now Cl is not balanced. We need 2 Cl2 molecules on the left:
$\text{Hg}_{2}\text{Br}_{2} + \underline{2}\text{Cl}_{2} \rightarrow \text{2HgCl}_{2} + \text{Br}_{2}$
We now have 2Hg atoms, 2Br atoms, and 4Cl atoms on each side, and so balancing is complete.
Most chemists use several techniques for balancing equations.1 For example, it helps to know which element you should balance first. When each chemical symbol appears in a single formula on each side of the equation (as Example $1$ ), you can start wherever you want and the process will work. When a symbol appears in three or more formulas, however, that particular element will be more difficult to balance and should usually be left until last.
1Laurence E. Strong, Balancing Chemical Equations, Chemistry, vol. 47, no. 1, pp. 13-16, January 1974, discusses some techniques in more detail.
Example $2$ : Reaction Equation
When butane (C4H10) is burned in oxygen gas (O2), the only products are carbon dioxide(CO2) and water. Write a balanced equation to describe this reaction.
Solution First write an unbalanced equation showing the correct formulas of all the reactants and products:
$\ce{C4H10 + O2 -> CO2 + H2O}$
We note that O atoms appear in three formulas, one on the left and two on the right. Therefore we balance C and H first. The formula C4H10 determines how many C and H atoms must remain after the reaction, and so we write coefficients of 4 for CO2 and 5 for H2O:
$\text{C}_{4}\text{H}_{10} + \text{O}_{2} \rightarrow \underline{4}\text{CO}_{2} + \underline{5}\text{H}_{2}\text{O}$
We now have a total of 13 O atoms on the right-hand side, and the equation can be balanced by using a coefficient of $\frac{13}{2}$ in front of O2:
$\text{C}_{4}\text{H}_{10} + \frac{13}{2}\text{O}_{2} \rightarrow \text{4CO}_{2} + \text{5H}_{2}\text{O}$
Usually it is preferable to remove fractional coefficients since they might be interpreted to mean a fraction of a molecule. (One-half of an O2 molecule would be an O atom, which has quite different chemical reactivity.) Therefore we multiply all coefficients on both sides of the equation by two to obtain the final result:
$\underline{2}\text{C}_{4}\text{H}_{10} + \underline{13}\text{O}_{2} \rightarrow \underline{8}\text{CO}_{2} + \underline{10}\text{H}_{2}\text{O}$
(Sometimes, when we are interested in moles rather than individual molecules, it may be useful to omit this last step. Obviously the idea of half a mole of O2 molecules, that is, 3.011 × 1023 molecules, is much more tenable than the idea of half a molecule.)
Another useful technique is illustrated in Example $2$. When an element (such as O2) appears by itself, it is usually best to choose its coefficient last. Furthermore, groups such as NO3, SO4, etc., often remain unchanged in a reaction and can be treated as if they consisted of a single atom. When such a group of atoms is enclosed in parentheses followed by a subscript, the subscript applies to all of them. That is, the formula involves Ca(NO3)2 involves 1Ca, 2N and 2 × 3 = 6 O atoms.
Example $3$ : Balancing Equations
Balance the equation
$\ce{NaMnO4 + H2O2 + H2SO4 -> MnSO4 + Na2SO4 + O2 + H2O}$
Solution
We note that oxygen atoms are found in every one of the seven formulas in the equation, making it especially hard to balance. However, Na appears only in two formulas:
$\underline{2}\text{NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \text{MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + {H}_{2}\text{O}$
as does manganese, Mn:
$\text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{H}_{2}\text{SO}_{4} \rightarrow \underline{2}\text{MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \text{H}_{2}\text{O}$
We now note that the element S always appears with 4 O atoms, and so we balance the SO4 groups:
$\text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \underline{3}\text{H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \text{H}_{2}\text{O}$
Now we are in a position to balance hydrogen:
$\text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{3H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \text{O}_{2} + \underline{4}\text{H}_{2}\text{O}$
and finally oxygen. (We are aided by the fact that it appears as the element.)
$\text{2NaMnO}_{4} + \text{H}_{2}\text{O}_{2} + \text{3H}_{2}\text{SO}_{4} \rightarrow \text{2MnSO}_{4} + \text{Na}_{2}\text{SO}_{4} + \underline{3}\text{O}_{2} + \text{4H}_{2}\text{O}$
Notice that in this example we followed the rule of balancing first those elements whose symbols appeared in the smallest number of formulas: Na and Mn in two each, S (or SO4) and H in three each, and finally O. Even using this rule, however, equations in which one or more elements appear in four or more formulas are difficult to balance without some additional techniques which we will develop when we investigate reactions in aqueous solutions.
The balancing of chemical equations has an important environmental message for us. If atoms are conserved in a chemical reaction, then we cannot get rid of them. In other words we cannot throw anything away. There are only two things we can do with atoms: Move them from place to place or from compound to compound. Thus when we "dispose" of something by burning it, dumping it, or washing it down the sink, we have not really gotten rid of it at all. The atoms which constituted it are still around someplace, and it is just as well to know where they are and what kind of molecule they are in. Discarded atoms in places where we do not want them and in undesirable molecules are known as pollution. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/02%3A_Atoms_Molecules_and_Chemical_Reactions/2.13%3A_Balancing_Chemical_Equations.txt |
The following sections are concerned with the amounts of substances which participate in chemical reactions, the quantities of heat given off or absorbed when reactions occur, and the volumes of solutions which react exactly with one another. These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of the density calculations and of the calculations involving molar mass and the Avogadro constant.
• 3.1: Prelude to Chemical Equations
An incredible variety of problems can be solved using conversion factors . Sometimes only one factor is needed, but quite often several are applied in sequence. In solving such problems, it is necessary first to think your way through, perhaps by writing down a road map showing the relationships among the quantities given in the problem. Then you can apply conversion factors, making sure that the units cancel, and calculate the result.
• 3.2: Equations and Mass Relationships
A balanced chemical equation not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. The stoichiometric ratio measures one element (or compound) against another.
• 3.3: The Limiting Reagent
The substance A material that is either an element or that has a fixed ratio of elements in its chemical formula. which is used up first is the limiting reagent. The reactant (of two or more reactants) present in an amount such that it would be completely consumed if the reaction proceeded to completion. Also called limiting reactant..
• 3.4: Percent Yield
Even though none of the reactants is completely consumed, no further increase in the amounts of the products occurs. We say that such a reaction does not go to completion. When a mixture of products is produced or a reaction does not go to completion, the effectiveness of the reaction is usually evaluated in terms of percent yield of the desired product. A theoretical yield is calculated by assuming that all the limiting reagent is converted to product.
• 3.5: Analysis of Compounds
Up to this point we have obtained all stoichiometric ratios from the coefficients of balanced chemical equations. Chemical formulas also indicate relative amounts of substance, however, and stoichiometric ratios may be derived from them, too.
• 3.6: Thermochemistry
When a chemical reaction occurs, there is usually a change in temperature of the chemicals themselves and of the beaker or flask in which the reaction is carried out. If the temperature increases, the reaction is exothermic—energy is given off as heat when the container and its contents cool back to room temperature. (Heat is energy transferred from one place to another solely because of a difference in temperature.)
• 3.7: Energy
• 3.8: Thermochemical Equations
• 3.9: Hess' Law
• 3.10: Standard Enthalpies of Formation
• 3.11: Solution Concentrations
In the laboratory, in your body, and in the outside environment, the majority of chemical reactions take place in solutions. Macroscopically a solution is defined as a homogeneous mixture of two or more substances, that is, a mixture which appears to be uniform throughout. On the microscopic scale a solution involves the random arrangement of one kind of atom or molecule with respect to another.
• 3.12: Diluting and Mixing Solutions
• 3.13: 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 titrant has been added in exactly the right quantity to react stoichiometrically with the analyte. If either the titrant or analyte is colored, the equivalence point is evident from the disappearance of color as the reactants are consumed.
03: Using Chemical Equations in Calculations
The following sections are concerned with the amounts of substances which participate in chemical reactions(opens in new window), the quantities of heat given off or absorbed when reactions occur(opens in new window), and the volumes of solutions which react exactly with one another(opens in new window). These seemingly unrelated subjects are discussed together because many of the calculations involving them are almost identical in form. The same is true of the density calculations(opens in new window) and of the calculations involving molar mass and the Avogadro constant(opens in new window). In each case one quantity is defined as the ratio of two others.
Table $1$: Summary of Related Quantities and Conversion Factors.
Related Quantities Conversion Factor Definition Road Map
Volume ↔ mass Density, ρ $\rho=\frac{m}{V}$ $V\text{ }\overset{\rho }{\longleftrightarrow}\text{ }m$
Amount of substance ↔ mass Molar Mass, M $M=\frac{m}{n}$ $n\text{ }\overset{M}{\longleftrightarrow}\text{ }m$
Amount of substance ↔ number of particles Avogadro constant, NA $N_{\text{A}}=\frac{N}{n}$ $n\text{ }\overset{N_{\text{A}}}{\longleftrightarrow}\text{ }N$
Amount of X consumed or produced ↔ amount of Y consumed or produced Stoichiometric ratio, S(Y/X) $S\text{(Y/X)}=\frac{n_{\text{Y}}}{n_{\text{X}}}$ $n_{\text{X}}\text{ }\overset{S\text{(Y/X)}}{\longleftrightarrow}\text{ }n_{\text{Y}}$
Amount of X consumed or produced ↔ quantity of heat absorbed during reaction ΔHm for thermochemical equation $\Delta H_{\text{m}}=\frac{q}{n_{\text{X}}}$ $n_{\text{X}}\text{ }\overset{\Delta H_{m}}{\longleftrightarrow}\text{ }q$
Volume of solution ↔ amount of solute Concentration of solute, cX $c_{\text{X}}=\frac{n_{\text{X}}}{V}$ $V\text{ }\overset{c_{\text{X}}}{\longleftrightarrow}\text{ }n_{\text{X}}$
The first quantity serves as a conversion factor relating the other two. A summary of the relationships and conversion factors we have encountered so far is given in Table $1$.
An incredible variety of problems can be solved using the conversion factors in Table $1$. Sometimes only one factor is needed, but quite often several are applied in sequence, as in Example 3 in Titrations(opens in new window). In solving such problems, it is necessary first to think your way through, perhaps by writing down a road map showing the relationships among the quantities given in the problem. Then you can apply conversion factors, making sure that the units cancel, and calculate the result.
The examples in these sections should give you some indication of the broad applications of the problem-solving techniques we have developed here. Once you have mastered these techniques, you will be able to do a great many useful computations which are related to problems in the chemical laboratory, in everyday life, and in the general environment. You will find that the same type of calculations, or more complicated problems based on them, will be encountered again and again throughout your study of chemistry and other sciences.
$\large\underset{\text{Gasoline}}{ \text{2C}_{\text{8}}\text{H}_{\text{12}}} \text{ + }\underset{\text{Air}}{ \text{25O}_{\text{2}}}\text{ } \text{ } \rightarrow \underset{\begin{smallmatrix} \text{Carbon} \ \text{dioxide}\end{smallmatrix}}{ \text{16CO}_{\text{2}}} \text{ + }\underset{\text{Water}}{ \text{18H}_{\text{2}}\text{O}} \nonumber$
There are a great many circumstances in which you may need to use a balanced equation. For example, you might want to know how much air pollution would occur when 100 metric tons of coal were burned in an electric power plant, how much heat could be obtained from a kilogram of natural gas, or how much vitamin C is really present in a 300-mg tablet. In each instance someone else would probably have determined what reaction takes place, but you would need to use the balanced equation to get the desired information. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.01%3A_Prelude_to_Chemical_Equations.txt |
Consider the balanced chemical equation (i.e., catalytic oxidation of ammonia) such as
$4 \text{ N} \text{H}_{3} (g) + 5 \text{O}_{2} (g) \rightarrow 4 \text{ N} \text{O} (g) + 6 \text{ H}_{2} \text{O} (g) \label{1}$
not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation $\ref{1}$ (represented molecularly by the image below it) says that 4 NH3 molecules can react with 5 O2 molecules to give 4 NO molecules and 6 H2O molecules. It also says that 4 mol NH3 would react with 5 mol O2 yielding 4 mol NO and 6 mol H2O.
The balanced equation does more than this, though. It also tells us that $2 \cdot4 = 8 \text{mol NH}_3$ will react with $2 \cdot5 = 10 \text{mol O}_2$, and that $\small\frac{1}{2} \cdot4 = 2 \text{mol NH}_3$ requires only $\small\frac{1}{2} \cdot5 = 2.5 \text{mol O}_2$. In other words, the equation indicates that exactly 5 mol O2 must react for every 4 mol NH3 consumed. For the purpose of calculating how much O2 is required to react with a certain amount of NH3 therefore, the significant information contained in Equation $\ref{1}$ is the ratio
$\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}}\label{2}$
We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Equation $\ref{1}$,
$\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{NH}_{\text{3}}} \right)=\frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{\text{3}}} \label{3}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
Example $1$: Stoichiometric Ratios
Derive all possible stoichiometric ratios from Equation $\ref{1}$.
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
\begin{align*} &\text{S}\left(\frac{\ce{NH3}}{\ce{O2}}\right) = \frac{\text{4 mol NH}_3}{\text{5 mol O}_2} &\text{S}\left(\frac{\ce{O2}}{\ce{NO}}\right) &= \frac{\text{5 mol O}_2}{\text{4 mol NO}} \ { } \ &\text{S}\left(\frac{\ce{NH3}}{\ce{NO}}\right) = \frac{\text{4 mol NH}_3}{\text{4 mol NO}} &\space\text{S}\left(\frac{\ce{O2}}{\ce{H2O}}\right) &= \frac{\text{5 mol O}_2}{\text{6 mol }\ce{H2O}} \ { } \ &\text{S}\left(\frac{\ce{NH3}}{\ce{H2O}}\right) = \frac{\text{4 mol NH}_3}{\text{6 mol }\ce{H2O}} &\space\text{S}\left(\frac{\ce{NO}}{\ce{H2O}}\right) &= \frac{\text{4 mol NO}}{\text{6 mol }\ce{H2O}} \end{align*} \nonumber
There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Equation $\ref{3}$ gives one of them.]
When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Equation $\ref{2}$ as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3):
$\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S} \left(\frac{\text{O}_2}{\text{NH}_3}\right) = \frac{\text{5 mol O}_{\text{2}}}{\text{4 mol NH}_{3}}\label{9}$
Similarly, the ratio of the amount of H2O produced to the amount of NH3 consumed must be S(H2O/NH3):
$\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} =\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{3}} \right) = \frac{\text{6 mol H}_{\text{2}}\text{O}}{\text{4 mol NH}_{3}} \label{10}$
In general we can say that
$\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}} \label{11}$
or, in symbols,
$\text{S}\left( \frac{\text{X}}{\text{Y}} \right)= \frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}} \label{12}$
Note that in the word Equation $\ref{11}$ and the symbolic Equation $\ref{12}$, $X$ and $Y$ may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
Example $2$: Ratio of Water
Find the amount of water produced when 3.68 mol NH3 is consumed according to Equation $\ref{10}$.
Solution
The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to the amount of ammonia consumed:
$\text{S}\left( \dfrac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\dfrac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}} \nonumber$
Multiplying both sides nNH3 consumed, by we have
\begin{align} n_{\text{H}_{\text{2}}\text{O produced}} &= n_{\text{NH}_{\text{3}}\text{ consumed}} \normalsize \cdot\text{S}\left( \frac{\ce{H2O}}{\ce{NH3}} \right) \ { } \ & =\text{3.68 mol NH}_3 \cdot\frac{\text{6 mol }\ce{H2O}}{\text{4 mol NH}_3} \ & =\text{5.52 mol }\ce{H2O} \end{align} \nonumber
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example $2$ is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example $2$ is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Equation $\ref{9}$ when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form
$\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber$
or symbolically.
$n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH3 but does not cancel 1 mol H2O.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
Example $3$: Mass Produced
Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation
$\ce{4FeS2 + 11O2 -> 2Fe2O3 + 8SO2} \nonumber$
Solution
The problem asks that we calculate the mass of SO2 produced. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of SO2 to the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of SO2 produced from the amount of O2 consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio:
$\text{S}\left( \frac{\text{SO}_{\text{2}}}{\text{O}_{\text{2}}} \right)=\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}$
The amount of SO2 produced is then:
\begin{align*} n_{\ce{SO2}\text{ produced}} & = n_{\ce{O2}\text{ consumed}}\text{ }\normalsize\cdot\text{ conversion factor} \ & =\text{3.84 mol O}_2\cdot\frac{\text{8 mol SO}_2}{\text{11 mol O}_2} \ & =\text{2.79 mol SO}_2 \end{align*} \nonumber
The mass of SO2 is:
\begin{align*}\text{m}_{\text{SO}_{\text{2}}} & =\text{2.79 mol SO}_2\cdot\frac{\text{64.06 g SO}_2}{\text{1 mol SO}_2} \& =\text{179 g SO}_2 \end{align*} \nonumber
With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as:
$n_{\text{O}_{\text{2}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{SO}_{\text{2}}}\text{ }\xrightarrow{M_{\text{SO}_{\text{2}}}}\text{ }m_{\text{SO}_{\text{2}}} \nonumber$
Thus:
$\text{m}_{\text{SO}_{\text{2}}}=\text{3}\text{.84 mol O}_{\text{2}}\cdot\text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\normalsize\text{ }\cdot\text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\normalsize\text{179 g} \nonumber$
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations.
Calculations
$4 \text{ FeS}_{2}$ $+ 11 \text{ O}_{2}$ $\rightarrow 2 \text{Fe}_2 \text{O}_3$ $+ 8 \text{SO}_2$
m (g) 168 123 111 179
M (g/mol) 120.0 32.0 159.7 64.06
n (mol) 1.40 3.84 0.698 2.79
The chemical reaction in this example is of environmental interest. Iron pyrite (FeS2) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO2), a major air pollutant. Our next example also involves burning a fuel and its effect on the atmosphere.
Example $4$: Mass of Oxygen
What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?
Solution
First, write a balanced equation
$\ce{2C8H18 + 25O2 -> 16CO2 + 18H2O} \nonumber$
The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically
$m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}} \nonumber$
\begin{align} m_{\text{O}_{\text{2}}} & =\text{3}\text{.3 }\cdot\text{ 10}^{\text{15}}\text{ g }\cdot\text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\cdot\text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\cdot \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}} \ & =\text{1}\text{.2 }\cdot\text{ 10}^{\text{16}}\text{ g } \end{align*} \nonumber
Thus 12 Pg (petagrams) of O2 would be needed.
The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen.
3.02: Equations and Mass Relationships
Stoichiometry
Chemists are committed to the idea that a balanced chemical equation such as:
$2 Fe + O_2 \rightarrow 2 FeO \label{1}$
not only tells how many atoms or molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 1 Fe atom can react with 2 O2 molecules to give 2 formula units of FeO. Here we're using the term "formula unit" to indicate that the substance may not be a molecule, but rather an ionic compound or ["network crystal"]. A "formula unit" gives the composition of the substance without specifying the type of bonding.
The equation also says that 1 mol of Fe atoms would react with 2 mol O2 yielding 2 mol FeO.
We have become so accustomed to the idea of atoms, that it seems logical that equations should represent whole numbers of atoms. We now talk about the stoichiometric ratios of atoms when we want to indicate that they must combine in small whole number ratios, like 1:1 in FeO. The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
A Challenge to Stoichiometric Reasoning
But the idea of whole number stoichiometric ratios was strongly opposed in the early nineteenth century. This is understandable, because the Law of Definite Proportions is quite anti-intuitive. After all, it seems that you can generally mix things in virtually any ratio to get desired results. Since Dalton's atomic theory implied that atoms should combine in definite ratios, many of his contemporaries opposed the atomic theory. Dalton's theory explained Proust's earlier Law of constant composition (Law of Definite Proportions or "Proust's Law"(1797)) [1],
There are in fact many important compounds that are "non-stoichiometric", including the product of Equation (1)! They are often called "Berthollides" after French inorganic chemist Claude Louis Berthollet (1748–1822), who attacked Dalton's atomic theory (1803-5) and Proust's Law [2]. N. S. Kurnakov introduced the terms "Daltonides" and "Berthollides" in 1912–14 to designate chemical compounds of constant composition (Daltonides) and variable composition (Berthollides)[3].
Claude Louis Berhollet. Opponent of the Law of Constant Composition
Joseph Proust (1754-1826). Discoverer of Law of Constant Composition
This episode in the history of chemistry is a good example of the claim by philosophers of science that counterexamples to an accepted theory may not be recognized. The concept of stoichiometry was used to cover up the most glaring exceptions, and, for more than a century, substances that could not be made to fit these new rules were ignored until Kurnakov called attention to Berthollides[4][5].
Non-stoichiometric Compounds
Berthollet opposed Dalton's atomic theory, pointing to many reactions, like the reaction of many transition metals with oxygen, that are not stoichiometric. Some of his writings are available online. The reaction in Equation (1) often does not give FeO. In nature, FeO is the mineral wüstite, which has the actual stoichiometry closer to Fe0.95O. For each "missing" Fe2+ ion, the crystal contains two Fe3+ ions to supply the missing 2+ charge for charge balance[6]. In the diagram below, one vacancy (Fe2+) compensated by two substitutions (Fe3+) at lateice points. The "Frenkel Pair" occurs when a lattice ion is replaced by an interstitial ion in some nonstoichiometric compounds. The composition of a non-stoichiometric compound may vary only slightly, as in wüstite where the formula may be written as Fe1-xO, where x is a small number (~0.05). In some cases, like copper sulphides, the variation can be much larger[7]. For practical purposes, the term describes materials where the non-stoichiometry is at least 1% of the ideal composition.
Types of defects in crystals leading to non-stoichiometry
Details of Stoichiometric Reasoning
To contrast stoichiometric with nonstoichiometric compounds, refer to Equation (2),
$3 Fe + 2 O_2 \rightarrow Fe_3O_4 \label{2}$
In addition to atom ratios, it also tells us that 2 × 3 mol = 6 mol Fe will react with 2 × 2 mol = 4 mol O2, and that ½ × 3 mol = 1.5 mol Fe requires only ½ × 2 mol = 1 mol O2. In other words, the equation indicates that exactly 2 mol O2 must react for every 3 mol Fe consumed. For the purpose of calculating how much O2 is required to react with a certain amount of Fe therefore, the significant information contained in Eq. (2) is the ratio
$\dfrac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}} \nonumber$
We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (2),
$\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}} \tag{3}$
Stoichiometric Reasoning and Non-Stoichiometric Compounds
Under normal circumstances, the stoichiometric ratio holds closely (Fe3O4 is a combination of Fe2O3 and FeO, so it may incorporate the nonstoichiometry of FeO described above). But unusual geological or synthetic conditions lead to other stoichiometries. For example, a technique called "sputter deposition" or "Sputtering" involves heating one reactant in a vacuum with a small amount of the second reactant present as a vapor, as shown in the Figure at right. The product is collected on a cold substrate.
This leads to berthollides when an iron sample is sputtered in the presence of water vapor. Oxide compositions ranging from Fe3O4 to Fe2O3, were obtained, depending on the temperature and pressure [8]. They would have stoichiometric ratios ranging from $\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}$ for Equation (2).
to $\frac{\text{1.5 mol O}_{\text{2}}}{\text{2 mol Fe}}$ for the reaction 2 Fe + 1.5 O2 → Fe2O3, and most of the ratios would not be easily reduced to whole number ratios.
It is remarkable that stoichiometric ratios are used to understand and guide the synthesis of even non-stoichiometric compounds, so this is truly an important area of chemistry.
Example $1$
Derive all possible stoichiometric ratios from Eq. (2)
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
$\text{S}\left( \frac{\text{Fe}}{\text{O}_{2}} \right)=\frac{\text{3 mol Fe}}{\text{2 mol O}_{\text{2}}}~~~~~~\text{S}\left( \frac{\text{Fe}_{3}\text{O}_{4}}{\text{Fe}} \right)=\frac{\text{1 mol Fe}_{3}\text{O}_{4}}{\text{3 mol Fe}}$ $\text{S}\left( \frac{\text{O}_{2}}{\text{Fe}_{3}\text{O}_{4}} \right)=\frac{\text{2 mol O}_{2}}{\text{1 mol Fe}_{2}\text{O}_{4}}~~~~~\text{S}\left( \frac{\text{Fe}}{\text{Fe}_{\text{3}}\text{O}_{4}} \right)=\frac{\text{3 mol Fe}}{\text{1 mol Fe}_{\text{3}}\text{O}_{4}}$
When any stoichiometric chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (2) as an example, this means that the ratio of the amount of O2 consumed to the amount of Fe consumed must be the stoichiometric ratio S(O2/Fe): $\frac{n_{\text{O}_{\text{2}}\text{ consumed}}}{n_{\text{Fe}_\text{ consumed}}}=\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}$ Similarly, the ratio of the amount of Fe3O4 produced to the amount of Fe consumed must be
S(Fe3O4/Fe):
$\frac{n_{\text{Fe}_{3}\text{O}_{4}}\text{ produced}}{n_{\text{Fe}\text{ consumed}}}$
Note that in the word Eq. (4a) and the symbolic Eq. (4b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
Example $2$
Find the amount of Fe3O4 produced when 3.68 mol Fe is consumed according to Eq. (2).
Solution
The amount of Fe3O4 produced must be in the stoichiometric ratio S(Fe3O4/Fe) to the amount of Fe consumed:
$\text{S}\left( \frac{\text{Fe}_{3}\text{O}_{4}}{\text{Fe}} \right)$
Multiplying both sides nFe consumed, by we have
$n_{\text{Fe}_{3}\text{O}_{4}\text{ produced}}$
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (4) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form $\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}$ or symbolically.
$n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol Fe cancels 1 mol Fe but does not cancel 1 mol Fe3O4.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
Example $3$
Add example text here.
Solution
Add example text here.
Calculate the mass of Oxygen (O2) consumed when 3.68 mol Fe reacts according to Equation (2).
Solution
The problem asks that we calculate the mass of O2 consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of O2 to the mass of O2. Therefore this problem in effect is asking that we calculate the amount of O2 consumed from the amount of Fe consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio
$\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{Fe}} \right)=\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}$ $n_{\text{O}_{\text{2}}}\text{ consumed}$
The mass of O2 is $\text{m}_{\text{O}_{\text{2}}}=\text{2}\text{.45 mol O}_{\text{2}}\times \frac{\text{32}\text{.0 g O}_{\text{2}}}{\text{1 mol O}_{\text{2}}}=\text{78.5 g O}_{\text{2}}$ With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of Fe to moles of O2 and the molar mass will convert moles of O2 to grams of O2. A schematic road map for the one-step calculation can be written as $n_{\text{Fe}}~~\xrightarrow{S\text{(O}_{\text{2}}\text{/Fe}\text{)}}~~n_{\text{O}_{\text{2}}}~~\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}$ Thus $\text{m}_{\text{O}_{\text{2}}}=\text{3.68 mol Fe}~~\times ~~\frac{\text{2 mol O}_{\text{2}}}{\text{3 mol Fe}}~~\times~~\frac{\text{32.0 g}}{\text{1 mol O}_{\text{2}}}=\text{78.5 g O}_{2}$
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations of the mass of product.
3 Fe + 2 O2 1 Fe3O4
m (g) 205.5 78.5 284.0
M (g/mol) 55.845 32.0 231.54
n (mol) 3.68 2.45 1.23
Example 4
We noted above that when iron is sputtered in the presence of water vapor, a range of products from Fe3O4 to Fe2O3 (common rust) may be produced. Prepare a table similar to the one above for the reaction in which 0.200 g of Fe is converted to Fe2O3.
Solution
First, write a balanced equation
2 Fe + 3 H2O → Fe2O3 + 3 H2 The problem gives the mass of Fe. Thinking the problem through before trying to solve it, we realize that the molar mass of Fe could be used to calculate the amount of Fe consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed and the amount of Fe2O3 and H2 produced. Finally, the molar masses of O2, Fe2O3, and H2 permit calculation of the mass of O2, Fe2O3, and H2.
We might start the table by entering the given mass, and the molar masses which we calculate from atomic weight tables:
2 Fe + 3 H2O → 2 Fe2O3 + 3 H2
m (g) 0.200
M (g/mol) 55.845 32.0 159.69 2.016
n (mol)
Now we can calculate the amount of Fe present:
$\text{n (mol)}~~ = ~~\frac{\text{m (g)}}{\text{M (g/mol)}}$ $\text{n (mol)} ~~=~~\frac{\text{0.200 g Fe}}{\text{55.847 g/mol}}~~=~~\text{0.00358 mol Fe}$
Then the stoichiometric ratios are used to calculate the amounts of water and products:
$\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{Fe}} \right)=\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{2 mol Fe}}$
So the amount of water required is
0.00358 mol Fe x (3 mol H2O / 2 mol Fe) = 0.00537 mol H2O
Similarly, we use the stoichiometric ratio to calculate the amount of product:
$\text{S}\left( \frac{\text{Fe}_{\text{2}}\text{O}_{3}}{\text{Fe}} \right)=\frac{\text{1 mol Fe}_{\text{2}}\text{O}_{3}}{\text{2 mol Fe}}$
So the amount of Fe2O3 produced is
0.00358 mol Fe x (1 mol Fe2O3 / 2 mol Fe) = 0.00179 mol Fe2O3.
The amount of H2 produced must be the same as the amount of water consumed, since they are in the ratio 3:3 from the equation.
We can add these to the table:
2 Fe + 3 H2O → 2 Fe2O3 + 3 H2
m (g) 0.200
M (g/mol) 55.845 32.0 159.69 2.016
n (mol) 0.00358 0.00537 0.00179 0.00537
Finally, we can use the molar masses to convert from amounts (in mol) to masses (in g):
2 Fe + 3 H2O → 2 Fe2O3 + 3 H2
m (g) 0.200 0.097 0.286 0.0108
M (g/mol) 55.845 18.015 159.69 2.016
n (mol) 0.00358 0.00537 0.00179 0.00537
Note that the sum of the masses of reactants equals the sum of the masses of the products. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.01%3A_Cultural_Connections-_Berthollides-_A_Challenge_to_Chemical_Stoichiometry.txt |
The Presidential Green Chemistry Challenge Awards [1] began in 1995 as an effort to recognize individuals and businesses for innovations in green chemistry.
IWC replica Réplica de reloj Typically five awards are given each year, one in each of five categories: Academic, Small Business, Greener Synthetic Pathways, Greener Reaction Conditions, and Designing Greener Chemicals. A committee of the American Chemical Society selects the awardees. Through 2006, a total of 57 technologies have been recognized for the award, and over 1000 nominations have been submitted.
Propylene glycol is a potential replacement for toxic ethylene glycol in antifreeze used in cars, solar heating systems, and deicing airplane wings. It is also used as a solvent in pharmaceuticals, or as a moisturizer in cosmetics, as well as the main ingredient in deodorant sticks, in fog machines, and a multitude of other uses.
It's an important enough industrial chemical that the 2006 Presidential Green Chemistry Award was given to Professor Galen J. Suppes of the University of Missouri/Columbia for an improved synthesis.
Propylene glycol is synthesized in the laboratory by oxidation of propylene (CH2CHCH3, see Figure above) with potassium permanganate:
$3 \ce{CH_2CHCH_3} + 2 \ce{KMnO_4} + 4 \ce{H_2O} \rightarrow 3 \ce{HO-CH_2-CHOH-CH_3} + 2 \ce{MnO_2} + 2 \ce{KOH} \label{1}$
This is considered a "good" synthesis, because it converts nearly all of the starting material to the desired product. But environmentally (and economically) it is extremely undesirable, for several reasons. It uses a non-renewable petrochemical as a reactant, uses an expensive oxidizing agent (KMnO4), produces a strong caustic (KOH) as a product and an undesirable byproduct (MnO2). It does have one advantage: It proceeds at low temperatures.
Stoichiometry
Let's abbreviate Equation $\ref{1}$ by symbolizing propylene as P and propylene glycol as PG to make the discussion easier:
$\ce{3 P + 2 KMnO_4 + 4 H_2O \rightarrow 3 PG + 2 MnO_2 + 2 KOH} \label{1a}$
A balanced chemical equation such as Equation (1) above not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 3 propylene (P) molecules react with 2 KMnO4 formula units and 4 H2O molecules to give 3 propylene glycol (PG) molecules, 2 MnO2 formula units and 2 KOH formula units. Here we use the term "formula unit" in cases where the substances are not necessarily molecules, but ionic or other species.
The equation also says that 3 mol of P reacts with 2 mol of KMnO4 and 4 mol of water yielding 2 mol of PG, 2 mol MnO2 and 2 mol of KOH.
The balanced equation does more than this, though. It also tells us that 2 × 3 = = 6 mol P will react with 2 x 2 = 4 mol KMnO4 and 2 x 4 = 8 mol water, and that ½ × 3 = 1.5 mol P requires only ½ × 2 =1 mol KMnO4 and ½ × 4 = 2 mol of H2O. In other words, the equation indicates that exactly 4 mol H2O must react for every 3 mol of P and 2 mol of KMnO4 consumed. For the purpose of calculating how much KMnO4 is required to react with a certain amount of P therefore, the significant information contained in Equation $\ref{1}$ is the ratio
$\dfrac{\text{2 mol KMnO}_{\text{4}}}{\text {3 mol P}} \nonumber$
We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Equation $\ref{1}$,
$\text{S}\left( \frac{\text{KMnO}_{\text{4}}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{\text{4}}}{\text{3 mol P}} \label{2}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
Example $1$
Derive all possible stoichiometric ratios from Equation $\ref{1}$.
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
$\text{S}\left( \frac{\text{KMnO}_{4}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{4 mol H}_{\text{2}}\text{O}} \nonumber$
$\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{P}} \right)=\frac{\text{4 mol H}_{\text{2}}\text{O}}{\text{2 mol P}}\nonumber$
$\text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{3 mol P}}\nonumber$
$\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{MnO}_{\text{2}}} \right)=\frac{\text{4 mol H}_{\text{2}}\text{O}}{\text{2 mol MnO}_{\text{2}}}\nonumber$
$\text{S}\left( \frac{\text{KMnO}_{4}}{\text{MnO}_{\text{2}}} \right)=\frac{\text{2 mol KMnO}_{4}}{\text{2 mol MnO}_{\text{2}}}\nonumber$
$\text{S}\left( \frac{\text{P}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{3 mol P}}{\text{4 mol H}_{\text{2}}\text{O}}\nonumber$
There are several more stoichiometric ratios, because they can link any reactant to any other reactant, any reactant to any product, or any product to any other product. When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Equation \ref{1} as an example, this means that the ratio of the amount of KMnO4 consumed to the amount of propylene (P) consumed must be the stoichiometric ratio S(KMnO4/P):
$\frac{n_{\text{KMnO}_{\text{4}}\text{ consumed}}}{n_{\text{P consumed}}} = \text{S}\left( \frac{\text{KMnO}_{\text{4}}}{\text{P}} \right)=\frac{\text{2 mol KMnO}_{\text{4}}}{\text{3 mol P}} \nonumber$
Similarly, the ratio of the amount of MnO2 produced to the amount of P consumed must be S(MnO2/P):
$\frac{n_{\text{MnO}_{\text{2}}\text{ produced}}}{n_{\text{P consumed}}} = \text{S}\left( \frac{\text{MnO}_{\text{2}}}{\text{P}} \right) = \frac{\text{2 mol MnO}_{\text{2}}}{\text{3 mol P}} \nonumber$
In general we can say that
$\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)} \nonumber \] or, in symbols, $\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)} \nonumber$ Note that in the word Equation \ref{3a} and the symbolic Equation \ref{3b}, X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios. Example \(2$
If 3.68 mol propylene (P) is consumed according to Equation \ref{1}, find the amounts of (a) potassium permanganate (KMnO4) required, and the amounts of
1. propylene glycol, PG, and
2. MnO2 produced (assuming plenty of the other reactant, water is available).
Solution
a. The amount of KMnO4 required must be in the stoichiometric ratio S(KMnO4 / P) to the amount of propylene consumed:
$\text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right) = \frac{n_{\text{KMnO}_{\text{4 consumed}}}}{n_{\text{P consumed}}} \nonumber$
Multiplying both sides nP consumed, by we have $n_{\text{KMnO}_{\text{4 consumed}}} = n_{\text{P consumed}}\times \text{S}\left( \frac{\text{KMnO}_{4}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{2 mol KMnO}_{4}}{\text{3 mol P}}=\text{2}\text{.45 mol KMnO}_{4}$
b. The amount of propylene glycol produced must be in the stoichiometric ratio S(PG/P) to the amount of propylene consumed:
$\text{S}\left( \frac{\text{PG}}{\text{P}} \right)=\frac{n_{\text{PG produced}}}{n_{\text{P consumed}}} \nonumber$
Multiplying both sides nP consumed, by we have
$n_{\text{PG produced}}=n_{\text{P consumed}}\times \text{S}\left( \frac{\text{PG}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{3 mol PG}}{\text{3 mol P}}=\text{3}\text{.68 mol PG} \nonumber$
c. Similarly, the amount of MnO2 produced must be in the stoichiometric ratio S(MnO2 / P) to the amount of propylene consumed:
$\text{S}\left( \frac{\text{MnO}_{2}}{\text{P}} \right) = \frac{n_{\text{MnO}_{\text{2 produced}}}}{n_{\text{P consumed}}} \nonumber$
Multiplying both sides nP consumed, by we have
$n_{\text{MnO}_{\text{2 produced}}} = n_{\text{P consumed}}\times \text{S}\left( \frac{\text{MnO}_{2}}{\text{P}} \right) = \text{3}\text{.68 mol P}\times \frac{\text{2 mol MnO}_{2}}{\text{3 mol P}}=\text{2}\text{.45 mol MnO}_{2} \nonumber$
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations.
Calculations from Example 2
3 CH2CHCH3 + 2 KMnO4 + 4 H2O → 3 CH2(OH)CH(OH)CH3 + 2 MnO2 + 2 KOH
n (mol) 3.68 2.45 4.91 3.68 2.45 2.45
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Equation \ref{3} when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form
$\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber$
or symbolically.
$n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol KMnO4 cancels 1 mol KMnO4 but does not cancel 1 mol H2O.
Masses of Reactants and Products
We can also calculate the masses of reactants and products. The molar masses can be determined from the formulas, and the masses are calculated as follows:
m (g)= n (mol) x M (g/mol)
For KMnO4,
mKMnO4 = nKMnO4 x MKMnO4
mKMnO4 = 2.45 mol x 158.03 g/mol = 387.17 g
This result is added to the table below, and you may want to see if you can verify the other results in the table.
Calculations of masses of Reactants and Products
3 CH2CHCH3 + 2 KMnO4 + 4 H2O → 3 CH2(OH)CH(OH)CH3 + 2 MnO2 + 2 KOH
n (mol) 3.68 2.45 4.91 3.68 2.45 2.45
M (g/mol) 42.08 158.03 18.02 76.10 86.94 56.11
m(g) 155 387 88.47 280 213 138
Atom Economy
Notice that there seems to be an aweful lot of extraneous stuff in this synthesis. It requires 387 g of KMnO4, but that results only in the addition of two -OH groups to each propylene molecule (and they may come,in part, from H2O).
As a measure of this inefficiency, Barry trost developed the concept of atom economy, and for this work he received the Presidential Green Chemistry Challenge Award in 1998[2]. The percentage atom economy is defined as
$\text{% atom economy} = \dfrac{\text{mass of atoms utilized}}{\text{mass of all reactant atoms}} \times 100\% \label{atomecon}$
So we see that the mass of reactants that actually end up in products is 155 g (3.68 mol) of propylene and 125.2 g of OH groups (2 x 3.68 mol, or 7.36 mol, with a molar mass of 17.01 g/mol OH), for a total of 280 g. The mass of all reactants is
155 g + 387 g + 88.5 g = 630 g,
so the percent atom efficiency is (via Equation \ref{atomecon})
$\text{% atom economy} = \dfrac{280 g}{630 g} x 100% = 44.4% \nonumber$
We see that a lot of atoms are wasted in this synthesis. For comparison, let's look at several industrial processes that have been developed.
The Suppes Green Chemistry Award
Professor Suppes’s system that won the Green Chemistry Award couples a new copper-chromite catalyst to convert glycerin to propylene glycol. First, it utilizes a byproduct of biodiesel synthesis, glycerine (rather than the propylene used above, which comes from petroleum). Second, it uses a catalyst, which is not consumed in a chemical reaction and thus does not reduce the atom economy. Finally, this process uses a lower temperature and lower pressure than do previous systems (428 °F versus 500 °F and <145 psi versus >2,170 psi), and produces less byproduct than do similar catalysts[3].
The synthesis of propylene glycol from glycerine
The reaction involves dehydration (removal of -H and -OH, or H2O) of glycerine, (yielding 2,3-dihydroxypropene or 3-hydroxypropanal) followed by hydrogenation of the resulting double bond to give the final product. The overall equation for the reaction is
$\ce{H2 + HO-CH2CH(OH)CH2OH → HO-CH2-CHOH-CH3 + H2O } \label{4}$
Example $3$
Calculate the mass of hydrogen that's required to react with 100 g of glycerine, and the masses of propylene glycol and water that result in the Suppes Green Synthesis of propylene glycol.
Solution
First, we'll start the same kind of table that we used above, but we'll start with the information we have: the mass of glycerine, and the molar masses that we can calculate from formulas:
HO-CH2CH(OH)CH2OH + H2 CH2(OH)CH(OH)CH3 + H2O
M (g/mol) 92.1 2.02 76.10 18.02
m (g) 100
n(mol)
We know that the stoichiometry of the reaction involves the amounts of reactants and products, not their masses, so we'll convert each mass to an amount. For example, the amount of glycerine is:
\begin{align*} n(mol) &= \dfrac{m(g)}{M\, (g/mol)} \[4pt] &= \dfrac{100\, g}{92.1\, g/mol} \[4pt] &= 1.09\, mol \end{align*}
HO-CH2CH(OH)CH2OH + H2 CH2(OH)CH(OH)CH3 + H2O
M (g/mol) 92.1 2.02 76.10 18.02
m (g) 100
n(mol) 1.09
We can easily calculate the other amounts, because the stoichiometric ratios are all 1. So for the amount of hydrogen consumed:
$\text{n}_{\text{H}_{2}} = \text{n}_{\text{HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} ~x~\frac{\text{1 mol H}_{2}}{\text{1 mol HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} \nonumber$
$\text{n}_{\text{H}_{2}} = \text{1.09 mol} ~x~\frac{\text{1 mol H}_{2}}{\text{1 mol HO-CH}_{2}\text{CH(OH)CH}_{2}\text{OH}} = 1.09 mol \nonumber$
HO-CH2CH(OH)CH2OH + H2 CH2(OH)CH(OH)CH3 + H2O
M (g/mol) 92.1 2.02 76.10 18.02
m (g) 100
n(mol) 1.09 1.09 1.09 1.09
Now we can use the amounts to calculate the masses of the other reactants and products, remembering that
m(g) = n(mol) x M(g/mol).
So for hydrogen,
mH2 = 1.09 mol x 2.02 g/mol = 2.20 g.
Verify the other table entries:
HO-CH2CH(OH)CH2OH + H2 CH2(OH)CH(OH)CH3 + H2O
M (g/mol) 92.1 2.02 76.10 18.02
m (g) 100 2.20 83.0 19.6
n(mol) 1.09 1.09 1.09 1.09
Now we can calculate the percent atom economy for this process:
% atom economy = $\frac{mass ~of ~atoms ~utilized}{mass ~of~all ~reactant ~atoms} x 100%$
All of the atoms in the reactants except one oxygen atom which is removed from the glycerine. Since we have 1.09 mol of glycerine, the mass of oxygen not incorporated into product is 1.09 mol O x 15.999 g/mol O = 17.44 g O. The mass of reactants utilized is therefore 100 g - 17.44 g = 82.56 g, and the mass of product formed is 83.0 g.
percent atom economy = $\frac{82.56}{83.0} ~x~ 100% = 99.5%$
Not only that, but the product, water, is innocuous, and it uses an overproduced reactant! The main cost is heating and pressurizing the reactor.
Example $4$
What mass of oxygen would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?
Solution
First, write a balanced equation
$\ce{2C8H18 + 25O2 → 16CO2 + 18H2O} \nonumber$
The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically
$m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}} \nonumber$
$m_{\text{O}_{\text{2}}}=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g } \nonumber$
Thus 12 Pg (petagrams) of O2 would be needed. The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.02%3A_Environment-_Atom_Efficiency_and_the_2006_Presidential_Green_Chemistry_Award.txt |
Equations and Mass Relationships in Everyday Life
If you're observant and pay attention to nutrition labels on foods, you may have noticed labels like the one here, where the fats don't seem to add up.
If one 12 g serving of Crisco® contains 3 g of saturated fat, 0g of trans fat, 6 g of polyunsaturated fat, and 2.5 g of monounsaturated fat[1] what happened to the missing 0.5 g of fat? 3 g + 0 g + 6 g + 2.5 g = 11.5 g!
In many cases there's a bigger disparity than this. The fats don't add up[2] because the weight of glycerol is not included in the separately listed components. Trans fatty acids are now recognized as a major dietary risk factor for cardiovascular diseases, and the US FDA has revised food labeling requirements to include trans fats.[3]
Are companies pulling the wool over our eyes? In order to understand what's going on, we need to look into the nature of vegetable fats and oils, which are triglycerides.
Triglycerides
Vegetable fats and oils are all triglycerides, which contain a glycerol () three carbon "backbone" with 3 long chain "fatty acids" attached through ester linkages, as in the figure below. The actual shape is shown in the Jmol model, which can be rotated with the mouse. Triglycerides are called "fats" when they're solids or semisolids, and "oils" when they're liquids.
A triglyceride, overall unsaturated, with the glycerol "backbone" on the left, and saturated palmitic acid, monounsaturated oleic acid, and polyunsaturated alpha-linolenic acid. The shape of the acids is not accurately represented.
The long chain fatty acids may be saturated with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are unsaturated and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is polyunsaturated, with multiple double bonds (it is linolenic acid). Various cooking oils have |known concentrations of saturated and unsaturated fatty acids.
As of 2010, Crisco consists of a blend of soybean oil, fully hydrogenated cottonseed oil, and partially hydrogenated soybean and cottonseed oils. Each of these oils is a complex mixture of triglycerides, all with different fatty acid substituents. (We've discussed the benefits and drawbacks of saturated and unsaturated oils elsewhere).
A triglyceride would be called "unsaturated" if it contained just 1 unsaturated fatty acid (and 2 unsaturated), so how can the actual amount of saturated and unsaturated fatty acids be reliably reported? The tryglycerides have to be decomposed into their component fatty acids, and the total amount of each kind (saturated, monounsaturated, and polyunsaturated) reported separately. But this leaves out the glycerol that results from the decomposition, as shown in the equation below for the triglyceride containing 2 palmitic acid (P = C16H32O2) and 1 linolenic acid (L = C18H30O2) substituents. The triglyceride can be abbreviated "GPPL" for glycerol (G) with 2 palmitic acid and 1 linolenic acid (L) substituents:
$\ce{(C16H32O2)}\textbf{CH}_\textbf{2}\textbf{CH} \ce{(C16H32O2)}\textbf{CH}_\textbf{2}\ce{-(C18H30O2) + 3 H2O} \xrightarrow{\ce{NaOH}} \ce{1 C3H6O3 + 2 C16H32O2 + 1 C18H30O2} \nonumber$
GPPL + 3 H2O Glycerol + 2 P + 1 L (1)
This is called a "hydrolysis" reaction, because water causes the decomposition. Sodium hydroxide (NaOH) written above the arrow indicates that NaOH is a catalyst, and isn't consumed or integrated into the products of the chemical reaction.
This balanced chemical equation can tell us where the missing mass from the nutrition label has gone.
The equation not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved, so it will allow us to find out how much water is consumed and how much glycerol is produced, and those will explain the "missing" masses on the nutrition label.
The equation says that 1 GPPL molecules can react with 3 H2O molecules to give 1 G molecule, 2 P molecules and 1 L molecule. It also says that 1 mol GPPL would react with 3 mol H2O yielding 1 mol G, 2 mol P, and 1 mol L.
The balanced equation does more than this, though. It also tells us that 2 × 1 = 2 mol GPPL will react with 2 × 3 = 6 mol H2O, to form 2 × 1 = 2 mol G, and that ½ × 1 = 0.5 mol GPPL requires only ½ × 3 = 1.5 mol H2O. In other words, the equation indicates that exactly 3 mol H2O must react for every 1 mol GPPL consumed, and for every 1 mol GPPL is consumed, 1 mol G, 2 mol P, and 1 mol L will be produced. For the purpose of calculating how much H2O is require to react with a certain amount of GPPL, the significant information contained in Eq. (1) is the ratio
$\dfrac {\text {3 mol H}_{\text{2}}{\text{O}}} {\text {1 mol GPPL}} \nonumber$
We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1), $\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{GPPL}} \right)~=~\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}\text{)}}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
Example $1$
Derive all possible stoichiometric ratios from Eq. (1)
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
$\text{S}\left( \frac{\text{GPPL}}{\text{G}} \right)=\frac{\text{1 mol GPPL}}{\text{1 mol G}}$
$\text{S}\left( \frac{\text{L}}{\text{P}} \right)=\frac{\text{1 mol L}}{\text{2 mol P}}$
$\text{S}\left( \frac{\text{GPPL}}{\text{P}} \right)=\frac{\text{1 mol GPPL}}{\text{2 mol P}}$
$\text{S}\left( \frac{\text{L}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{1 mol L}}{\text{3 mol H}_{\text{2}}\text{O}}$
$\text{S}\left( \frac{\text{P}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol P}}{\text{3 mol H}_{\text{2}}\text{O}}$
$\text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{1 mol G}}{\text{3 mol H}_{\text{2}}\text{O}}$
When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of H2O consumed to the amount of GPPL consumed must be the stoichiometric ratio S(H2O/GPPL):
$\frac{n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}}{n_{\text{GPPL}\text{ consumed}}}$$=\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{GPPL}} \right)=\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}}$ Similarly, the ratio of the amount of G produced to the amount of GPPL consumed must be
S(G/GPPL):
$\frac{n_{\text{G produced}}}{n_{\text{GPPL}\text{ consumed}}}~=~\text{S}\left( \frac{\text{G}}{\text{GPPL}} \right)=\frac{\text{1 mol G}}{\text{1 mol GPPL}}$
In general we can say that
$\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)}$
or, in symbols,
$\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)}$
Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
Example $2$
Find the amount of glycerol produced when 3.68 mol H2O is consumed according to Eq. (1).
(C16H32O2)CH2CH(C16H32O2)CH2-(C18H30O2) + 3 H2O $\xrightarrow{\text{NaOH}}$1 C3H6O3 + 2 C16H32O2 + 1 C18H30O2
GPPL + 3 H2O $\xrightarrow{\text{NaOH}}$ Glycerol + 2 P + 1 L (1)
Solution
The amount of glycerol produced must be in the stoichiometric ratio S(G/H2O)to the amount of water consumed:
$\text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{n_{\text{G produced}}}{n_{\text{H}_{\text{2}}\text{O consumed}}}$
Multiplying both sides nH2O consumed, by we have
$n_{\text{G}\text{O produced}}=n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}\times \text{S}\left( \frac{\text{G}}{\text{H}_{\text{2}}\text{O}} \right)=\text{3}\text{.68 mol H}_{\text{2}}\text{O}\times \frac{\text{1 mol G}}{\text{3 mol H}_{\text{2}}\text{O}}=\text{1}\text{0.23 mol H}$
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form $\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}$ or symbolically. $n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NH3 cancels 1 mol NH3 but does not cancel 1 mol H2O
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
Example $\PageIndex{3$
1. Calculate the mass of glycerol (G) produced when 3.84 mol GPPL is reacted with adequate H2O according to Equation (1).
2. Show that a nutritional label for this fat would have trans, unsaturated, and polyunsaturated fats that don't add up to the total fat.
Solution
The problem asks that we calculate the mass of glycerol produced. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of glycerol to the mass of glycerol. Therefore this problem in effect is asking that we calculate the amount of glycerol produced from the amount of GPPL consumed. This is the same problem as in Example 2. It requires the stoichiometric ratio
$\text{S}\left( \frac{\text{G}}{\text{GPPL}} \right)=\frac{\text{1 mol G}}{\text{1 mol GPPL}}$
The amount glycerol, G, produced is then
$n_{\text{G produced}}=n_{\text{GPPL consumed}}\text{ }\!\!\times\!\!\text{ conversion factor}=\text{3}\text{.84 mol GPPL}\times \frac{\text{1 mol G}}{\text{1 mol GPPL}}=\text{3}\text{.84 mol G}$
The mass of glycerol, (G = C3H6O3) is $\text{m}_{\text{G}}~=~\text{3}\text{.84 mol G}\times \frac{\text{92}\text{.1 g G}}{\text{1 mol G}}=\text{354 g G}$ With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as $n_{\text{G}} ~ \xrightarrow{S\text{(G}\text{/GPPL}\text{)}}~ n_{\text{G}}~\xrightarrow{M_{\text{G}}} ~ m_{\text{G}}$ Thus $\text{m}_{\text{G}}=\text{3}\text{.84 mol GPPL}\times ~ \frac{\text{1 mol G}}{\text{1 mol GPPL}} ~ \times ~ \frac{\text{92}\text{.1 g}}{\text{1 mol G}}=\text{354 g}$ These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations.
C16H32O2)CH2CH(C16H32O2)CH2-(C18H30O2)
GPPL
+ 3 H2O → C3H6O3
G
+ 2 C16H32O2
P
+ 1 C18H30O2
L
m (g) 3185 208 354 1969 1069
M (g/mol) 829.3 18.02 92.1 256.4 278.4
n (mol) 3.84 11.52 3.84 7.68 3.84
b. When we calculate the masses of all reactants and products, we get the results shown in the table. In this case, the total mass of fat is 3185 g. It is a mixed saturated (palmitic acid) and polyunsaturated (linolenic acid) fat, that is degraded to 1969 g of saturated fat and 1069 g of polyunsaturated fat. We assume that no trans fat was produced, and no monounsaturated fatty acids are present. So the label would have numbers proportional to:
Total Fat 3185
Saturated Fat 1969
Trans Fat 0
Unsaturated Fat 0
Polyunsaturated Fat 1069
The components add up to 3038 g, which is less than the total fat because the mass of glycerol (354 g) and water (207.6) are not accounted for.
Example $1$
Show that the calculation of the mass of water required to react with 3.68 mol of GPPL in the example above is correct.
Solution
Symbolically
$n_{\text{GPPL}} ~ \xrightarrow{S\text{(H}_{\text{2}}\text{O}\text{/GPPL}\text{)}} ~ n_{\text{H}_{\text{2}}\text{O}}~\xrightarrow{M_{\text{H}_{\text{2}}\text{O}}} ~ m_{\text{H}_{\text{2}}\text{O}}$
$3.84 \text{mol GPPL}~\times~\frac{\text{3 mol H}_{\text{2}}\text{O}}{\text{1 mol GPPL}} ~ \times ~ \frac{\text{18}\text{.02 g}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{208 g }$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.03%3A_Everyday_Life-_Why_Fats_Don%27t_Add_Up_on_Food_Nutrition_Labels.txt |
Everyday Examples using Chemical Calculations
What do and have to do with chemistry?
We often use a stoichiometry calculation in everyday life without even realizing it, especially while cooking, preparing a grocery list or building something!
You: "Stoichiometry? What is that?"
Me: "It's using unit conversions in order to answer a question."
You: "Oh, that's all? Besides not being able to pronounce the word very well, I thought it was some scary type of calculation."
Me: "Believe it or not, you've been doing these types of calculations for a long time already. It is just a bunch of unit conversions!"
The purpose of this page is to help you identify calculations you already use in your everyday life that you probably don't even realize you're doing.
Then once you recognize that you are in fact using the same calculations in your everyday life as in this chemistry course, hopefully this page will help you relate what you are "thinking" to what it looks like when you write it out as a math equation.
The thing that takes getting used to is how we can relate what we use every day to something that we are learning.
Read through these first three examples to get a feeling for how we actually use these calculations in our everyday life. Then move on to examples 4 and 5 to see how we relate these everyday calculations to chemistry.
If you get stumped on Examples 4 and 5, refer back to Examples 1, 2 and 3. Keep in mind that you will be using the EXACT same calculation, you are just replacing words like "eggs, ham, cheese, and cookies", with chemical names or chemical formulas. If you get confused, try replacing the words of a chemistry problem with more familiar words and see if that helps you learn these chemical calculations. For instance, you could replace NH3 with the word "egg" to help you feel more comfortable.
Example $1$:
I want to have friends over for lunch on Saturday and make grilled cheese sandwiches that require two slices of bread and one slice of cheese. I open the refrigerator to find that I have 5 slices of cheese. I look in the bread box to find that I have 10 slices of bread.
How many sandwiches can I make?
Answer
These three steps are the first steps to solving this type of question, in chemistry, it is called a "stoichiometry" problem.
Step 1) Write out the recipe, also known as an equation
+
2 slices of bread + 1 slice of cheese → 1 grilled cheese sandwich
Step 2) Find Quantity (moles) and Identify useful unit conversions and/or molar ratios
What we know:
1) We know we have 5 slices of cheese
2) We know we have 10 slices of bread.
According to the equation written in step 1, ratios or unit conversions can be written by using the number in front of the ingredient, also known as a co-efficient.
2 slices of bread is required for one slice of cheese.
Remember, the powerful thing about ratios is that we can also write them "upside-down".
The following ratios or unit conversions that can be written from this equation are:
\begin{align} \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 slice of cheese}} \ & \ \frac{\text{1 sandwich}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 sandwich}} \ & \ \frac{\text{1 sandwich}}{\text{1 slice of cheese}}&=\frac{\text{1 slice of cheese}}{\text{1 sandwich}} \ \end{align}
Step 3) Use these ratios as a unit conversion to obtain the unit you are interested in
Before we start a calculation, let us visualize what we have to start with.
Can you tell by looking at the picture how many sandwiches we can make?
You may even get an answer without even have to "think" about.
Let's re-organize this picture to represent the ratio that we determined in step 2.
Is it easier to see how many sandwiches we can make now?
Now that we have visualized this process, let's set up an equation to give us the same answer.
Remember, the purpose of this page is to help you identify calculations you do in your everyday life that you probably don't even realize you're doing.
If you did come to an answer in your head, this is what the equation looks like when written out.
\begin{align} & mol_{\text{sandwiches}}=\text{5 slices of cheese}\times \frac{\text{1 sandwich}}{\text{1 slice of cheese}}=\text{5 sandwiches} \ & \ & mol_{\text{sandwiches}}=\text{10 slices of bread}\times \frac{\text{1 sandwich}}{\text{2 slices of bread}}=\text{5 sandwiches} \ \end{align} In this case it doesn't matter which one starting material or reagent that you start with. As long as you know the relationship or the ratio between what you are looking for and what you know, then you will always be able to come to an answer.
To re-cap the steps we used to solve this problem:
To re-cap the steps we used to solve this problem:
Step 1) Write equation
Step 2) Write or find moles
Step 3) Use molar ratio
Example $2$: Let's go shopping!
You just volunteered to provide cookies for your school's bake sale. They ask you to bring 100 cookies. You go home and pull out this recipe from your recipe box.
Cookies Recipe - Ingredients
4 Tablespoons Unsalted Butter
1/3 Cup Brown Sugar
1/3 Cup White, Granulated Sugar
1 Egg
1 Cup All Purpose Flour
1 Tablespoon Cornstarch
1/2 teaspoon Baking Soda
1/4 teaspoon Kosher Salt
Makes 20 cookies
You don't have any of the ingredients at home so we will need to go to the grocery store.
We need to make a list so we make sure to get enough ingredients to make 100 cookies.
Step 1
Write an equation.
We will use our recipe as our equation.
Step 2
Identify useful unit conversions.
If we used our recipe we could relate any one ingredient on the recipe to any other ingredient on the recipe. Similar to making sandwiches.
In chemistry terms this is called a molar ratio. A molar ratio is just a unit conversion. It allows us to get from what we know to what we want.
Another helpful unit conversion that I will highlight is the following: 1 recipe = 20 cookies
If we need 100 cookies, how many batches of cookies do we need to make?
You might have already answered five?
Step 3
Let's set up an equation using the unit conversions from Step 1 and 2,
$batches_{\text{cookies}}=\text{100 cookies}\times \frac{\text{1 batch}}{\text{20 cookies}}=\text{5 batches}$ Now that I know how many batches of cookies I need, I can write my grocery list.
Grocery List
Ingredients for one batch Batches
Total Amount
4 Tablespoons Unsalted Butter x 5 20 Tbsp Unsalted Butter
1/3 Cup Brown Sugar x 5 5/3 Cup Brown Sugar
1/3 Cup White, Granulated Sugar x 5 5/3 Cup White, Granulated Sugar
1 Egg x 5 5 Eggs
1 Cup All Purpose Flour x 5 5 Cups All Purpose Flour
1 Tablespoon Cornstarch x 5 5 Tbsp Cornstarch
1/2 teaspoon Baking Soda x 5 2.5 tsp Baking Soda
1/4 teaspoon Kosher Salt x 5 1.25 tsp Kosher Salt
Makes 20 cookies x 5 100 Cookies
To re-cap the steps we used to solve this problem:
Step 1) Write equation
Step 2) Write or find moles
Step 3) Use molar ratio
Example $3$: I'm on a diet!
One of my favorite breakfast foods are cheese omelets in the morning. I am on a diet so I always measure the amount of each ingredient I use. I always make two in case someone else wants one. Below is the recipe for my 'perfect' omelet.
Recipe:
6 Large eggs - 200. g per one egg
1 cups of shredded cheese - 50. g per one cup
I open the refrigerator this morning to find an excess of large eggs and 250. g of cheese that is about to spoil.
• Question 1: How many eggs do I need if I want to use all of the cheese?
• Question 2: How many omelets can I make for breakfast with all of the ingredients I pulled out of the refrigerator?
• Question 3: What is the total mass of my cooked omelets?
Step 1) Write out the recipe in the form of a balanced equation
+ +
6 Large eggs + 1 cups of cheese → 2 omelets
Unit Conversions
According to the equation written in step 1, these molar ratios can be written by using the number in front of the ingredient, also known as a co-efficient.
Once again, when you are first starting out, it is helpful to write the ratios or the stoichiometric ratios connecting all of the components of the recipe.
It makes it easier to pick out which one ratio you will use as a unit conversion to help you calculate what the question is asking.
As you can see, there are more molar ratios this time because there are more starting materials. Additionally, the question gave us information about the mass per unit food item. Therefore there are 8 unit conversions in this question.
$\frac{\text{6 Large eggs}}{\text{1 cups of cheese}}=\frac{\text{1 cups of cheese}}{\text{6 Large eggs}}$ $\frac{\text{1 cups of cheese}}{\text{2 omelets}}=\frac{\text{2 omelets}}{\text{2 cups of cheese}}$
$\frac{\text{6 Large eggs}}{\text{2 omelets}}=\frac{\text{2 omelets}}{\text{6 Large eggs}}$
$\frac{\text{1 Large egg}}{\text{200. g egg}}=\frac{\text{200. g egg}}{\text{1 Large egg}}$ $\frac{\text{1 cup of cheese}}{\text{50. g cheese}}=\frac{\text{50. g cheese}}{\text{1 cup of cheese}}$
$\frac{\text{2 omelet}}{\text{1250 g omelet}}=\frac{\text{1250 g omelets}}{\text{2 omelets}}$
We can determine the mass of 2 omelets by using the law of conservation of mass. Therefore, I will add the mass of 6 eggs and 1 cup of cheese to calculate the last unit conversion listed above.
+ +
1200. g egg + 50. g cheese 625 g omelet + 625 g omelet
Find moles
In order for us to be able to compare the starting materials and answer question 2 and 3, we need to convert all of the starting materials from mass into the quantity associated with recipe or equation .
The actual quantity of cheese can be calculated using the mass per one unit of material. In this case, 1 cup of cheese is 50. grams.
In chemistry terms, this is called the molar mass. Molar mass can be used as a unit conversion. Refer above for useful unit conversions
To relate the idea of molar mass to everyday life is something we do all of the time without even thinking about it.
For instance, you have an assortment of candy in a bucket. Three kids come up to you and ask you for some candy.
Do you weigh the candy and give the three kids equivalent masses? No. That would be unrealistic and probably a little messy.
Instead, we just decide that one candy bar should be given to each kid. Kid #1 receives one Snickers bar, kid #2 receives one Milky Way and kid #3 receives one Kit Kat bar. They are all happy because they received their OWN candy bar; they don't care that one might weigh a little bit more than the other candy bars.
To summarize, Snickers, Milky Way, and Kit Kat all represent ONE candy bar, however they have different masses...but that's okay. To relate it even more to chemistry, one mole of any compound is equivalent to one mole of any other compound, they just have a different mass depending on what substance you are talking about; just substitute these words and you are back to our everyday example; 'mole' = 'candy bar' and 'compound' = 'type of candy bar'.
You can use this idea in reverse, if you know the mass, then the amount of candy bars or 'moles' can be calculated.
$\text{ }mol_{\text{cheese (actual)}}=\text{250. g of cheese}\times \frac{\text{1 cup of cheese}}{\text{50. g of cheese}}=\text{5.0 cups of cheese}$
Step 3 - Use Molar Ratio
Now that we know how much cheese we have, we can calculate how many eggs we need based on our recipe in Step 1. We can also calculate how many omelets we will make.
We can represent it visually before writing an actual calculation.
Based on these calculations, we just answered question 1 and 2. We will use 30 eggs from the refrigerator in order to use up all of the cheese in order to make 10 omelets.
We will need one more step in order to answer question 3, how much will all of our omelets weigh?
Step 4 - Convert to Answer
Let's find out how many 10 omelets will weigh using the useful unit conversion that we identified in Step 2.
$\text{ }mass_{\text{omelet}}=\text{10 omelets}\times \frac{\text{1250 g omelet}}{\text{2 omelets}}=\text{6250 g of omelet made}$
The thing that takes getting used to is how we can relate what we use every day to something that we are learning. If you get stumped on the next two examples, refer back to Examples 1, 2 and 3. Keep in mind that you will be using the EXACT same calculation, you are just replacing words like "eggs, ham, cheese, and cookies", with chemical names or chemical formulas. If you get confused, try replacing the words of a chemistry problem with more familiar words and see if that helps you learn these chemical calculations.
In the examples above you saw a lot of whole numbers. In chemistry you will be seeing more decimal numbers. Imagine that you could make 2.5 omelets and that is the same thing as producing or using 2.5 mol of a chemical substance.
Example $4$
Find the amount of water produced when 3.68 mol NH3 is consumed according to Equation $\ref{3}$.
Solution
The amount of water produced must be in the stoichiometric ratio S(H2O/NH3) to the amount of ammonia consumed:
$\large\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NH}_{\text{3}}} \right)=\Large\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{NH}_{\text{3}}\text{ consumed}}}$
Multiplying both sides nNH3 consumed, by we have
\begin{align} \large n_{\text{H}_{\text{2}}\text{O produced}} &= \large n_{\text{NH}_{\text{3}}\text{ consumed}} \normalsize \times\text{S}\left( \frac{\ce{H2O}}{\ce{NH3}} \right) \ { } \ & =\text{3.68 mol NH}_3 \times \frac{\text{6 mol }\ce{H2O}}{\text{4 mol NH}_3} \ & =\text{5.52 mol }\ce{H2O} \end{align}
This is a typical illustration of the use of a molar ratio as a conversion factor. Example 4 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Step 3 when using the molar ratio. Simply remember that the coefficients in a balanced chemical equation give molar ratios, and that the proper choice results in cancellation of units. In road-map form
$\large \text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced} \nonumber$
or symbolically.
$\large n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}} \nonumber$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the SAME substance. In other words, 1 mol NH3 cancels 1 mol NH3 but does not cancel 1 mol H2O.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
Example $5$
The chemical reaction in this example is of environmental interest. Iron pyrite (FeS2) is often an impurity in coal, and so burning this fuel in a power plant produces sulfur dioxide (SO2), a major air pollutant.
Calculate the mass of sulfur dioxide (SO2) produced when 3.84 mol O2 is reacted with FeS2 according to the equation
$\ce{4 FeS2 (s) + 11 O2 (g) → 2 Fe2O3 (s) + 8 SO2 (g)} \nonumber$
Solution
The problem asks that we calculate the mass of SO2 produced. As we learned above in Example 3 - I'm on a Diet! and in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of SO2 to the mass of SO2. Therefore this problem in effect is asking that we calculate the amount of SO2 produced from the amount of O2 consumed. We will start this problem the same as in Example 4. It requires the molar ratio
Step 1 - Write an Equation
The equation was given to us in the question.
Unit Conversions
The question is asking us about the amount of SO2 produced from the amount of O2 consumed so I want both of those substances in the molar ratio.
$\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}=\frac{\text{11 mol O}_{\text{2}}}{\text{8 mol SO}_{\text{2}}}$
The question is also asking about the mass of SO2, so I want to write the molar mass.
$\frac{\text{1 mol SO}_{\text{2}}}{\text{64.06 g SO}_{\text{2}}}=\frac{\text{64.06 g SO}_{\text{2}}}{\text{1 mol SO}_{\text{2}}}$
Find Moles
The problem gives us the amount of oxygen gas in moles, so we do not need to use the molar mass equation like we did in Example 3 - I am on a Diet!
Step 3 - Use Molar Ratio
The amount of SO2 produced is then
\begin{align} n_{\text{SO}_{\text{2}}}\text{ produced}&=n_{\text{O}_{\text{2}}\text{ actual}}\text{ }\!\!\times\!\!\text{ conversion factor} \n_{\text{SO}_{\text{2}}}\text{ produced}&=\text{3}\text{.84 mol O}_{\text{2}}\times \frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}=\text{2}\text{.79 mol SO}_{\text{2}}\text{ produced} \end{align}
Step 4 - Convert to Answer
The question is asking about the mass of SO2 not the amount of SO2. We can use the unit conversion or molar mass written in Step 2.
The mass of SO2 is
\begin{align} \text{mass}_{\text{SO}_{\text{2}}}&=\text{2}\text{.79 mol SO}_{\text{2}}\times \frac{\text{64}\text{.06 g SO}_{\text{2}}}{\text{1 mol SO}_{\text{2}}}=\text{179 g SO}_{\text{2}} \end{align} With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of O2 to moles of SO2 and the molar mass will convert moles of SO2 to grams of SO2. A schematic road map for the one-step calculation can be written as $mol_{\text{(O}_{\text{2}}\text{)}}\text{ }\xrightarrow{ratio\text{(SO}_{\text{2}}\text{/O}_{\text{2}}\text{)}}\text{ }mol_{\text{(SO}_{\text{2}}\text{)}}\text{ }\xrightarrow{M.M._{\text{SO}_{\text{2}}}}\text{ }mass_{\text{(SO}_{\text{2}}\text{)}}$ This is how the equation would look if it were all done in one step:
$\text{mass}_{\text{(SO}_{\text{2}}\text{)}}=\text{3}\text{.84 mol O}_{\text{2}}\times \text{ }\frac{\text{8 mol SO}_{\text{2}}}{\text{11 mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol SO}_{\text{2}}}=\text{179 g}$
Notice that all of the units that are the same on the top and the bottom of the ratios can cancel, leaving just the units that you want at the end. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.04%3A_Food-_Let%27s_Cook.txt |
We eat a lot of sugars, and in the next few sections we'll explore some of the body chemistry that explains why they may (or may not) lead to weight gain, why they're a good source of energy, and why astronauts excrete more water than they drink. As we've seen, there are many sugars, but one of the most common is sucrose, C12H22O11. A balanced overall chemical equation for the metabolism of sucrose will help us understand the questions above, and many others. A balanced chemical equation such as:
2 C12H22O11(s) + 35 O2(g) → 12 CO2(g) + 11 H2O(l) (1) not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 2 C12H22O11 molecules can react with 35 O2 molecules to give 12 CO2 molecules and 11 H2O molecules. It also says that 2 mol 2C12H22O11 would react with 35 mol O2 yielding 12 mol CO2 and 11 mol H2O.
In other words, the equation indicates that exactly 35 mol O2 must react for every 2 mol C12H22O11consumed. For the purpose of calculating how much O2 is required to react with a certain amount of C12H22O11 therefore, the significant information contained in Eq. (1) is the ratio
$\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}$ We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1), $\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{35 mol O}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
EXAMPLE 1 Derive all possible stoichiometric ratios from Eq. (1)
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used, so in addition to (2) above, we have:
$\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}$
$\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{CO}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{12 mol CO}_{\text{2}}}=\frac{\text{1 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{6 mol CO}_{\text{2}}}$
$\text{S}\left( \frac{\text{CO}_{\text{2}}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}\right)=\frac{\text{12 mol CO}_{\text{2}}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}$
$\text{S}\left( \frac{\text{CO}_{2}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{12 mol CO}_{2}}{\text{11 mol H}_{\text{2}}\text{O}}$
$\text{S}\left( \frac{\text{CO}_{2}}{\text{O}_{\text{2}}} \right)=\frac{\text{12 mol CO}_{2}}{\text{35 mol O}_{\text{2}}}$ $\text{S}\left( \frac{\text{O}_{\text{2}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{35 mol O}_{\text{2}}}{\text{11 mol H}_{\text{2}}\text{O}}$
$\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{11 mol H}_{\text{2}}\text{O}}$
There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.]
When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of O2 consumed to the amount of NH3 consumed must be the stoichiometric ratio S(O2/NH3): $\frac{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}}{n_{\text{O}_{\text{2}}\text{ consumed}}}=\text{S}\left( \frac{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{O}_{\text{2}}}\right)=\frac{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}{\text{35 mol O}_{\text{2}}}$ Similarly, the ratio of the amount of H2O produced to the amount of NH3 consumed must be
S(H2O/C12H22O11):
$\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}$ In general we can say that $\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}\text{ (3}\text{a)}$ or, in symbols, $\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}\text{ (3}\text{b)}$
Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
EXAMPLE 2 Find the amount of water produced when 1 cup (roughly 8 oz or 240 g) C12H22O11 is consumed according to Eq. (1).
Solution
The amount of water produced must be in the stoichiometric ratio S(H2O/C12H22O11) to the amount of sugar consumed, and the amount is n = m/M = 240 g /342.3 g mol-1 = .70 mol.
$\frac{n_{\text{H}_{\text{2}}\text{O produced}}}{n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}} = \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right) =\frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}$ Multiplying both sides by nC12H22O11 consumed, we have $n_{\text{H}_{\text{2}}\text{O produced}}=n_{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\text{ consumed}}\times \text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}} \right)=\text{0.70 mol}\text{C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}\times \frac{\text{11 mol H}_{\text{2}}\text{O}}{\text{2 mol C}_{\text{12}}\text{H}_{\text{22}}\text{O}_{\text{11}}}=\text{3}\text{.85 mol H}_{\text{2}}\text{O}$
This calculation shows why an astronaut dirnks about 2 L of H2 per day, but excretes about 2.4 L of H2 per day! Think about it, and check your answer[1]
This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form
$\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}$ or symbolically. &nbnbsp; $n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol C12H22O11 cancels 1 mol C12H22O11 but does not cancel 1 mol H2O.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
EXAMPLE 3 It is estimated that each human exhales about 1 kg (2.2 lb)[2][3] of carbon dioxide per day. If that came entirely from glucose, what mass of glucose must be metabolized according to the equation:
C6H12O6 + 6 O2 → 6 CO2 + 6 H2O
Solution
The problem asks that we calculate the mass of C6H12O6 consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the amount of CO2 to the mass of CO2. Then we can calculate the amount of C6H12O6 consumed from the amount of CO2 produced with a stoichiometric ratio, just as in Example 2. Finally, we can convert the amount of glucose to its mass using the molar mass as a conversion factor. It requires the stoichiometric ratio.
$\text{S}\left( \frac{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}$
The amount of CO2 produced is n = m/M = 1000 g/44 g mol-1 = 22.7 mol
The amount of C6H12O6 consumed is then
$n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}= n_{\text{CO}_{\text{2}}\text{ produced}}\text{ }\!\!\times\!\!\text{ conversion factor}=\text{n mol CO}_{\text{2}}\times \frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}$
= 22.7 mol CO2 $\times \frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{6 mol CO}_{\text{2}}}$ = 3.79 mol
The mass of C6H12O6 is
$\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{3.79 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}\times \frac{\text{180 g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{682 g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}$
By similar calculations, we could show that to produce 6 mol of carbon dioxide, it takes the same amount of oxygen (22.7 mol) and the same amount (22.7 mol)of water will be produced. That's 22.7 mol x 32 g/mol = 726 g of oxygen consumed and 22.7 mol x 18 g/mol = 409 g of water produced. The reactants weigh 682 + 726 = 1408 g and the products weigh the same (within error), 1000 + 409 = 1409 g.
We notice two things: First, both products are eliminated (CO2 in breath, water in urine, so how do we gain weight? Only sugar that isn't metabolized goes into weight gain, and we'll see soon how to calculate that.
Second, this accounts for .409 kg of the total 2.4 L of water excreted per day. It's a rough estimate of the part that results from metabolism (not from ingested water).
With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of CO2 to amount (in moles) of C6H12O6 and the molar mass will convert amount of C6H12O6 to mass (in grams) of SO2. A schematic road map for the one-step calculation can be written as
$n_{\text{CO}_{\text{2}}}\text{ }\xrightarrow{S\text{(C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}\text{/O}_{\text{2}}\text{)}}\text{ }n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{M_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}}\text{ }m_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}$ Thus $\text{m}_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{x}\text{xx mol CO}_{\text{2}}\times \text{ }\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{11 mol O}_{\text{2}}}\text{ }\times \text{ }\frac{\text{64}\text{.06 g}}{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=\text{179 g}$ The chemical reaction in this example is of environmental interest. If each person on Earth exhales 1000 g of CO2/day, and there are 6.7 billion people, this accounts for 6.7 x 1012 (6.7 quadrillion) g or 6.2 Tg (teragrams) of the greenhouse gas CO2. How does this compare with the amount produced by burning fuels in its effect on the atmosphere? In 2000, fossil fuel burning released 7 x 109 tons of carbon dioxide [4]. That's 7 x 109 tons x 907 kg/ton x 1000 g/kg or 6.3 x 1015. Our breath contributes about 0.1%. The mass of atmospheric carbon dioxide comes from calculations like the following:
EXAMPLE 4 What mass of oxygen and carbon dioxide would be consumed when 3.3 × 1015 g, 3.3 Pg (petagrams), of octane (C8H18) is burned to produce CO2 and H2O?
Solution
First, write a balanced equation
2C8H18 + 25O2 → 16CO2 + 18H2O The problem gives the mass of C8H18 burned and asks for the mass of O2 required to combine with it. Thinking the problem through before trying to solve it, we realize that the molar mass of octane could be used to calculate the amount of octane consumed. Then we need a stoichiometric ratio to get the amount of O2 consumed. Finally, the molar mass of O2 permits calculation of the mass of O2. Symbolically
$m_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{M_{\text{C}_{\text{8}}\text{H}_{\text{18}}}}\text{ }n_{\text{C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\xrightarrow{S\text{(SO}_{\text{2}}\text{/C}_{\text{8}}\text{H}_{\text{18}}\text{)}}\text{ }n_{\text{O}_{\text{2}}}\xrightarrow{M_{\text{O}_{\text{2}}}}\text{ }m_{\text{O}_{\text{2}}}$
$m_{\text{O}_{\text{2}}}=\text{3}\text{.3 }\times \text{ 10}^{\text{15}}\text{ g }\times \text{ }\frac{\text{1 mol C}_{\text{8}}\text{H}_{\text{18}}}{\text{114 g}}\text{ }\times \text{ }\frac{\text{25 mol O}_{\text{2}}}{\text{2 mol C}_{\text{8}}\text{H}_{\text{18}}}\text{ }\times \text{ }\frac{\text{32}\text{.00 g}}{\text{1 mol O}_{\text{2}}}=\text{1}\text{.2 }\times \text{ 10}^{\text{16}}\text{ g }$ Thus 12 Pg (petagrams) of O2 would be needed. By a similar calculation, we see that the amount of carbon dioxide would be 16/2 = 8 times the amount of octane consumed, or 2.3 x 1014 mol, which, multiplied by the molar mass 44 g/mol, is 1 x 1016 g or 10 Pg. The large mass of oxygen obtained in this example is an estimate of how much O2 is removed from the earth’s atmosphere each year by human activities. Octane, a component of gasoline, was chosen to represent coal, gas, and other fossil fuels. Fortunately, the total mass of oxygen in the air (1.2 × 1021 g) is much larger than the yearly consumption. If we were to go on burning fuel at the present rate, it would take about 100 000 years to use up all the O2. Actually we will consume the fossil fuels long before that! One of the least of our environmental worries is running out of atmospheric oxygen. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.05%3A_Foods-_Metabolism_of_Dietary_Sugar.txt |
There are many demonstrations that purport to exemplify mass relationships and stoichiometry, but they deal with solutions or gases.
When reactions are demonstrated with solutions[1], students usually must (1) trust that a reported molarity is correct, and (2) deal with a layer of abstraction (reactant masses must be calculated from volumes, using the reported concentrations and molar masses. If gases are used (with a eudiometer, etc.)[2], the layer of abstraction exists through the ideal gas law.
In reactions between weighable reactants, (1) the product must not be lost as a smoke, etc., and (2) one reactant must be easily removed, so that the mass of a reactant can be determined from the mass of product less the mass of the other, initially weighed reactant. There are several possibilities, but none gives truly quantitative data, so they are best carried out (possibly with openly "idealized" data) to illustrate how a more carefully controlled experiment could yield more accurate results.
1. Al (or Zn, Mn) + I2 Powdered aluminum and powdered iodine are mixed together and then one or two drops of water are added to the mixture. Shortly after the water is added, a vigorous reaction occurs. There is flame and lots of iodine vapor given off. Some of the iodine vapor can be seen to be reacting to form some other substance because the color is no longer violet but kind of a reddish brown. This vigorous reaction shows that aluminum metal is a fairly strong reducing agent and iodine solid is a fairly strong oxidizing agent.[3]
2. Al + Br2 (here some AlBr3 may be lost along with vaporizing Br2) Must be done in a hood.[4]
3. Fe (or Zn) + S (here the excess sulfur burns in air, without much loss of ZnS). Must be done in a hood. [5]Metal powder and sulfur are mixed. A heated metal rod causes the mixture to react vigorously.
[6][7] FeS is nonstoichiometric, but close to 1:1[8]
1. Mg + S: In a mortar and pestle, grind together precisely weighed portions of ~0.5 g of powdered magnesium and about ~0.7 g powdered sulfur (Sulfur is flammable; it may be irritating to the eyes and to the respiratory system if inhaled as a dust. Pour the powder on a fireproof weighed ceramic plate, and ignite with a propane torch or butane torch lighter. Combustions forms a toxic and corrosive gas. Weigh the product.
2. P + Br2 (PBr4+ Br decomposes to PBr3 which may escape as smoke or vapor). Care! the reaction is quite violent with either red or white phosphorus. Must be done in a hood. [9][10] [11]
3. pyrophoric iron in air. Need not be done in a hood. Pyrophoric iron is not pure Fe, and the reaction isn't simply Fe + 3/2 O2 --> Fe2O3). But the reaction of pyrophoric iron with air is spectacular, and if a known mass of pyrophoric iron is spilled through air and collected, a calculated value for the mass of the product can be used, with appropriate caviates, to illustrate stoichiometric concepts.u
3.2.07: Sports Physiology and Health- Hydrogen Powered Bicycles Run on Water
Electric vehicles of all kinds are attracting increased interest everywhere, but electric bicycles are very popular in China (where there are over 120 million of them), the Netherlands, and India. They typically have a rechargeable battery pack and electric hub motor.
Electric hub motors on front or back wheels
A new electricity source combines a hydrogen fuel cell with a "sodium silicide" fuel cartridge, winner of a "Green Chemistry Challenge Award"[1]. The sodium silicyde reacts with water to make the hydrogen fuel [2][3][4]:
2 NaSi(s) + 5H2O(l) → Na2Si2O5(s) + 5H2(g) (1)
The composition of sodium silicide may depend on the method of synthesis. Silicides can be made by the reaction of active metals (like Mg) with sand, or by heating sodium with silicon. Dye et al [5] prepare sodium silicide by the reaction of sodium metal with silica gel, obtaining black powders of (hypothetically) Na4Si4 nanoparticles.
Equation (1) not only tells how many molecules of each kind are involved in a reaction, it also indicates the amount of each substance that is involved. Equation (1) says that 2 NaSi formula units can react with 5 H2O molecules to give 1 Na2Si2O5(s) formula unit and 5 H2 molecules. Here we're using the term "formula unit" to indicate that the substance may not be a molecule, but rather an ionic compound or ["network crystal"]. A "formula unit" gives the composition of the substance without specifying the type of bonding.
Equation (1) also says that 1 mol NaSi would react with 5 mol H2O yielding 1 mol Na2Si2O5(s) and 5 mol H2.
The balanced equation does more than this, though. It also tells us that 2 × 2 mol = 4 mol NaSi will react with 2 × 5 mol = 10 mol H2O, and that ½ × 2 mol = 1 mol NaSi requires only ½ × 5 = 2.5 mol H2O. In other words, the equation indicates that exactly 5 mol H2O must react for every 2 mol NaSi consumed. For the purpose of calculating how much H2O is required to react with a certain amount of NaSi therefore, the significant information contained in Eq. (1) is the ratio
$\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}$ We shall call such a ratio derived from a balanced chemical equation a stoichiometric ratio and give it the symbol S. Thus, for Eq. (1), $\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NaSi}} \right)=\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}~~~~~ \text{(2)}$
The word stoichiometric comes from the Greek words stoicheion, “element,“ and metron, “measure.“ Hence the stoichiometric ratio measures one element (or compound) against another.
EXAMPLE 1
Derive all possible stoichiometric ratios from Eq. (1)
Solution
Any ratio of amounts of substance given by coefficients in the equation may be used:
$\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}\text{O}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}}$ $\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}} \right) =\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{2}}$
$\text{S}\left( \frac{\text{NaSi}}{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}} \right) =\frac{\text{2 mol NaSi}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}$ $\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{H}_{\text{2}}} \right)=\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{5 mol H}_{\text{2}}}$ $\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}}$ $\text{S}\left( \frac{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{H}_{\text{2}}} \right) =\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{5 mol H}_{\text{2}}}$ There are six more stoichiometric ratios, each of which is the reciprocal of one of these. [Eq. (2) gives one of them.] When any chemical reaction occurs, the amounts of substances consumed or produced are related by the appropriate stoichiometric ratios. Using Eq. (1) as an example, this means that the ratio of the amount of H2O consumed to the amount of NaSi consumed must be the stoichiometric ratio S(H2O/NaSi): $\frac{n_{\text{H}_{\text{2}}\text{O}\text{ consumed}}}{n_{\text{NaSi}\text{ consumed}}}=\text{S}\left( \frac{\text{H}_{\text{2}}\text{O}}{\text{NaSi}} \right) =\frac{\text{5 mol H}_{\text{2}}\text{O}}{\text{2 mol NaSi}}$ Similarly, the ratio of the amount of H2 produced to the amount of NaSi consumed must be
S(H2/NaSi):
$\frac{n_{\text{H}_{\text{2}}\text{ produced}}}{n_{\text{NaSi}\text{ consumed}}} =\text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\frac{\text{5 mol H}_{\text{2}}}{\text{2 mol NaSi}}$ In general we can say that $\text{Stoichiometric ratio }\left( \frac{\text{X}}{\text{Y}} \right)=\frac{\text{amount of X consumed or produced}}{\text{amount of Y consumed or produced}}~~~~~\text{(3a)}$ or, in symbols, $\text{S}\left( \frac{\text{X}}{\text{Y}} \right)=\frac{n_{\text{X consumed or produced}}}{n_{\text{Y consumed or produced}}}~~~~~\text{(3b)}$
Note that in the word Eq. (3a) and the symbolic Eq. (3b), X and Y may represent any reactant or any product in the balanced chemical equation from which the stoichiometric ratio was derived. No matter how much of each reactant we have, the amounts of reactants consumed and the amounts of products produced will be in appropriate stoichiometric ratios.
EXAMPLE 2
Find the amount of hydrogen produced when 3.68 mol NaSi is consumed according to Eq. (1).
Solution
The amount of hydrogen produced must be in the stoichiometric ratio S(H2/NaSi) to the amount of ammonia consumed:
$\text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\frac{n_{\text{H}_{\text{2}}\text{ produced}}}{n_{\text{NaSi}\text{ consumed}}}$
Multiplying both sides nNaSi consumed, by we have $n_{\text{H}_{\text{2}}\text{ produced}} =n_{\text{NaSi}\text{ consumed}}\times \text{S}\left( \frac{\text{H}_{\text{2}}}{\text{NaSi}} \right) =\text{3}\text{.68 mol NaSi}\times \frac{\text{5 mol H}_{\text{2}}}{\text{2 mol NaSi}}=\text{9}\text{.20 mol H}_{\text{2}}$ This is a typical illustration of the use of a stoichiometric ratio as a conversion factor. Example 2 is analogous to Examples 1 and 2 from Conversion Factors and Functions, where density was employed as a conversion factor between mass and volume. Example 2 is also analogous to Examples 2.4 and 2.6, in which the Avogadro constant and molar mass were used as conversion factors. As in these previous cases, there is no need to memorize or do algebraic manipulations with Eq. (3) when using the stoichiometric ratio. Simply remember that the coefficients in a balanced chemical equation give stoichiometric ratios, and that the proper choice results in cancellation of units. In road-map form $\text{amount of X consumed or produced}\overset{\begin{smallmatrix} \text{stoichiometric} \ \text{ ratio X/Y} \end{smallmatrix}}{\longleftrightarrow}\text{amount of Y consumed or produced}$ or symbolically. $n_{\text{X consumed or produced}}\text{ }\overset{S\text{(X/Y)}}{\longleftrightarrow}\text{ }n_{\text{Y consumed or produced}}$
When using stoichiometric ratios, be sure you always indicate moles of what. You can only cancel moles of the same substance. In other words, 1 mol NaSi cancels 1 mol NaSi but does not cancel 1 mol H2.
The next example shows that stoichiometric ratios are also useful in problems involving the mass of a reactant or product.
EXAMPLE 3
Suppose it is reasonable to carry about 5 pounds (2268 g) of water on a bicycle for "fuel". Calculate the mass of NaSi that needs to be supplied by a cartridge on the bicycle to completely react with the water.
The problem asks that we calculate the mass of NaSi consumed from the mass of H2O consumed. As we learned in Example 2 of The Molar Mass, the molar mass can be used to convert from the mass of water to the amount of water. We can then use the appropriate stoichiometric ratio to calculate the amount of NaSi that will react, and finally, use the molar mass to calculate the mass of NaSi.
We require the stoichiometric ratio
$\text{S}\left( \frac{\text{NaSi}}{\text{H}_{\text{2}}} \right) =\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}}$
The amount of H2 present is
$\text{n(mol)}~=~\frac{\text{m(g)}}{\text{M(g/mol)}}$
= 2268 g/18.015 g/mol = 125.9 mol H2O
The amount of NaSi required is then
$n_{\text{NaSi consumed}} ~=~n_{\text{H}_{2}\text{O consumed}}~~\times~~\text{ conversion factor}$
$=\text{125.9 mol H}_{2}\text{O}~~\times~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}} =\text{50.36 mol NaSi}$
The mass of NaSi is $\text{m}_{\text{NaSi}}=\text{50}\text{.36 mol NaSi}\times \frac{\text{51}\text{.08 g NaSi}}{\text{1 mol NaSI}}=\text{2572 g NaSi}$
This is a reasonably sized cartridge (about 5.67 lb).
With practice this kind of problem can be solved in one step by concentrating on the units. The appropriate stoichiometric ratio will convert moles of H2O to moles of NaSi and the molar mass will convert moles of NaSi to grams of NaSi. A schematic road map for the one-step calculation can be written as
$n_{\text{H}_{2}\text{O}}~~\xrightarrow{S\text{(NaSi}\text{/H}_{\text{2}}\text{O)}}~~n_{\text{NaSi}}~~\xrightarrow{M_{\text{NaSi}}}~~m_{\text{NaSi}}$ Thus $\text{m}_{\text{NaSi}}=\text{125.9 mol H}_{\text{2}}\text{O}\times~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{\text{2}}\text{O}}~~\times~~\frac{\text{51.06 g}}{\text{1 mol NaSi}}=\text{2572 g NaSi}$ These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. You may verify the additional calculations that have been done to show the masses of hydrogen product that would be expected. We'll fill in the remaining spots below.
2 NaSi + 5 H2O → 1 Na2Si3O5 + 5 H2
m (g) 2572 2268 253.8
M (g/mol) 51.08 18.015 182.15 2.016
n (mol) 50.36 125.9 125.9
EXAMPLE 4
Suppose the 2572 g cannister of NaSi in Example 3 is completely depleted, which means completely converted to Na2Si2O5. What mass of the product results?
The problem gives the mass of NaSi and asks for the mass of Na2Si2O5 that would result from it's complete reaction with water. Thinking the problem through before trying to solve it, we realize that the molar mass of NaSi could be used to calculate the amount of NaSi consumed. Then we need a stoichiometric ratio to get the amount of Na2Si2O5 produced. Finally, the molar mass of Na2Si2O5 permits calculation of the mass of Na2Si2O5. Symbolically
$m_{\text{NaSi}}~~$ $\xrightarrow{M_{\text{NaSi}}}~~$ nNaSi $~~\xrightarrow{S\text{(Na}_{2}\text{Si}_{2}\text{O}_{5}\text{/NaSi)}}$ $~~n_{\text{Na}_{\text{2}}\text{Si}_{2}\text{O}_{5}}$ $~~\xrightarrow{M_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}}$ $~~m_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}$
$m_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}$ $=\text{2572 g }~~\times~~\frac{\text{1 mol NaSi}}{\text{58.08 g}}$ $~~\times~~\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{2 mol NaSi}}$ $~~\times~~\frac{\text{182.14 g}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}} =\text{4033 g }$
Now we can complete the table above by adding the amount of Na2Si2O5 (25.18 mol, half the amount of NaSi) and its mass, 4033 g (or about 8.9 lb). Will the bike have gained weight, since the cartridge went from 2572 g of NaSi to 4033 g of Na2Si2O5? | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.02%3A_Equations_and_Mass_Relationships/3.2.06%3A_Lecture_Demonstrations.txt |
Example 4 from Equations and Mass Relationships(opens in new window) also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the laboratory as well as the environment, inexpensive reagents like atmospheric O2 are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent, or the reactant (of two or more reactants) present in an amount such that it would be completely consumed if the reaction proceeded to completion. Also called limiting reactant.
For further understanding of limiting reagents, check out the video below or fiddle with the Phet simulation linked below the video.
Choose one of the games above to learn more about how limiting reactants work.
Example $1$: Mass of Product
When 100.0 g mercury is reacted with 100.0 g bromine to form mercuric bromide, which is the limiting reagent? What mass of product will be formed?
Solution: A mixture of one or more substances dissolved in a solvent to give a homogeneous mixture. The balanced equation A representation of a chemical reaction that has values of the stoichiometric coefficients of reactants and products such that the number of atoms of each element is the same before and after the reaction.
$\text{Hg } + \text{ Br}_{2} \rightarrow \text{Hg} \text{Br}_{2} \nonumber$
tells us that according to the atomic theory, 1 mol Hg is required for each mole. That chemical amount of a substance containing the same number of units as 12 g of carbon-12. of Br2. That is, the stoichiometric ratio S(Hg/Br2) = 1 mol Hg/ 1 mol Br2. Let us see how many moles of each we actually have
\begin{align} n_{\text{Hg}} &=\text{100.0 g}\times \dfrac{\text{1 mol Hg}}{\text{200.59 g}}=\text{0.4985 mol Hg} \ n_{\ce{Br2}}&=\text{100}\text{.0 g}\times \dfrac{\text{1 mol Br}_2}{\text{159.80 g}}=\text{0.6258 mol Br}_2 \ \end{align}
When the reaction ends, 0.4985 mol Hg will have reacted with 0.4985 mol Br2 and there will be
$( 0.6258 – 0.4985 ) \text{ mol Br}_2 = 0.1273 \text{ mol Br}_2 \nonumber$
left over. Mercury is therefore the limiting reagent.
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example $1$ this ratio of initial amounts
$\dfrac{n_{\text{Hg}}\text{(initial)}}{n_{\text{Br}_{\text{2}}}\text{(initial)}}=\dfrac{\text{0}\text{0.4985 mol Hg}}{\text{0}\text{0.6258 mol Br}_{\text{2}}}=\dfrac{\text{0}\text{0.7966 mol Hg}}{\text{1 mol Br}_{\text{2}}} \nonumber$
was less than the stoichiometric ratio
$\text{S}\left( \dfrac{\text{Hg}}{\text{Br}_{\text{2}}} \right)= \dfrac{\text{1 mol Hg}}{\text{1 mol Br}_{\text{2}}} \nonumber$
This indicated that there was not enough Hg to react with all the bromine and mercury was the limiting reagent. The corresponding general rule, for any reagents X and Y, is
\begin{align} & \text{If} ~ \dfrac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}} ~ \text{is less than S}\left( \dfrac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\dfrac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \dfrac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align} \nonumber
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Calculations of respective reactants and products in chemical equation
$\text{Hg}$ $+ \text{Br}_{2}$ $\rightarrow \text{HgBr}_{2}$
m (g) 100 100
M (g/mol) 200.59 159.80 360.398
n (mol) 0.4985 0.6259 --
if all Hg reacts -0.4985 -0.4985 +0.4985
if all Br2 reacts -0.6258 -0.6258
Actual Reaction
Amounts
-0.4985 -0.4985 +0.4985
Actual Reaction
Masses
-100 -79.66 +179.7
(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example.
Example $2$: Limiting Reagent
Iron can be obtained by reacting the ore hematite (Fe2O3) with coke (C). The latter is converted to CO2. As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have?
Solution:
a) Write a balanced equation $2 \text{Fe}_{2} \text{O}_{3} + 3 \text{C} \rightarrow 3\text{CO}_{2} + 4 \text{Fe}$
The stoichiometric ratio connecting C and Fe2O3 is:
$\text{S}\left( \dfrac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)= \dfrac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}} = \dfrac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}} \nonumber$
The initial amounts of C and Fe2O3 are calculated using appropriate molar masses
\begin{align} \text{ }n_{\text{C}}\text{(initial)} &=\text{2.84}\times \text{10}^{6}\text{g}\times \dfrac{\text{1 mol C}}{\text{12.01 g}}=\text{2.36}\times \text{10}^{5}\text{mol C} \ & \ n_{\ce{Fe2O3}}\text{(initial)}&=\text{20.5}\times \text{10}^{6}\text{g}\times \dfrac{\text{1 mol }\ce{Fe2O3}}{\text{159.69 g}}=\text{1.28}\times \text{10}^{5}\text{mol }\ce{Fe2O3} \end{align}
Their ratio is:
$\dfrac{n_{\text{C}}\text{(initial)}}{n_{\ce{Fe2O3}}\text{(initial)}} = \dfrac{\text{2.36}\times \text{10}^{5}\text{mol C}}{\text{1.28}\times \text{10}^{5}\text{mol }\ce{Fe2O3}} = \dfrac{\text{1.84 mol C}}{\text{1 mol }\ce{Fe2O3}} \nonumber$
Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first since it will be consumed first.
b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed. Therefore we use nFe2O3 (initial) to calculate how much Fe can be obtained.
$n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}} \nonumber$
$m_{\text{Fe}}=\text{1.28 }\times 10^5\text{ mol }\ce{Fe2O3}\text{ }\times \text{ } \dfrac{\text{4 mol Fe}}{\text{2 mol } \ce{Fe2O3}} \text{ }\times \text{ } \dfrac{\text{55.85 g}}{\text{mol Fe}} =\text{1.43 }\times 10^7\text{ g Fe} \nonumber$
This is 1.43 × 106 g, or 14.3 Mg, Fe.
As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
Liebig’s law of the minimum
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group, those elements that comprise a single column of the periodic table, also called family, of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams.
3.03: The Limiting Reagent
We have examined the idea of amino acid complementarity, where different foods in the diet supply the correct balance of amino acids to construct Human protein, in another section. Human proteins require stoichiometric amounts of about ten "essential" amino acids, so if only one is missing, the protein cannot be synthesized, and protein malnourishment results. This illustrates the concept of a "limiting reactant", which is a reactant present in less than the stoichiometric amount relative to other reactants.
A diverse group of surviving cultures worldwide have adopted diets, different as they may be, that supply the correct amino acid balance[1]. Anthropologists believe that fortuitous adoption of these diets provided survival value for the cultures. In areas where food is scarce, cultures that do not adopt diets with correct amino acid balance may not survive. Here are some examples of diets that supply approximately the correct stoichiometric amounts of amino acids:
Tostadas
In Latin America, diets combining corn tortillas or other corn products with beans, like the tostada shown here, are common.
Tostada with corn tortilla, bean topping, and vegetables[2]
Beans eaten alone provide limiting amounts of sulfur containing amino acids like methionine and cysteine, so these amino acids limit the amount of Human protein that can be synthesized. Beans have large amounts of the amino acids lysine and tryptophan, so these are "excess reactants" when used to synthesize Human protein, and will be degraded to urea and wasted. If wheat, rice, or corn are eaten alone, they typically provide amounts of lysine and tryptophan which limit the amount of Human protein that can be synthesized. But if beans are eaten with grains or corn, the excess reactants of the beans complement the limiting reactants of the grain (and vice versa). Human protein can then be synthesized efficiently, and very small amounts of amino acids are simply excreted as urea.
The origins of this diet are ancient. The Tehuacan Archaeological-Botanical Project (1961–1964) investigated the beginnings of agriculture in the New World and the concomitant rise of autochthonous (indigenous) civilization [3]. The project is described in six volumes, the first of which explores subsistence and the environment [4]. The domestication of maize after 8000 B.C.E. and beans (3000–4000 B.C.E.) in the Tehuacan Valley (the “Valley of Mexico”) of south central Mexico, gave inhabitants a nutritional advantage in the face of progressively deteriorating resources. Maize cob size increased rapidly under domestication from only a centimeter or two to current size, while Phaseolus vulgaris (kidney and pinto beans) similar to modern ones probably existed in preagricultural times. “Wherever these two food crops met, an immediate adaptive combination favored by human selection was formed.”[5].
Falafel and Pita
In the Middle East, a meal may consist of falafel and hummus made from chickpeas eaten with pita bread made from wheat.
Pita with humus and falafel balls[6]
Dal (lintil soup) Bhat (grain or maize)[7]
Dal and Chapatis
In India, rice or chapatis are eaten with dal (lentils) to provide amino acid balance.
Corn Deficiencies
We'll invesigate experiments dealing with corn, and the protein malnutrition that can result if it's eaten alone. It's well known that corn is a poor protein food by itself, being low in lysine and tryptophan [8], and even more so when it is nixtamalized. These deficiencies motivated researchers to develop the QPM (Quality Protein Maze) to increase concentrations of these essential amino acids in its protein.
Experiments on rats, whose amino acid requirements are similar to Humans', provide a basis for some stoichiometric calculations [9]. As the table below shows, there was little weight gain improvement when LYS or TRP were added separately to a "Base" diet. That means that neither alone is a limiting amino acid, preventing by itself the synthesis of rat protein. But when both were added, a significant increase was measured, along with a decrease in serum urea. That means that both were limiting, and when both were added, much less amino acid was wasted. The amino acids went into protein synthesis, rather than being simply metabolized and excreted as urea. While the addition of isoleucine in diet 5 made little difference (so it must be supplied adequately by corn), the addition of threonine (THR) alone to Diet 4 increased body weight gain, so it must be limiting in a diet of corn + LYS + TRP. Addition of ILE, methionine (MET), histidine (HIS) and valine (VAL) finally constitutes a nearly balanced diet, as evidenced by the large weight gains and low serum urea in Diet 8.
Table I: Effect of Protein Diet on Rat Nutrition
Diet weight gain
g/day
Serum Urea
mg/100 mL
1. Basea 1.75 36.2
2. Base + LYS 1.53 36.0
3. Base + TRP 1.56 42.3
4. Base + LYS + TRP 2.62 25.8
5. Diet 4 + ILE 2.49 24.9
6. Diet 4 + THR 3.29 20.4
7. Diet 4 + ILE + THR 4.20 23.7
8. Diet 7 + HIS + MET + VAL 5.95 7.4
9. Corn + Soybean meal 6.75 38.5b
aThe base diet consisted of 920.2 g corn, 30.0 g corn oil, 35.0 g mineral mix, 10 g vitamin mix, 2.5 g limestone, and 2.3 g choline. b This large increase is due to the doubling of total dietary protein, rather than protein inefficiency.
The same paper[10] provides the sequence, from most limiting to least, of the amino acids in corn, and the requirement of the rat (similar to Human) published by the NRC[11]. We've added the amino acid amounts in lentils to represent beans[12]
Table II: Amino Acid Profiles of Nutritional Requirements and Foods
Amino Acid Requirement
(g/kg)
Amount in corn
(g/kg)
Amount in lentils
(g/kg)
Lysine (LYS) 7.0 2.4 6.3
Tryptophan (TRP) 1.5 0.7 0.81
Methionine (MET)
and Cysteine (CYS)
6.0 3.8 0.77 + 1.18
Histidine (HIS) 3.0 2.0 2.54
Valine (VAL) 6.0 4.6 4.48
Isoleucine (ISO) 5.0 3.9 3.90
Threonine (THR) 5.0 3.9 3.23
Arginine (ARG) 6.0 5.0 6.97
Phenylalanine (PHN)
and Tyrosine (TYR)
8.0 9.6 4.45 + 2.41
Leucine (LEU) 7.5 11.0 6.54
How Amino Acids Form Proteins
Amino acids are joined by a very simple "condensation" reaction where two molecules join by eliminating water, as shown in the figure below. The portion of the molecule labeled "R" stands for a variable part of the molecule that distinguishes different amino acids; the rest of the molecule is common to all amino acids.
Formation of a peptide bond [13]
This process can continue with different amino acids adding to either end of the protein chain. The H-N of an amino acid adds to the C-OH end of the protein, eliminating water (H-OH) and forming a C-N bond where the water was eliminated (or the C-OH end of an amino acid adds to the N-H end of the protein). This process continues until the protein has hundreds of different amino acids.
Example 1: Stoichiometry of Corn Malnutrition
a. Correct, or Stoichiometric Amino Acid Ratios:
From data in the Table II above, calculate the optimal molar ratio of lysine (LYS, C6H14N2O2, Molar mass 146.19 g/mol) to Arginine (ARG, C6H14N4O2, Molar mass 174.2 g/mol) in the diet.
b. Limiting Amino Acid:
From data in the Table above, calculate which, LYS or ARG, is the limiting reactant in a corn diet?
c. How much of the excess amino acid will be wasted in 1 kg of corn diet?
Solution
a. The balanced equation will have coefficients that we need to determine:
x LYS + y ARG + other amino acids → Protein + water
According to the atomic theory, x mol LYS is required for each y mole of ARG to make useful protein. If the diet is optimal when 7.0 g of LYS is eaten with every 6.0 g of ARG (this ratio might be provided by eggs, for example), we can calculate the optimal stoichiometric ratio:
$\text{n}_{\text{LYS}}=\text{7.0 g}\times \frac{\text{1 mol LYS}}{\text{146}\text{.19 g}}=\text{0}\text{.048 mol LYS}$
$\text{n}_{\text{ARG}}=\text{6.0 g}\times \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.034 mol ARG}$
The ratio is $\frac{\text{LYS}}{ARG}~=~\frac{0.048}{0.034}~=~ \frac{1.4}{1}$
And since 1.4 is about 1-2/5 or 7/5,
$\frac{\text{7/5}}{1}~=~ \frac{7}{5}$
This exact amino acid ratio may not be found in any particular rat protein, but it is the average ratio for all the proteins in the rat. The (partial) chemical equation is
7 LYS + 5 ARG + other amino acids → Protein + water
That is, the stoichiometric ratio S(LYS/ARG) = 7 mol LYS / 5 mol ARG.
b. Let's see what the mole ratio is in the corn diet.
$\text{n}_{\text{LYS}}=\text{2.4 g}\times \frac{\text{1 mol LYS}}{\text{146}\text{.19 g}}=\text{0}\text{.016 mol LYS}$
$\text{n}_{\text{ARG}}=\text{5.0 g}\times \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.029 mol ARG}$
The ratio is $\frac{LYS}{ARG}~=~\frac{0.016}{0.029}~=~ \frac{0.55}{1}$
Clearly, LYS is the limiting amino acid because the ratio initial amounts of LYS to ARG (0.55:1) is much less than the required stoichiometric ratio (1.4 : 1).
General Approach to Limiting Reactant Problems
From the corn diet example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together.
The corresponding general rule, for any reagents X and Y, is
\begin{align} & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align}
Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.
We can verify that LYS is limiting by calculating the amount of LYS that would be required if all the ARG reacted, using a stoichiometric ratio:
$n_{\text{ARG}}\text{ }\xrightarrow{S\text{(LYS/ARG}\text{)}}\text{ }n_{\text{LYS}}$
$\text{0.029 mol ARG}~\times~\frac{\text{7 mol LYS}}{\text{5 mol ARG}}~=~\text{0.041 mol LYS}$
Similarly, if all the LYS reacts, the required amount of ARG is
$\text{0.016 mol LYS}~\times~\frac{\text{5 mol ARG}}{\text{7 mol LYS}}~=~\text{0.018 mol ARG}$
This is much less than the amount present, so ARG is the Excess Reactant.
c. We can easily use the molar masses to convert the amounts of each amino acid to the masses.
Let's set up a table:
7 LYS + 5 ARG other amino acids → Protein + water
M (g/mol) 146.19 174.2
m (g) 2.4 5.0
n (mol)present 0.016 0.029
if all ARG reacts -0.041 -0.029
if all LYS reacts -0.016 -0.011
Remaining amount 0 0.018
Remaining mass (g) 0 3.14
When the reaction ends, 3.14 g of the original 5.0 g (63%) remains, and will be metabolized to urea and excreted. Of the entire diet of 2.4 g LYS + 5.0 g ARG, we see that 3.14 g or 42% is wasted. What a waste of food and the resources to produce it! Notice that the corn serving will also have to be larger than the serving of optimal diet for the same amount of protein, because corn has only 5 g/kg ARG, while the optimal diet has 6.0 g/kg. The optimal diet might be provided by meat or eggs, but they are environmentally demanding and present health concerns.
Example 2: Complementary Amino Acids
A subsistence bean diet may also lead to protein malnutrition, as inspection of Table II above shows. Although the Lysine content is much greater than in beans, the methionine (MET) and Cystine (CYS) levels are low, as are the Phenylalanine (PHE) and Tyrosine (TYR).
But suppose beans and corn are eaten in the same day, in a diet that mixed equal masses of corn and lentils. Now recalculate the limiting amino acid in a diet of 2.4 + 6.3 = 8.7 g of LYS and 5.0 + 6.97 = 11.97 g of ARG.
b. In this case, how much of the excess amino acid is wasted?
Solution
$\text{n}_{\text{LYS}}~=~\text{8.7 g}~\times~ \frac{\text{7 mol LYS}}{\text{146.19 g}}=\text{0}\text{.060 mol LYS}$
$\text{n}_{\text{ARG}}~=~\text{11.97 g}~\times~ \frac{\text{1 mol ARG}}{\text{174.2 g}}=\text{0}\text{.069 mol ARG}$
Now we see that the ratio of amounts present is
0.060 mol LYS / 0.069 mol ARG = 0.869,
which is still less than the stoichiometric ratio,
7 mol LYS / 5 mol ARG = 1.4
So LYS is once again the limiting amino acid. Recalculating the values in a table as above, we get
7 LYS + 5 ARG other amino acids → Protein + water
M (g/mol) 146.19 174.2
m (g) 8.7 11.97
n (mol)present 0.060 0.069
if all ARG reacts -0.080 -0.069
if all LYS reacts -0.060 -0.043
Remaining amount 0 0.027
Remaining mass (g) 0 4.61
In this case, 4.61 g of the excess ARG remains from a 11.97 g portion, or 38% of the ARG. The total diet is 8.7 g LYS + 11.97 g ARG = 20.7 g, and only 22% of that is wasted, a big improvement over the corn alone case above.
In order to design a proper diet, complementary protein foods would have to be chosen from tables of amino acid contents of foods.
Is Nixtamalization an Anthropological Counterargument?
In Central America, field corn or "maize" is traditionally boiled in lime (CaO) to loosen the hulls and soften the kernels in the preparation of tortilla flour. The process is shown beautifully in a Culinary Institute of America film. The Southern US breakfast "grits" are also made this way[14].
But this process converts some of the already limiting tryptophan to 2-aminoacetophenone, exacerbating the poor nutritional, and survival value, of the corn. But the nixtamalization also makes niacin in the kernels more bioavailable, reducing the incidence of Pellagra, and more than offsetting the loss of amino acids if plenty of beans are also included in the diet. The beans & corn diet, even with nixtamalization, has survival value for cultures.
Example 3: Amino Acid Detection
To detect an amino acid (even in a fingerprint in forensic chemistry), the ninhydrin test is often used.
Ninhydrin
In the ninhydrin test, two ninhydrin molecules (C9H6O4, shown to the left) become linked by the N attached to the first carbon of the amino acid chain, producing the blue/purple ion shown at right.
Blue/Purple Product[15]
The balanced chemical equation is:
2 C9H6O4 + C11H12N2O2 → (C9H5O2)-N=(C9H4O2) + C10H9NO + CO2 + 3 H2O
If 2.00 mg of ninhydrin (Nin) is used to detect 2 mg of TRP, has enough ninhydrin been added to react with all the TRP? Which is the limiting reagent, and what mass of H2O will be formed?
Solution
The stoichiometric ratio connecting Nin and TRP is
$\text{S}\left( \frac{\text{Nin}}{\text{TRP}} \right)=\frac{\text{2 mol Nin}}{\text{1 mol TRP}}$ The initial amounts of Nin and TRP are calculated using appropriate molar masses(160.13 g mol−1 for Nin and 204.23 g mol−1 for TRP: \begin{align} & \text{ }n_{\text{Nin}}\text{(initial)}=\text{2}\text{.00}\times \text{10}^{\text{-3}}\text{g}\times \frac{\text{1 mol Nin}}{\text{178}\text{.14 g}}=\text{1}\text{.12}\times \text{10}^{\text{-5}}\text{mol Nin} \ & \ & n_{\text{TRP}}\text{(initial)}=\text{2}\text{.00}\times \text{10}^{\text{-3}}\text{g}\times \frac{\text{1 mol TRP}}{\text{204}\text{.23 g}}=\text{9}\text{.79}\times \text{10}^{\text{-6}}\text{mol TRP} \ \end{align} Their ratio is $\frac{n_{\text{Nin}}\text{(initial)}}{n_{\text{TRP}}\text{(initial)}}=\frac{\text{1}\text{.123}\times \text{10}^{\text{-5}}\text{mol Nin}}{\text{9}\text{.79}\times \text{10}^{\text{-6}}\text{mol TRP}}=\frac{\text{1}\text{.15 mol Nin}}{\text{1 mol TRP}}$ Since this ratio is smaller than the 2:1 stoichiometric ratio, you have insufficient Nin to react with all the TRP, so Nin is the limiting reagent. To ensure detection, in would be better to add excess Ninhydrin.
b) The amount of water product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant TRP will be left over, but all the initial amount of Nin will be consumed. Therefore we use nNin (initial) to calculate how much H2O can be obtained
$n_{\text{Nin}}\text{ }\xrightarrow{S\text{(water/Nin}\text{)}}\text{ }n_{\text{water}}\xrightarrow{M_{\text{water}}}\text{ }m_{\text{water}}$ $m_{\text{water}}=\text{1}\text{.12 }\times \text{ 10}^{\text{-5}}\text{ mol Nin }\text{ }\times \text{ }$ $\frac{\text{3 mol water}}{\text{2 mol Nin}}\text{ }\times \text{ }\frac{\text{18}\text{.016 g}}{\text{mol water}}=\text{3}\text{.02 }\times \text{ 10}^{\text{-4}}\text{ g water}$
This is 0.302 mg water.
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. Calculations are shown for each possible case, assuming that one reactant is completely consumed and determining if enough of the other reactants is present to consume it. If not, that scenario is discarded.
2 C9H6O4 + C11H12N2O2 (C9H5O2)-N=(C9H4O2)) + C10H9NO + CO2 + 3 H2O
m (g) 0.0020 0.0020
M (g/mol) 178.1 204.23 303.3 159.2 44.0 18.0
n (mol) 1.12 x 10-5 9.79 x 10-6 -- -- -- --
if all C9H6O4 reacts -1.12 x 10-5 -5.62 x 10-6
if all C11H12N2O2 reacts -1.96 x 10-5 -9.79 x 10-6
Actual Reaction
Amounts
-1.12 x 10-5 -5.62 x 10-6 5.62 x 10-6 5.62 x 10-6 5.62 x 10-6 1.68 x 10-5
Actual Reaction
Masses
-0.0020 -0.00115 1.71 x 10-3 8.93 x 10-4 2.47 x 10-4 3.02 x 10-4
As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.01%3A_Cultural_Connections-_Anthropology_and_Protein_Stoichiometry.txt |
Algal Blooms
Algal blooms like the ones in the photos below may be harmful when the algae are toxic, or if they reduce the oxygen concentration enough to emperil other organisms. [1]. Blooms are visible because algae concentrations may reach millions of cells per milliliter. Blooms often result when one limiting reagent is supplied to the environment, either naturally, or through Human activities. A limiting reagent is one of several reactants that is necessary for a reaction to occur, but which is present in low concentration, so no reaction occurs even though there is an excess of all other reactants.
Coccolithophore algal bloom in the Bering Sea in 1998[2]
A "red tide" which may poison seafood and cause human illness or death, caused by a dinoflagellate species.[3]
The limiting reagent that prevents uncontrolled algae growth is often phosphorus, and it may be in low concentrations because phosphate mineral sources ("phosphate rock", like apatite) are insoluble.
Solubilizing Phosphate Rock: H3PO4
Phosphate often limits growth of foodcrops, so producing soluble phosphate is a significant sector of the fertilizer industry. Runoff from agricultural fields is often the cause of algal blooms.
Phosphate rock is solubilized for fertilizer by treatment with sulfuric acid, giving phosphoric acid and, as a byproduct, gypsum (CaSO4· 2 H2O used in Plaster of Paris and "drywall").
The reaction is:
Ca5(PO4)3Cl + 5 H2SO4 + 10 H2O → 3 H3PO4 + 5 CaSO4 · 2 H2O + HCl
Phosphoric Acid[4]
Limiting Reagent Example 1
EXAMPLE 1 When 100.0 g of chloroapetite rock is reacted with 100.0 g of sulfuric acid to form phosphoric acid and gypsum, which is the limiting reagent? --- Solution The balanced equation
Ca5(PO4)3Cl + 5 H2SO4 + 2 H2O → 3 H3PO4 + 5 CaSO4 · 2 H2O + HCl
tells us that according to the atomic theory, 1 mol Ca5(PO4)3Cl is required for every 5 moles of H2SO4. That is, the stoichiometric ratio S(Ca5(PO4)3Cl /H2SO4) = 1 mol Ca5(PO4)3Cl/ 5 mol H2SO4. Let us see how many moles of each we actually have
\begin{align} & n_{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}=\text{100}\text{.0 g}\times \frac{\text{1 mol}}{\text{520}\text{.8 g}}=\text{0}\text{.192 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl} \ & n_{\text{H}_{\text{2}}\text{SO}_{\text{4}}}=\text{100}\text{.0 g}\times \frac{\text{1 mol}_{\text{2}}}{\text{98}\text{.1 g}}=\text{1}\text{.02 mol H}_{\text{2}}\text{SO}_{\text{4}} \ \end{align} If all the H2SO4 were to react, it would require 1.02 mol H2SO4 x (1 mol Ca5(PO4)3Cl / 5 mol H2SO4) = 0.204 mol Ca5(PO4)3Cl, but only 0.192 mol is present. So Ca5(PO4)3Cl is the limiting reagent, an all of the H2SO4 cannot react.
When the reaction ends, 0.960 mol H2SO4 will have reacted with 0.192 mol Ca5(PO4)3Cl and there will be (1.02 – 0.960) mol H2SO4 = 0.06 mol H2SO4 left over. Ca5(PO4)3Cl is therefore the limiting reagent.
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. Calculations are shown for each possible case, assuming that one reactant is completely consumed and determining if enough of the other reactants is present to consume it. If not, that scenario is discarded.
Solutions to Example 1
Ca5(PO4)3Cl + 5 H2SO4 + 10 H2O → 3 H3PO4 + 5 CaSO4 · 2 H2O + HCl
m (g) 100 100
M (g/mol) 521 98.1 18.0 98.0 172.2 36.5
n (mol) 0.192 1.02 -- -- -- --
if all Ca5(PO4)3Cl reacts -0.192 -0.960 -1.92 +0.576 +0.960 +0.192
if all H2SO4 reacts -0.204 -1.02
Actual Reaction
Amounts
0.192 0.960 1.92 0.576 0.960 0.192
Actual Reaction
Masses
100 94.2 34.2 56.5 165.2 7.0
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts $\frac{n_{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}\text{(initial)}}}{n_{\text{H}_{\text{2}}\text{SO}_{\text{4}}}\text{(initial)}}=\frac{\text{0.192 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{1.02 mol H}_{\text{2}}\text{SO}_{\text{4}}}= \frac{\text{0.188 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{mol H}_{\text{2}}\text{SO}_{\text{4}} }$ was less than the stoichiometric ratio $\text{S}\left( \frac{\text{Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{H}_{\text{2}}\text{SO}_{\text{4}}} \right)=\frac{\text{1 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{5 mol H}_{\text{2}}\text{SO}_{\text{4}}}= \frac{\text{0.200 mol Ca}_{\text{5}}\text{(PO}_{\text{4}}\text{)}_{\text{3}}\text{Cl}}{\text{mol H}_{\text{2}}\text{SO}_{\text{4}} }$ This indicated that there was not enough Ca5(PO4)3Cl to react with all the H2SO4 and Ca5(PO4)3Cl was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If}\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align} (Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example.
TriSodium phosphate, TSP
The phosphoric acid may be applied as a fertilizer solution, or converted to trisodium phosphate (TSP), a solid:
H3PO4 + 3 NaOH → 3 Na3PO4 + 3 H2O
Trisodium Phosphate, TSP. Note that the Na+ ions are ionically bonded.[5]
TSP is used as a cleaning and degreasing agent, and was used extensively in household detergents in the US until its deleterious effects on the environment were appreciated in the 1970s.[6]
Limiting Reagents and Ecological Stoichiometry
Since release of TSP and phosphoric acid may not be controlled in all countries, and in all activities, they often provide a source for phosphorus which is a limiting reagent in nature because of the low solubility of mineral phosphates.
As we explained when we discussed the significance of formulas, ecological stoichiometry examines the stoichiometric relationship between the nutritional demands of a species and the food available to the species.[7] [8]. Proponents say that "Ecological stoichiometry recognizes that organisms themselves are outcomes of chemical reactions and thus their growth and reproduction can be constrained by supplies of key chemical elements [especially carbon (C), nitrogen (N) and phosphorus (P)]".[9]
For example, by writing an approximate chemical equation for photosynthesis in oceanic algae, we can predict which nutrients (nitrogen as potassium nitrate, KNO3-, phosphorus as phosphoric acid, H3PO42-, etc.) are required for algae growth, and what products result from algal respiration.
106 CO2(g) + 16 KNO3(aq) + H3PO4(aq) + 122 H2O(l) + 16 H+(aq) ↔
C106H263O110N16P1(s) + 138 O2(g) + 16 K+ (1)
The "formula" for algae, (C106H263O110N16P1)(M = 3553.259 g/mol) does not represent a single molecule, but just the overall composition of the algae (one might call it an "average" molecular formula)[10].
Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the environment, inexpensive reagents like atmospheric O2 are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent.
Limiting Reagent Example 2
EXAMPLE 2 In a small scale experiment to model fertilizer runoff, water containing 10 g of H3PO4 contaminates a small pond which already contains 300 g of KNO3. If the pond contains stable algae which forms according to the equation above, and plenty of CO2 and other reactants are available, (a) what is the limiting reagent, and (b) how much algae can form as a result of the runoff?
Solution
a) The stoichiometric ratio connecting KNO3 and H3PO4 is
$\text{S}\left( \frac{\text{KNO}_{\text{3}}}{\text{H}_{\text{3}}\text{PO}_{\text{4}}} \right)=\frac{\text{16 mol KNO}_{\text{3}}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}$ The initial amounts of KNO3 and H3PO4 are calculated using appropriate molar masses \begin{align} & \text{ }n_{\text{KNO}_{\text{3}}}\text{(initial)}=\text{300 g}\times \frac{\text{1 mol KNO}_{\text{3}}}{\text{101}\text{.1 g}}=\text{2.96}\text{mol KNO}_{\text{3}} \ & \ & n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{(initial)}=\text{10}\text{.0 g}\times \frac{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}{\text{98.0 g}}=\text{0}\text{.102}\text{mol H}_{\text{3}}\text{PO}_{\text{4}} \ \end{align} Their ratio is $\frac{n_{\text{KNO}_{\text{3}}}\text{(initial)}}{n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{(initial)}}=\frac{\text{2}\text{.96}\text{mol KNO}_{\text{3}}}{\text{0}\text{.102}\text{mol H}_{\text{3}}\text{PO}_{\text{4}}}=\frac{\text{29}\text{.0 mol KNO}_{\text{3}}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}$ Since this ratio is larger than the stoichiometric ratio, you have more than enough KNO3 to react with all the H3PO4. H3PO4 is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant KNO3 will be left over, but all the initial amount of H3PO4 will be consumed. Therefore we use nH3PO4 (initial) to calculate how much algae can be obtained $n_{\text{H}_{\text{3}}\text{PO}_{\text{4}}}\text{ }\xrightarrow{S\text{(Algae/H}_{\text{3}}\text{PO}_{\text{4}}\text{)}}\text{ }n_{\text{Algae}}\xrightarrow{M_{\text{Algae}}}\text{ }m_{\text{Algae}}$ $m_{\text{Algae}}=\text{0.102}\text{ mol H}_{\text{3}}\text{PO}_{\text{4}}\text{ }\times \text{ }\frac{\text{1 mol Algae}}{\text{1 mol H}_{\text{3}}\text{PO}_{\text{4}}}\text{ }\times \text{ }\frac{\text{3553 g}}{\text{mol Algae}}=\text{362 g Algae}$ Note that only 10 g of H3PO4 allowed the consumption of 456 g of carbon dioxide to make 362 g of Algae!
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. The H+ and K+ have been omitted.
Solutions to Example 2
106 CO2 + 16 KNO3 + H3PO4 + 122 H2O ↔ C106H263O110N16P1 + 138 O2
m (g) 300 10
M (g/mol) 44 101.1 98.0 18.0 3553.259 32
n (mol) present 2.96 0.102 -- -- --
if all KNO3 reacts -2.96 -0.185
if all H3PO4 reacts -1.63 -0.102
Actual Reaction
Amounts
-10.81 -1.63 -0.102 -12.44 +0.102 +14.1
Actual Reaction
Masses
-476 -165 -10 -224 362
450
As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.02%3A_Environment-_TSP_Ecological_Stoichiometry_and_Algal_Blooms.txt |
Which video is more exciting?!?
Make sure your volume is turned up!
If you liked the one on the bottom, you have just experienced the excitement of how different amounts of chemicals can affect things. We often use the concept of limiting reagents in everyday life without even realizing it, especially while cooking or preparing a grocery list!
What do and have to do with chemistry?
When you cook, often times you can encounter a limiting reagent calculation that you have been doing in your head before you can remember. This page will use the familiarity of an everyday example to teach the steps of a limiting reagent problem.
Here are two things to look for when trying to identify a limiting reagent problem: 1) it should be a task that you start with at least two starting materials and 2) it should form at least one new product. Another main concept found in all limiting reagent questions is there will be starting material that remains when the task you are doing is complete. There are three methods (A, B, and C) that can be used to identify the limiting reagent. This is the most important step of this type of question. As you progress through this page, choose the method that is most intuitive to you.
An Example with Two Starting Materials
EXAMPLE 1 I want to have friends over for lunch on Saturday and make grilled cheese sandwiches that require two slices of bread and one slice of cheese. I open the refrigerator to find that I have 40 slices of cheese. I look in the bread box to find that I have 16 slices of bread.
Question 1: Which of my ingredients is the limiting the number of sandwiches I can make?
Question 2: How many sandwiches can I make?
Question 3: How much of my starting material is left over once I am done making sandwiches?
Jump to Step 3 - Identify a Limiting Reagent
Answers
These three steps are the first steps to solving this type of question and will be used in conjunction with ALL three methods A, B, and C.
Step 1) Write out the recipe, also known as an equation
+
2 slices of bread + 1 slice of cheese → 1 grilled cheese sandwich
Step 2) Find Quantity (moles) and Identify useful unit conversions and/or molar ratios
What we know:
1) We know we have 40 slices of cheese
2) We know we have 16 slices of bread.
According to the equation written in step 1, these ratios can be written by using the number in front of the ingredient, also known as a co-efficient.
2 slices of bread is required for one slice of cheese.
Remember, the powerful thing about ratios is that we can also write them "upside-down".
The following ratios or unit conversions that can be written from this equation are: \begin{align} \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 slice of cheese}} \ & \ \frac{\text{1 sandwich}}{\text{2 slices of bread}}&=\frac{\text{2 slices of bread}}{\text{1 sandwich}} \ & \ \frac{\text{1 sandwich}}{\text{1 slice of cheese}}&=\frac{\text{1 slice of cheese}}{\text{1 sandwich}} \ \end{align}
Step 3) Use these ratios as a unit conversion to identify the limiting reagent.
Useful Math Review: Units cancel out just like numbers. If there is a unit in the numerator (on top of the ratio) and the same unit in the denominator (on the bottom of the ratio), they will cancel each other out.
Question 1: Which of my ingredients is the limiting the number of sandwiches I can make? You maybe just did this calculation in your head without even realizing the math that you did.
Use Method A, Method B, and/or Method C to answer question 1.
Method A - Step 3: Identify the Limiting Reagent
Let's use what we know and our unit conversion to lead us to what we want to know. Never start a math problem with a unit conversion if you don't have to.
What we know:
1) We know we have 40 slices of cheese
2) We know we have 16 slices of bread.
Let's set up equations using the ratios from Step 2 to figure out how much of the other starting material we would need to make our sandwiches. \begin{align} & n_{\text{bread needed}}=\text{40 slices of cheese}\times \frac{\text{2 slices of bread}}{\text{1 slice of cheese}}=\text{80 slices of bread} \ & \ & n_{\text{cheese needed}}=\text{16 slices of bread}\times \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}=\text{8 slices of cheese} \ \end{align} These calculations show us that we need 80 slices of bread . . . do we have that many? No.
These calculations show us that we need 16 slices of cheese . . . do we have that many? Yes.
Because we don't have enough bread, bread is considered to be our limiting factor, also known as the limiting reagent.
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to Step 4, to answer question 2 and 3.
Method B - Step 3: Identify the Limiting Reagent
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting before moving on to the final step of calculating the amount of product you can make.
We must compare the theoretical stoichiometric ratio X/Y with the actual ratio of amounts of X and Y which were initially mixed together. Whichever ratio is LESS than the theoretical stoichiometric ratio given by the co-efficients in the equation is considered to be the limiting reagent.
In Example 1 this ratio of initial amounts Actual ratio (using the materials we actually have)
$\frac{n_{\text{slices of bread (actual)}}}{n_{\text{slices of cheese (actual)}}}=\frac{\text{16 slices of cheese}}{\text{40 slices of bread}}=\frac{\text{0.4 slices of bread}}{\text{1 slice of cheese}}$ was LESS than the theoretical stoichiometric ratio Theoretical ratio (in an ideal world we would have all the materials necessary)
$\text{Theoretical}\left( \frac{\text{1 slices of bread}}{\text{2 slices of cheese}} \right)=\frac{\text{0.5 slice of bread}}{\text{1 slices of cheese}}$ $\frac{\text{0.4 slices of bread}}{\text{1 slice of cheese}}<\frac{\text{0.5 slice of bread}}{\text{1 slices of cheese}}$
This indicates that there is not enough bread to react with all the cheese and the bread is the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \ & \ & \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \ \end{align} Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to Step 4, to answer question 2 and 3.
Method C - Step 3: Identify the Limiting Reagent
These calculations can also be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Don't forget about the ratios when using the table method.
Option 1: Use all of the cheese
cheese + 2 bread → sandwich
Starting material (quantity) 40 16 --
if all cheese is used -40 -80 n/a
Actual Product
(quantity)
0 -64
n/a
There can't be a negative amount of something
Option 2: Use all of the bread
cheese + 2 bread → sandwich
Starting material (quantity) 40 16 --
if all bread is used -8 -16 +8
Actual Product
(quantity)
32 0 +8
Because Option 2 leaves all of the products with positive numbers, the bread is the limiting reagent. Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We can now move on to Step 4, to answer question 2 and 3.
Step 4) Find the answer
Now we can answer question 2 and 3 using the Limiting Reagent that we found using Method A, B or C.
Question 2: Calculate how many sandwiches we can make (must start with limiting reagent)
Answer 2:
\begin{align} & n_{\text{sandwiches made}}=\text{16 slices of bread}\times \frac{\text{1 sandwich}}{\text{2 slice of bread}}=\text{8 sandwiches} \ \end{align} Question 3: When we are finished making sandwiches, how much cheese will be left over? (This is a two-step calculation)
Answer 3:
\begin{align} & n_{\text{cheese USED}}=\text{16 slices of bread}\times \frac{\text{1 slice of cheese}}{\text{2 slices of bread}}=\text{8 slices of cheese USED} \ \end{align} Next, simply subtract how much cheese you used from how much you started with.
\begin{align} & \text{40 slices of cheese to START}-\text{8 slices of cheese USED}=\text{32 slices of cheese left over} \ \end{align}
A follow up question could be:
Question 4: If I wanted to make sure that I used up all of the left over cheese and make a special trip to the store. How much bread should I get at the store.
(This calculation would be similar to Question/Answer 2.)
Answer 4:
\begin{align} & n_{\text{bread needed}}=\text{32 slices of cheese}\times \frac{\text{2 slices of bread}}{\text{1 slice of cheese}}=\text{64 more slices of bread need to be purchased} \ \end{align}
(Of course, when the actual and theoretical molar ratio of X and Y are equivalent, both reagents will be completely consumed at the same time, and neither is in excess.). Back to top of page
An Example with more than Two Starting Materials
EXAMPLE 2 One of my favorite breakfast foods are ham and cheese omelets in the morning. I am on a diet so I always measure the amount of each ingredient I use. I always make two in case someone else wants one. Below is the recipe for my 'perfect' omelet.
Recipe:
6 Large eggs - 200. g per one egg
1 cup of ham - 125. g per one cup
2 cups of shredded cheese - 50. g per one cup
I open the refrigerator this morning to find an excess of large eggs (50 eggs), 400. g of ham and 250. g of cheese.
Question 1: How many omelets can I make for breakfast with all of the ingredients I pulled out of the refrigerator?
Question 2: What remains when I am done cooking all of the omelets?
Question 3: What is the total mass of our cooked omelets?
Jump to Step 3
Answers
Step 1) Write out the recipe in the form of a balanced equation
+ + + +
6 Large eggs + 2 cups of cheese + 1 cup of ham → 2 omelets
Step 2) Find Quantity (moles) and Identify useful unit conversions and/or molar ratios
Unit Conversions
According to the equation written in step 1, these molar ratios can be written by using the number in front of the ingredient, also known as a co-efficient.
Once again, when you are first starting out, it is helpful to write the ratios or the stoichiometric ratios connecting all of the components of the recipe.
It makes it easier to pick out which one ratio you will use as a unit conversion to help you calculate what the question is asking.
As you can see, there are more molar ratios this time because there are more starting materials. Additionally, the question gave us information about the mass per unit food item. Therefore there are 8 unit conversions in this question.
$\frac{\text{6 Large eggs}}{\text{2 cups of cheese}}=\frac{\text{2 cups of cheese}}{\text{6 Large eggs}}$ $\frac{\text{6 Large eggs}}{\text{1 cup of ham}}=\frac{\text{1 cup of ham}}{\text{6 Large eggs}}$
$\frac{\text{6 Large eggs}}{\text{2 omelets}}=\frac{\text{2 omelets}}{\text{6 Large eggs}}$ $\frac{\text{2 cups of cheese}}{\text{2 omelets}}=\frac{\text{2 omelets}}{\text{2 cups of cheese}}$
$\frac{\text{2 cups of cheese}}{\text{1 cup of ham}}=\frac{\text{1 cup of ham}}{\text{6 Large eggs}}$ $\frac{\text{1 cup of ham}}{\text{2 omelets}}=\frac{\text{2 omelets}}{\text{1 cup of ham}}$
$\frac{\text{1 cup of ham}}{\text{125 g ham}}=\frac{\text{125 g ham}}{\text{1 cup of ham}}$ $\frac{\text{1 cup of cheese}}{\text{50. g cheese}}=\frac{\text{50. g cheese}}{\text{1 cup of cheese}}$
Find Moles
Then, we need to convert all of the starting materials from mass into the quantity associated with recipe or equation to be able to compare the starting materials.
The actual quantity of ham and cheese can be calculated using the mass per one unit of material. For instance, in this case, 1 cup of cheese is 50. grams.
In chemistry terms, this is called the molar mass. Molar mass can be used as a unit conversion. Refer above for useful unit conversions
To relate the idea of molar mass to everyday life is something we do all of the time without even thinking about it.
For instance, you have an assortment of candy in a bucket. Three kids come up to you and ask you for some candy.
Do you weigh the candy and give the three kids equivalent masses? No. That would be unrealistic and probably a little messy.
Instead, we just decide that one candy bar should be given to each kid. Kid #1 receives one Snickers bar, kid #2 receives one Milky Way and kid #3 receives one Kit Kat bar. They are all happy because they received their OWN candy bar; they don't care that one might weigh a little bit more than the other candy bars.
To summarize, Snickers, Milky Way, and Kit Kat all represent ONE candy bar, however they have different masses...but that's okay. To relate it even more to chemistry, one mole of any compound is equivalent to one mole of any other compound, they just have a different mass depending on what substance you are talking about; just substitute these words and you are back to our everyday example; 'mole' = 'candy bar' and 'compound' = 'type of candy bar'.
You can use this idea in reverse, if you know the mass, then the amount of candy bars or 'moles' can be calculated. \begin{align} & \text{ }n_{\text{cheese (actual)}}=\text{250. g of cheese}\times \frac{\text{1 cup of cheese}}{\text{50. g of cheese}}=\text{5.00 cups of cheese} \ \ & \text{ }n_{\text{ham (actual)}}=\text{400. g of ham}\times \frac{\text{1 cup of ham}}{\text{125 g of ham}}=\text{3.2 cups of ham} \ \end{align}
Step 3) Find the Limiting Reagent
Method A - Step 3: Identify the Limiting Reagent
Let's use what we know and our unit conversion to lead us to what we want to know. Never start a math problem with a unit conversion if you don't have to.
What we know:
1) We know we have 5.0 cups of cheese
2) We know we have 3.2 cups of ham.
Let's set up equations using the ratios from Step 2 to figure out how much of the other starting material we would need to make our sandwiches. \begin{align} & n_{\text{cheese needed}}=\text{3.2 cups of ham}\times \frac{\text{2 cups of cheese}}{\text{1 cup of ham}}=\text{6.4 cups of cheese} \ & \ & n_{\text{ham needed}}=\text{5.0 cups of cheese}\times \frac{\text{1 cup of ham}}{\text{2 cups of cheese}}=\text{2.5 cups of ham} \ \end{align} These calculations show us that we need 6.4 cups of cheese . . . do we have that much? No.
These calculations show us that we need 2.5 cups of ham . . . do we have that many? Yes.
Because we don't have enough cheese, cheese is considered to be our limiting factor, also known as the limiting reagent.
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to Step 4, to answer Example 2, questions 1, 2, and 3.
Method B - Step 3: Identify the Limiting Reagent
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting before moving on to the final step of calculating the amount of product you can make.
Since the eggs, or one starting material is in excess it doesn't need to be involved in finding the limiting reagent.
We must compare the theoretical stoichiometric ratio X/Y with the actual ratio of amounts of X and Y which were initially mixed together. Whichever ratio is LESS than the theoretical stoichiometric ratio given by the co-efficients in the equation is considered to be the limiting reagent.
In Example 1 this ratio of initial amounts Actual ratio (using the materials we actually have)
$\frac{n_{\text{cups of cheese (actual)}}}{n_{\text{cups of ham (actual)}}}=\frac{\text{5.0 cups of cheese}}{\text{3.2 cups of ham}}=\frac{\text{1.5625 cups of cheese}}{\text{1 cup of ham}}$ was LESS than the theoretical stoichiometric ratio Theoretical ratio (in an ideal world we would have all the materials necessary)
$\text{Theoretical}\left( \frac{\text{2 cups of cheese}}{\text{1 cup of ham}} \right)=\frac{\text{2.0 cups of cheese}}{\text{1 cup of ham}}$ $\frac{\text{1.5625 cups of cheese}}{\text{1 cup of ham}}<\frac{\text{2.0 cups of cheese}}{\text{1 cup of ham}}$
This indicates that there is not enough cups of cheese to react with all the ham and the cheese is the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \ & \ & \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \ \end{align} Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to Step 4, to answer Example 2, questions 1, 2, and 3.
Method C - Step 3: Identify the Limiting Reagent
These calculations can also be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Note: Don't forget about the ratios when using the table method.
Option 1: Use all of the ham
6 eggs + 1 cup of ham + 2 cups of cheese → 2 omelets
Starting material (quantity) 50 5 3.2 -
if all ham is used -30 -5 -6.4 n/a
Actual Product
(quantity)
20 0 -3.2
n/a
There can't be a negative amount of something
Option 2: Use all of the cheese
6 eggs + 1 cup of ham + 2 cups of cheese → 2 omelets
Starting material (quantity) 50 5 3.2 -
if all cheese is used -19.2 -1.6 -3.2 n/a
Actual Product
(quantity)
30.8 3.4 0 3.2
Option 3: Use all of the bread
6 eggs + 1 cup of ham + 2 cups of cheese → 2 omelets
Starting material (quantity) 50 5 3.2 -
if all cheese is used -50 -8.3 -16.7 n/a
Actual Product
(quantity)
0 -3.3 -13.5
n/a
There can't be a negative amount of something
Because Option 2 leaves us with all positive numbers, the cheese is the limiting reagent. Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We can now move on to Step 4, to answer Example 2, questions 1, 2, and 3.
Step 4) Find the answer
Now we can answer Example 2, questions 1 2 and 3 using the Limiting Reagent that we found using Method A, B or C.
Question 1: How many omelets can I make for breakfast with all of the ingredients I pulled out of the refrigerator? (Start with limiting reagent)
Answer 1: \begin{align} & n_{\text{omelets made}}=\text{5.0 cups of cheese}\times \frac{\text{2 omelets}}{\text{2 cups of cheese}}=\text{5 omelets} \ \end{align}
Question 2: What remains when I am done cooking the omelets? (This is a two-step calculation)
Answer 2:
\begin{align} & n_{\text{ham USED}}=\text{5.0 cups of cheese}\times \frac{\text{1 cup of ham}}{\text{2 cups of cheese}}=\text{2.5 cups of ham USED} \ \end{align}
\begin{align} & n_{\text{eggs USED}}=\text{5.0 cups of cheese}\times \frac{\text{6 Large eggs}}{\text{2 cups of cheese}}=\text{15 Large eggs USED} \ \end{align} Next, simply subtract how much ham and eggs you used from how much you started with.
\begin{align} & \text{3.2 cups of ham to START}-\text{2.5 cups of ham USED}=\text{0.7 cups of ham left over} \ \end{align}
\begin{align} & \text{50 Large eggs to START}-\text{15 Large eggs USED}=\text{35 Large eggs left over} \ \end{align}
Question 3: What is the total weight of one omelet
Answer 3: Use the answer to Question 1 in order to start this problem. \begin{align} & {\text{Mass of all omelets}}=\text{5.0 omelets}\times \frac{\text{(1200 g of egg + 125 g. of ham + 100. g of cheese)}}{\text{2 omelets}}=\text{3562.5 g of omelets} \ \end{align}
A follow up question could be:
Question 4: How much does one omelet weigh?
(This calculation would be similar to the molar mass explained above.)
Answer 4:
\begin{align} & {\text{Mass of one omelet}}=\frac{\text{3562.5 g of omelet}}{\text{5 omelets}}=\text{712.5 g per one omelet} \ \end{align} Back to top of page
Example 3 - Relating Real World examples to Chemical Equations
Ore hematite Pure Coke (carbon source)
Ore hematite, Fe2O3, is the chief iron ore used in production of iron metal.
Reacting the ore hematite (Fe2O3) with coke (C) produces iron metal (Fe) and CO2. As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have?
If you get stuck, go back to Example 2 and review each step using terms you are familiar with. You will use the exact same process and calculation to solve Example 3 as you used in Example 2.
Answer
Step 1
Write a balanced equation 2 Fe2O3 (s) + 3 C(s) → 3 CO2 (g) + 4 Fe(s)
Step 2
Find Moles and Identify useful Unit Conversions
The initial amounts of C and Fe2O3 are calculated using appropriate molar masses found on the periodic table
\begin{align} & \text{ }n_{\text{Carbon (actual)}}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g C}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g C}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \ & \ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{ (actual)}}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g Fe}_{\text{2}}\text{O}_{\text{3}}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \ \end{align}
The stoichiometric ratio connecting C and Fe2O3 is
$\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}= \frac{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{3 mol C}}$ Compare to Step 2 in the Omelet Example
Step 3
Find the limiting reagent
Method A
Let's use what we know and our unit conversion to lead us to what we want to know. Never start a math problem with a unit conversion if you don't have to.
What we know:
1) We know we have 1.28 x 105 mol of Fe2O3
2) We know we have 2.36 x 105 mol C.
Let's set up equations using the the ratio from Step 2 to figure out how much of the other starting material we would need to make our sandwiches. \begin{align} & \text{ }n_{\text{Carbon needed}}=\text{1.28}\times \text{10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\times \frac{\text{3 mol C}}{\text{2}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{1.92}\times \text{10}^{\text{5}}\text{ mol C needed} \ & \ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}\text{ needed}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{ mol C}\times \frac{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{3 mol C}}=\text{1.57}\times \text{10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ needed} \ \end{align}
These calculations show us that we need 1.92 x 105 mol C . . . do we have that ? Yes. We have 2.36 x 105 mol C
These calculations show us that we need 1.57 x 105 mol Fe2O3 . . . do we have that? No. We only have 1.28 x 105 mol Fe2O3
Because we don't have enough Fe2O3, Fe2O3 is considered to be our limiting factor, also known as the limiting reagent.
Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction. We could now move on to Step 4, to answer Question 1 and 2.
Method B
Their ratio is
$\frac{n_{\text{C}}\text{(actual)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(actual)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ was MORE than the Theoretical stoichiometric ratio
$\text{Theoretical}\left( \frac{\text{3 C}}{\text{2 Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{1.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ $\frac{\text{1.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}>\frac{\text{1.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$
This indicates that there is not enough Fe2O3 to react with all of the carbon (C). Therefore Fe2O3 is the limiting reagent.
In other words, you have more than enough C to react with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first since it will be consumed first.
The corresponding general rule, for any reagents X and Y, is
\begin{align} & \text{If Actual}~ \frac{\text{X}}{\text{Y}}<\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then X is limiting}\text{.} \ & \ & \text{If Actual}~ \frac{\text{X}}{\text{Y}}>\text{Theoretical}~ \frac{\text{X}}{\text{Y}}\text{, then Y is limiting}\text{.} \ \end{align}
We could now move on to Step 4, to answer Question 1 and 2.
Method C
2 Fe2O3 (s) + 3 C(s) → 3 CO2 (g) + 4 Fe(s)
Option 1: Use all of the iron (III) oxide
2 Fe2O3 (s) + 3 C (s) → 3 CO2 (g) + 4 Fe (s)
Starting material (mol) 1.28 x 105 2.36 x 105 -- --
if all Fe2O3 is used -1.28 x 105 -1.92 x 105 +1.92 x 105 +2.56 x 105
Actual Product
(quantity)
0 4.40 x 104 1.92 x 105 2.56 x 105
Option 2: Use all of the carbon
2 Fe2O3 (s) + 3 C (s) → 3 CO2 (g) + 4 Fe (s)
Starting material (mol) 1.28 x 105 2.36 x 105 -- --
if all carbon is used -1.57 x 105 -2.36 x 105 +2.36 x 105 +3.15 x 105
Actual Product
(quantity)
-2.9 x 104 0
n/a
Amounts cannot be negative
Because Option 1 leaves all of the amounts with positive numbers, the Fe2O3 is the limiting reagent. Once the limiting reagent is identified, only the initial amount of the limiting reagent can be used to start any calculations regarding this reaction.
Step 4
b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed. Therefore we use the actual $n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}$ to calculate how much Fe can be obtained
Four general steps:
Step 1) Write and equation
Step 2) Find moles
Step 3) Use molar ratio to identify limiting reagent
Step 4) Convert to the answer.
$m_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\xrightarrow{divide M_{\text{Fe}}}n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{times \text{ (Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{times M_{\text{Fe}}}\text{ }m_{\text{Fe}}$ $n_{\text{mol Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\text{2.56}\times \text{ 10}^{\text{5}}\text{ mol Fe}$
$m_{\text{ g Fe}}=\text{2}\text{.56 }\times \text{ 10}^{\text{5}}\text{ mol Fe}\times \frac{\text{55}\text{.85 g}}{\text{1 mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}$
We will make 1.43 × 106 g Fe, or 14.3 Mg, Fe with this amount of reagents.
Compare to Step 4 in the Omelet Example
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Let's Practice!
Check out these simulations to help you with this concept. Follow the four steps below.
1) Write a BALANCED chemical equation. The equation in the simulation is not balanced.
2) Choose two starting masses for your starting materials.
3) Work through the steps to calculate how much of each product will be left.
4) Work through the steps to calculate how much of a reactant will be left once the reaction is complete.
5) Click "Run Trial" to check if you are right!
As you can see from all of these examples, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams.
Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. In the laboratory as well as the environment, inexpensive reagents like atmospheric O2 are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent.
Back to top of page | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.03%3A_Everyday_Life-_Grilled_Cheese_Sandwiches_and_Omelets.txt |
Previously, we explored the Stoichiometry of hydrogen powered bicycles that "run on water". They typically have a rechargeable battery pack and electric hub motor. The company that makes the electrical fuel cell won a Presidential Green Chemistry Award[1].
Electric hub motors on front or back wheels
Now we'll see how the water cannister and fuel canister need to be matched for maximum range.
In the laboratory as well as the environment, inexpensive reagents like atmospheric O2 or water are often supplied in excess. Some portion of such a reagent will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent.
A new electricity source combines a hydrogen fuel cell with a "sodium silicide" fuel cartridge, winner of a "Green Chemistry Challenge Award"[2]. The sodium silicyde reacts with water to make the hydrogen fuel [3][4][5]:
2 NaSi(s) + 5H2O(l) → Na2Si2O5(s) + 5H2(g) (1)
The composition of sodium silicide may depend on the method of synthesis. Silicides can be made by the reaction of active metals (like Mg) with sand, or by heating sodium with silicon. Dye et al [6] prepare sodium silicide by the reaction of sodium metal with silica gel, obtaining black powders of (hypothetically) Na4Si4 nanoparticles.
Example 1
Sodium silicide/water bikes produce about 200W to run a bicycle for about 30 miles with a NaSi cartridge weighing about 1.5 lb [7]. If a bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent? What mass of solid product will be formed?
Solution
The balanced equation
2 NaSi(s) + 5H2O(l) → Na2Si2O5(s) + 5H2(g) (1)
tells us that according to the atomic theory, 2 mol NaSi is required for each 5 moles of 5H2O. That is, the stoichiometric ratio S(NaSi/H2O) = 2 mol NaSi/ 5 mol H2O. Let us calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL:
$\text{m}_{\text{H}_{2}\text{O}}~=~$ $\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}$
$\text{m}_{\text{NaSi}}~=~ [itex]\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}$
$\text{n}_{\text{NaSi}}=\text{680 g}\times \frac{\text{1 mol NaSi}}{\text{51.075 g}}=\text{13.3 mol NaSi}$
$\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{\text{2}\text{O}}$
If all the H2O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required:
$\text{n}_{\text{NaSi}}~=~ \text{n}_{\text{H}_{2}\text{O}}~~\times~~$ $~~\frac{\text{2 mol NaSi}}{\text{5 mol H}_{2}\text{O}}~~$ $~=~\text{23.5 mol NaSi}$
This is more than the amount of NaSi present,so NaSi is the limiting reactant and H2O is present in excess.
If all the NaSi reacts, the stoichiometric ratio allows us to calculate the amount of H2O that would be required:
$\text{n}_{\text{H}_{2}\text{O}}~=~ \text{n}_{\text{NaSi}}~~\times~~$ $~~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~~$
$~=~\text{13.3 mol NaSi}$ $~\times~\frac{\text{5 mol H}_{2}\text{O}}{\text{2 mol NaSi}}~=~$ 33.3 mol H2O
We require less than the amount of H2O present, so it is the excess reactant.
When the reaction ends, 13.3 mol of NaSi will have reacted with 33.3 mol H2O and there will be
(58.57 mol - 33.3 mol) = 25.4 mol H2O left over. NaSi is therefore the limiting reagent.
b. Since the water doesn't all react, we need to calculate the amount of solid product produced from the amount of NaSi consumed, by using the stoichiometric ratio:
$\text{n}_{\text{Na}_{2}\text{Si}_{2}\text{O}_{5}}~=~ \text{n}_{\text{NaSi}}~~\times~~$ $~~\frac{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}{\text{2 mol NaSi}}~~$ $~=~\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}$
The mass of water is then calculated by using the molar mass:
$\text{6.65 mol Na}_{2}\text{Si}_{2}\text{O}_{5}~\times~$ $\frac{\text{182.148 g}}{\text{1 mol Na}_{2}\text{Si}_{2}\text{O}_{5}}$ $~=~\text{1211 g Na}_{2}\text{Si}_{2}\text{O}_{5}$
c. About a pound (458 g) of water will react, and 1057-458 = 599 g of water will remain. That's enough for another 1.5 lb canister of NaSi.
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
2 NaSi + 5 H2O → 1 Na2Si2O5 + 5 H2
m (g) 680 1057
M (g/mol) 51.075 18.015 182.148 2.016
n (present,mol) 13.3 58.57
if all H2O reacts -58.67 -23.50 +11.73
if all NaSi reacts -13.3 -33.3 +6.65 +33.3
Actual Reaction
Amounts
-13.3 -33.3 +6.65 +33.3
Actual Reaction
Masses
-680 -600 +1211 +67.2
In the end, 1057 - 599 = 458 g of H2O will remain, along with 1211 g of sodium silicate. The 67.2 g of hydrogen produced gives a total of 1736 g. The mass of the reactants was also 1057 + 680 = 1737 g, equal within the error of measurement.
General Strategy for Limiting Reactant Problems
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together.
The general rule, for any reagents X and Y, is
\begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align}
(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example.
Example 2
Suppose the sodium silicide made by a different method is Na2Si [8][9], and it produces ordinary "water glass", or sodium silicate (Na2SiO3(s)[10]) rather than Na2Si2O5(s)[11]. In that case, if the bicyclist wants to travel about 30 miles and brings 1 quart of water and one 1.5 lb cartridge, which is the limiting reagent?
Solution
The balanced equation is:
Na2Si(s) + 3H2O(l) → Na2SiO3(s) + 3H2(g) (1)
Now the stoichiometric ratio S(NaSi/H2O) = 1 mol Na2Si/ 3 mol H2O. Again, let's calculate the amount in moles of each we actually have, assuming the density of water is 1.00 g/mL:
$\text{m}_{\text{H}_{2}\text{O}}~=~$ $\text{1 quart}~\times~\frac{\text{1 L}}{\text{1.05668821 quart}}~\times~\frac{\text{1000 g}}{\text{1 L} }~ =~\text{1057 g}$
$\text{m}_{\text{NaSi}}~=~ [itex]\text{1.5 lb}~\times~\frac{\text{453.59 g}}{\text{ quart}}~ =~\text{680 g}$
$\text{n}_{\text{Na}_{2}\text{Si}}~=~$ $\text{680 g}\times \frac{\text{1 mol Na}_{2}\text{Si}}{\text{74.065 g}}~=~$ 9.18 mol Na2Si
$\text{n}_{\text{H}_{\text{2}}\text{O}}=\text{1057 g}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18.015 g}}=\text{58.57 mol H}_{2}\text{O}$
If all the H2O were to react, the stoichiometric ratio allows us to calculate the amount of NaSi that would be required:
$\text{n}_{\text{Na}_{2}\text{Si}}~=~$ $\text{n}_{\text{H}_{2}\text{O}}~~\times~~$ $~~\frac{\text{1 mol Na}_{2}\text{Si}}{\text{3 mol H}_{2}\text{O}}~~$ $~=~\text{19.52 mol Na}_{2}\text{Si}$
This is more than the amount of Na2Si present,so Na2Si is the limiting reactant and H2O is present in excess.
If all the Na2Si reacts, the stoichiometric ratio allows us to calculate the amount of H2O that would be required:
$\text{n}_{\text{H}_{2}\text{O}}~=~$ $\text{n}_{\text{Na}_{2}\text{Si}}~~\times~~$ $~~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~~$
$~=~\text{13.3 mol Na}_{2}\text{Si}$ $~\times~\frac{\text{3 mol H}_{2}\text{O}}{\text{1 mol Na}_{2}\text{Si}}~=~$ 27.54 mol H2O
We require less than the amount of H2O present, so it is the excess reactant.
When the reaction ends, 9.18 mol of Na2Si will have reacted with 27.54 mol H2O and there will be:
(58.57 mol - 27.54 mol) = 31.03 mol H2O left over. Na2Si is therefore the limiting reagent.
These calculations can again be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
1 Na2Si + 3 H2O → 1 Na2SiO3 + 3 H2
m (g) 680 1057
M (g/mol) 74.065 18.015 122.06 2.016
n (present,mol) 9.18 58.57
if all H2O reacts -58.67 -19.52 +58.67 +19.52
if all NaSi reacts -9.18 -27.54 +27.54 +9.18
Actual Reaction
Amounts
-9.18 -27.54 +27.54 +9.18
Actual Reaction
Masses
-680 -496 +1121 +55.5
In the end, 1057 - 496 = 561 g of H2O will remain, along with 1121 g of sodium silicate. The 55.5 g of hydrogen produced gives a total of 1176 g. The mass of the reactants was also 496 + 680 = 1176 g, equal within the error of measurement.
This process seems less efficient than the one in Example 1, because only 55.5 g of hydrogen is produced, compared to 67.2 g in Ex. 1. Also, 496 g of water and 680 g of silicide are consumed (totalling 1176 g), while in Ex. 1, 599 g water and 680 g of silicide (1280 g total) were consumed, so we get 0.047 g hydrogen/g reactants in Ex. 2, while we get 0.053 g hydrogen/g reactants in Ex. 1. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.04%3A_Everyday_Life-_Sodium_Silicide_Fueled_Bicycles.txt |
Various formulations of gunpowder were apparently discovered and used before 1000 AD in China, and its military use is documented during the Jin Dynasty(1115–1234). Rockets, guns, cannons, grenades, and bombs were used against invading Mongols. Since the late 19th century, the original formulation has been called "black powder" to distinguish it from modern smokeless varieties[1]. Knowledge of gunpowder formulations, and of the products of their explosions, is essential in gunshot residue (GSR) analysis.
Solids and gases resulting from explosion of gunpowder[2]
A modern black powder substitute for muzzleloading rifles in FFG size[3]
Black powder is usually 75% potassium nitrate (KNO3, known as saltpeter or saltpetre), 15% softwood charcoal , and 10% sulfur (elemental S). Charcoal is made by heating wood with limited air, and is mostly carbon (elemental C), but contains trace minerals (such as potassium carbonate, K2CO3) and some partially decomposed wood chemicals like lignin C9H10O2, cellulose (C6H10O5)n.
There is no simple equation for the combustion of black powder because the products, as well as the reactants, are numerous and varied, as shown in this table:
55.91% solid products (in decending order of quantities) 42.98% gaseous products (in decending order of quantities)
K2CO3, K2SO4, K2S, S, KNO3, KSCN, C, NH4CO3, CO2, N2, CO, H2S, H2, CH4, H2O
The main products are K2CO3, CO2, and N2, so the equation for the combustion can be given as[4]
10 KNO3 + 3 S + 8 C → 2 K2CO3 + 3 K2SO4 + 6 CO2 + 5 N2 (1)
But it is often simplified to[5]:
2 KNO3 + S + 3 C → K2S + N2 + 3 CO2 (2)
Sometimes, formulas for charcoal (like C7H4O) that approximate it's composition, but don't represent any actual compound in the charcoal, are used in place of C:
6 KNO3 + C7H4O + 2 S → 2 K2S + 4 CO2 + 3 CO + 2 H2O + 2 N2 (3)
Example 4 from Equations and Mass Relationships we noted that one reactant in a chemical equation may be completely consumed without using up all of another. A mixture like gunpowder is formulated for "average conditions", and some portion of reactants may be left unchanged after the reaction. Conversely, at least one reactant is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent.
EXAMPLE 1 If 100 g of black powder is made from the general recipe above (75 g KNO3, 15 g C, 10 g S), and the combustion reaction is given by equation (2), which is the limiting reagent? What mass of solid product will be formed?
Solution
The balanced equation
2 KNO3 + S + 3 C → K2S + N2 + 3 CO2 (2)
Let's see how many moles of each we actually have
\begin{align} & n_{\text{KNO}_{\text{3}}}=\text{75.0 g}\times \frac{\text{1 mol KNO}_{\text{3}}}{\text{101.1 g}}=\text{0.742 mol KNO}_{\text{3}} \ & n_{\text{S}}=\text{10.0 g}\times \frac{\text{1 mol S}}{\text{32.1 g}}=\text{0.312 mol S} \ & n_{\text{C}}=\text{15.0 g}\times \frac{\text{1 mol C}}{\text{12.01 g}}=\text{1.249 mol C} \ \end{align}
Now we can use stoichiometric ratios to determine how much C and S would be required to react with all of the KNO3:
$n_{\text{S}}=\text{0.742 mol KNO}_{\text{3}}~\times~\frac{\text{ mol S}}{\text{2 mol KNO}_{\text{3}}}~=~\text{0.371 mol S}$
$n_{\text{C}}=\text{0.742 mol KNO}_{\text{3}}~\times~\frac{\text{3 mol C}}{\text{2 mol KNO}_{\text{3}}}~=~\text{1.13 mol C}$
Since only 0.312 mol S are present, and 0.371 mol S would be required to react with all of the KNO3, it is clear that this can't happen, and KNO3 must be present in excess. One of the other reactants must be limiting.
We can use stoichiometric ratios to discover how much KNO3 and S would be required if all the C reacts:
$n_{\text{KNO}_{\text{3}}}=\text{1.25 mol C}~\times~\frac{\text{2 mol KNO}_{\text{3}}}{\text{3 mol C}}~=~\text{0.833 mol KNO}_{\text{3}}$
$n_{\text{S}}=\text{1.25 mol C}~\times~\frac{\text{1 mol S}}{\text{3 mol C}}~=~\text{0.416 mol S}$
We see that C is also present in excess, so S must be the limiting reactant. We can prove it by using stoichiometric ratios to find out that there is plenty of C and KNO3 to react with all the S:
$n_{\text{C}}=\text{0.312 mol S}~\times~\frac{\text{3 mol C}}{\text{1 mol S}}~=~\text{0.936 mol C}$
$n_{\text{KNO}_{\text{3}}}=\text{0.312 mol S}~\times~\frac{\text{2 mol KNO}_{\text{3}}}{\text{1 mol S}}~=~\text{0.624 mol KNO}_{\text{3}}$
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting.
2 KNO3 + S +3 C → K2S + N2 + 3 CO2
m (g) 75.0 10.0 15.0
M (g/mol) 101.1 32.1 12.01 110.3 28.01 44.01
n (mol) 0.742 0.312 1.25
if all KNO3 reacts -0.742 -0.371 -1.13
if all S reacts -0.624 -0.312 -0.936
if all C reacts -0.833 -0.416 -1.25
Actual Reaction
Amounts
-0.624 -0.312 -0.936 + 0.312 +0.312 +0.936
Actual Reaction
Masses
-63.1 -10.0 -11.24 +34.4 +8.74 +41.2
We use the amount of limiting reagent to calculate the amount of product formed. $n_{\text{S}}~\xrightarrow{S\text{(K}_{\text{2}}\text{S/S)}}~n_{\text{K}_{\text{2}}\text{S}}\xrightarrow{M_{\text{K}_{\text{2}}\text{S}}} ~ m_{\text{K}_{\text{2}}\text{S}}$
$m_{\text{K}_{\text{2}}\text{S}} ~=~ \text{0.312 mol S} ~\times~ \frac{\text{1 mol K}_{\text{2}}\text{S}}{\text{1 mol S}} ~ \times ~ \frac{\text{110.3 g}}{\text{mol K}_{\text{2}}\text{S}}=\text{34.4 g K}_{\text{2}}\text{S}$
When the reaction ends, there will be (0.742 – 0.624) mol KNO3 = 0.118 mol KNO3, or 11.9 g left over. There will also be (1.25 - 0.936) = 0.314 mol C, or 3.76 g left over. S is therefore the limiting reagent.
The left over solids in GSR (gunshot residue) are detected by swiping areas with adhesive coated samplers, which are then viewed with a scanning electron microscope to indentify the particles.
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts
$\frac{n_{\text{S}}\text{(initial)}}{n_{\text{KNO}_{\text{3}}}\text{(initial)}}=\frac{\text{0.312 mol S}}{\text{0.742 mol KNO}_{\text{3}}}=\frac{\text{0.420 mol S}}{\text{1 mol KNO}_{\text{3}}}$ was less than the stoichiometric ratio $\text{S}\left( \frac{\text{S}}{\text{KNO}_{\text{3}}} \right)=\frac{\text{1 mol S}}{\text{2 mol KNO}_{\text{3}}}~=~ 0.5$ This indicated that there was not enough S to react with all the KNO3 and sulfur was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align}
(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example.
EXAMPLE 2 Iron can be obtained by reacting the ore hematite (Fe2O3) with coke (C). The latter is converted to CO2. As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have?
Solution
a) Write a balanced equation 2Fe2O3 + 3C → 3CO2 + 4Fe
The stoichiometric ratio connecting C and Fe2O3 is $\text{S}\left( \frac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ The initial amounts of C and Fe2O3 are calculated using appropriate molar masses \begin{align} & \text{ }n_{\text{C}}\text{(initial)}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \ & \ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \ \end{align} Their ratio is $\frac{n_{\text{C}}\text{(initial)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed. Therefore we use nFe2O3 (initial) to calculate how much Fe can be obtained $n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}}$ $m_{\text{Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\times \text{ }\frac{\text{55}\text{.85 g}}{\text{mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}$ This is 1.43 × 106 g, or 14.3 Mg, Fe.
As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.06%3A_Forensics-_Gunpowder_Stoichiometry.txt |
Calamine is an obsolete name for what is now known to be a mixture of two distinct minerals: zinc carbonate (ZnCO3 or smithsonite) and zinc silicate (Zn4Si2O7(OH)2·H2O, or hemimorphite). the name "Calamine" is now used only for calomine lotion, which is a suspension of ZnO and Fe2O3[1].
Smithsonite and hemimorphite may be similar in appearance to one another, but their appearance may be quite variable depending on location, so two samples of smithsonite (or hemimorphite) may look quite different as shown in the figures:
Hemimorphite from Mapimi, Durango, Mexico[2]
Hemimorphite[3]
Smithsonite from Tsumeb, Namibia[4]
Smithsonite from Tsumeb, Namibia[5]
The two minerals can only be reliably distinguished through chemical analysis.
Carbonate minerals like calcite or smithsonite react with acids to efforvesce (fizz) while dissolving and producing CO2(see equation (1) below). This test can be done with 1 M HCl, or household vinegar (crushing the sample will help if vinegar is used). While calcite (CaCO3) bubbles strongly in cold dilute acid, dolomite CaMg(CO3)2) and rhodochrosite (MnCO3) bubble weakly. Smithsonite (along with Siderite, FeCO3 and Magnesite, MgCO3) require heating to react.
Silicates, like hemimorphite, don't generally react with cold, dilute acids at all. So we could tell if a sample contained just smithsonite, because it would ocmpletely dissolve in acid. hemimorphite would not react, and a mixture of the two would partially dissolve.
It would be necessary to add an excess of HCl to the sample, otherwise it might not all dissolve because there isn't enough HCl, not because it's partially hemimorphite. If we have a 100 g sample that may contain smithsonite, hemimorphite, or both, we need to add enough acid to react with the sample, assuming it's all smithsonite, just to make sure.
Example 4 from Equations and Mass Relationships also illustrates the idea that one reactant in a chemical equation may be completely consumed without using up all of another. Here we want the smithsonite to be completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first (the smithsonite) is the limiting reagent.
EXAMPLE 1 When 100.0 g of smithsonite is reacted with 100.0 g of HCl to form carbon dioxide gas, which is the limiting reagent? What mass of product will be formed? (Note: HCl is provided as a solution with a concentration of 1-5% HCl for this purpose. The mass of HCl solution would be much (20-100 times) greater than the mass of HCl given here).
Solution
The balanced equation
ZnCO3 + 2 HCl → ZnCl2 + CO2 + H2O (1) tells us that according to the atomic theory, 2 mol HCl are required for each mole of ZnCO3. That is, the stoichiometric ratio S(HCl/ZnCO3) = 2 mol HCl/ 1 mol ZnCO3. Let us see how many moles of each we actually have \begin{align} & n_{\text{HCl}}=\text{100}\text{.0 g}\times \frac{\text{1 mol HCl}}{\text{36.5 g}}=\text{2}\text{.74 mol HCl} \ & n_{\text{ZnCO}_{\text{3}}}=\text{100}\text{.0 g}\times \frac{\text{1 mol ZnCO}_{\text{3}}}{\text{125}\text{.4 g}}=\text{0}\text{.798 mol ZnCO}_{\text{3}} \ \end{align}
Calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting.
For example, if all the HCl were to react, it would require
$\frac{\text{1 mol ZnCO}_3}{\text{2 mol HCl}}~~ x ~~ \text{2.74 mol HCl} = \text{1.37 mol ZnCO}_3$
Since there is not this much ZnCO3 present, this is impossible. HCl is in excess, and ZnCO3 is the limiting reactant. In the table, we've crossed out this calculation, and proceeded to calculate how much HCl would be required if all the ZnCO3 reacts (which is what happens).
calculations of how much HCl would be required if all the ZnCO3 reacts
ZnCO3 + HCl → ZnCl2 + CO2 + H2O
m (g) 100 100
M (g/mol) 125.1 36.5 136.3 44.0 18.0
n (mol) 0.798 2.74
if all ZnCO3 reacts -0.798 -1.60 +0.798 +0.798 +0.798
if all HCl reacts -2.74 -1.37
Actual Reaction
Amounts
-0.798 -1.60 +0.798 +0.798 +0.798
Actual Reaction
Masses
-100 -58.4 +108.8 +35.1 +14.4
We use the amount of limiting reagent to calculate the amount of product formed.
$\frac{\text{1 mol CO}_{2}}{\text{1 mol ZnCO}_{3}}~~ x ~~ \text{0.798 mol ZnCO}_{3} = \text{0.798 mol CO}_2$
$\text{0.798 mol CO}_2 ~~ x ~~ \frac{\text{44.0 g CO}_{2}}{\text{1 mol CO}_{2}} = \text{35.1 g CO}_2$
When the reaction ends, 1.60 mol HCl will have reacted with 0.798 mol ZnCO3 and there will be
(2.74 – 1.60) mol HCl = 1.14 mol HCl left over. ZnCO3 is therefore the limiting reagent. The left over HCl will ensure that if any material remains in a mineral test of a 100 g sample, that it can't be a carbonate.
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts $\frac{n_{\text{ZnCO}_{3}}\text{(initial)}}{n_{\text{HCl}}\text{(initial)}}=\frac{\text{0.798 mol ZnCO}_{3}}{\text{2.74 mol HCl}}=\frac{\text{0}\text{.291 mol ZnCO}_{3}}{\text{1 mol HCl}}$ was less than the stoichiometric ratio $\text{S}\left( \frac{\text{ZnCO}_{3}}{\text{HCl}} \right)=\frac{\text{1 mol ZnCO}_{3}}{\text{2 mol HCl}}~=~\frac{\text{0.5 mol ZnCO}_{3}}{\text{1 mol HCl}}$ This indicated that there was not enough Hg to react with all the bromine and mercury was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align}
(Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.). This general rule for determining the limiting reagent is applied in the next example.
EXAMPLE 2 Iron can be obtained by reacting the ore hematite (Fe2O3) with coke (C). The latter is converted to CO2. As manager of a blast furnace you are told that you have 20.5 Mg (megagrams) of Fe2O3 and 2.84 Mg of coke on hand. (a) Which should you order first—another shipment of iron ore or one of coke? (b) How many megagrams of iron can you make with the materials you have?
Solution
a) Write a balanced equation 2Fe2O3 + 3C → 3CO2 + 4Fe
The stoichiometric ratio connecting C and Fe2O3 is $\text{S}\left( \frac{\text{C}}{\text{Fe}_{\text{2}}\text{O}_{\text{3}}} \right)=\frac{\text{3 mol C}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.5 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ The initial amounts of C and Fe2O3 are calculated using appropriate molar masses \begin{align} & \text{ }n_{\text{C}}\text{(initial)}=\text{2}\text{.84}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol C}}{\text{12}\text{.01 g}}=\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C} \ & \ & n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}=\text{20}\text{.5}\times \text{10}^{\text{6}}\text{g}\times \frac{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}{\text{159}\text{.69 g}}=\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}} \ \end{align} Their ratio is $\frac{n_{\text{C}}\text{(initial)}}{n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{(initial)}}=\frac{\text{2}\text{.36}\times \text{10}^{\text{5}}\text{mol C}}{\text{1}\text{.28}\times \text{10}^{\text{5}}\text{mol Fe}_{\text{2}}\text{O}_{\text{3}}}=\frac{\text{1}\text{.84 mol C}}{\text{1 mol Fe}_{\text{2}}\text{O}_{\text{3}}}$ Since this ratio is larger than the stoichiometric ratio, you have more than enough C to react with all the Fe2O3. Fe2O3 is the limiting reagent, and you will want to order more of it first since it will be consumed first. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant C will be left over, but all the initial amount of Fe2O3 will be consumed. Therefore we use nFe2O3 (initial) to calculate how much Fe can be obtained $n_{\text{Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\xrightarrow{S\text{(Fe/Fe}_{\text{2}}\text{O}_{\text{3}}\text{)}}\text{ }n_{\text{Fe}}\xrightarrow{M_{\text{Fe}}}\text{ }m_{\text{Fe}}$ $m_{\text{Fe}}=\text{1}\text{.28 }\times \text{ 10}^{\text{5}}\text{ mol Fe}_{\text{2}}\text{O}_{\text{3}}\text{ }\times \text{ }\frac{\text{4 mol Fe}}{\text{2 mol Fe}_{\text{2}}\text{O}_{\text{3}}}\text{ }\times \text{ }\frac{\text{55}\text{.85 g}}{\text{mol Fe}}=\text{1}\text{.43 }\times \text{ 10}^{\text{7}}\text{ g Fe}$ This is 1.43 × 106 g, or 14.3 Mg, Fe. As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
The concept of a limiting reagent was used by the nineteenth century German chemist Justus von Liebig (1807 to 1873) to derive an important biological and ecological law. Liebig’s law of the minimum states that the essential substance available in the smallest amount relative to some critical minimum will control growth and reproduction of any species of plant or animal life. When a group of organisms runs out of that essential limiting reagent, the chemical reactions needed for growth and reproduction must stop. Vitamins, protein, and other nutrients are essential for growth of the human body and of human populations. Similarly, the growth of algae in natural bodies of water such as Lake Erie can be inhibited by reducing the supply of nutrients such as phosphorus in the form of phosphates. It is for this reason that many states have regulated or banned the use of phosphates in detergents and are constructing treatment plants which can remove phosphates from municipal sewage before they enter lakes or streams.
3.3.08: Lecture Demonstrations
Generation of Hydrogen
Artdej, Romklao; Thongpanchang, Tienthong. J. Chem. Educ. 2008, 85, 1382. Clyde R. Dillard; J. Chem. Educ., 1972, 49 (12), p A694. forums.jce.divched.org:8000/[email protected]
Equal volumes of vinegar or other acid are added to a series of volumetric flasks, and increasing masses of NaHCO3 are added to balloons. The balloons are attached to the necks of the flasks, then lifted to dump the NaHCO3 into the vinegar. After the vinegar becomes limiting, the size of the CO2-filled balloons no longer increases.
Nuts and Bolts Model
Use bolts and a mismatched number of nuts (and washers) to model limiting reactants. Craig Blankenship, J. Chem. Educ., 1987, 64 (2), p 134. forums.jce.divched.org:8000/[email protected]
Nylon Synthesis
The "nylon rope trick" can be used as a conceptual demonstration of limiting reagents, especially if different numbers of red and blue paper strips (of different lengths) are used to model the acid and amine. This can be related to amino acid nutrition and protein synthesis. Ed Vitz;J. Chem. Educ., 2005, 82 (7), p 1013l | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.07%3A_Geology-_Using_the_Acid_Test_to_Distinguish_the_Minerals_in_Calomine.txt |
The space shuttle used 2 Solid Propellant Boosters (SRBs, white) and a tank of LOX/LH2 (large orange tank)
Introduction
Jet aircraft designed to fly in Earth's atmosphere carry just fuel, and rely on atmospheric O2, which is supplied in excess, to burn the fuel. Rockets designed for interplanetary flight need to supply both a fuel and an oxidant for their propulsion, and the ratio of the two has to be exactly right for maximum propulsion and minimum mass. If the wrong size tanks for fuel and oxidizer are designed, some portion of such a reactant will be left unchanged after the reaction. Conversely, at least one reagent is usually completely consumed. When it is gone, the other excess reactants have nothing to react with and they cannot be converted to products. The substance which is used up first is the limiting reagent.
Principle of Rocket Operation
Newton's Third Law of Motion is "every action is accompanied by an equal and opposite reaction." A rocket operates on this principle. The force exerted by the rocket on the ejected reaction products equals the force of the ejected particles on the rocket. So the "impulses" (the product of the force and time that it acts) are equal in magnitude and opposite in direction. Since the impulse (F × t) equals the momentum (m × v), the momentum of the spent fuel in one direction equals the momentum of the rocket in the other. The fuel and oxidant must be designed to rapidly heat and eject combustion products in one direction, causing motion of the rocket in the opposite direction.
The gauge of efficiency for rocket propellants is specific impulse, stated in seconds. The higher the number, the "hotter" the propellant.
Specific Impulse (Isp is the period in seconds for which a 1-pound (0.45-kilogram) mass of propellant (total of fuel and oxidizer) will produce a thrust of 1 pound (0.45- kilogram) of force. The specific impulse for a fuel may vary somewhat due to conditions[2]
NASA and commercial launch vehicles use four types of propellants: (1) petroleum; (2) cryogenics; (3) hypergolics; and (4) solids[3]. Examples involving cryogenics (liquid hydrogen and liquid oxygen) and solids (aluminum and ammonium perchlorate, or Ammonium Perchlorate Composite Propellants, APCPs) are give below. Hypergolics are compounds that react upon mixing (without an ignition source), like nitrogen tetroxide and hydrazine[4] which we'll discuss later. Petroleum derivatives include RP-1 (Rocket Propellant or Refined Petroleum-1), which is similar to kerosene and used with an oxidant like liquid oxygen.
Examples with cryogenics and solids
LOX/RP-1[5] Hydrazine/
dinitrogen tetroxide[6]
ACPC[7] LOX/LH2
Max Specific Impulse ~353 258 ~250 444 [8]
Oxidizer to Fuel Ratio 2.56:1 0.77:1 calculate below calculate below
Example 1: LH2 + LOX
The Space Shuttle's large rust-orange booster fuel tank shown in the picture above holds liquid oxygen (LOX, 629,340 kg) and liquid hydrogen (LH2, 106 261 kg)[9].
a. Which is the limiting reagent?
b. What mass of product will be formed?
c. Did NASA make a mistake?
Solution
The balanced equation
2 H2 + O2 → H2O
tells us that according to the atomic theory, 2 mol H2 is required for each mole of O2. That is, the stoichiometric ratio S(H2/O2) = 2 mol H2/ 1 mol O2. Let us see how many moles of each we actually have $\text{n}_{\text{H}_{2}}=\text{1.06261}\times10^{8}\text{g}~\times~$ $\frac{\text{1 mol H}_{2}}{\text{2.016 g}}$ $~=~5.271\times~10^{7}\text{ mol H}_{2}$
$\text{n}_{\text{O}_{2}}$ $~=~\text{6.293}~\times~10^{8}\text{g}\times \frac{\text{1 mol O}_{2}}{\text{31.999 g}}$ $=\text{1.967}~\times~10^{7}\text{ mol O}_{\text{2}}$
If all the H2 were to react, the stoichiometric ratio allows us to calculate the amount of O2 that would be required:
$\text{n}_{\text{O}_{2}}~=~ \text{n}_{\text{H}_{2}}~~\times~~$ $~~\frac{\text{1 mol O}_{2}}{\text{2 mol H}_{2}}~~$ $~=~2.63\times10^{7}\text{mol O}_{2}$
This is more than the amount of oxygen present,so oxygen is the limiting reactant and H2 is present in excess.
If all the O2 reacts, the stoichiometric ratio allows us to calculate the amount of H2 that would be required:
$\text{n}_{\text{H}_{2}}~=~ \text{n}_{\text{O}_{2}}~~\times~~$ $~~\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}~~$
$~=~1.967\times10^{7}\text{mol O}_{2}~\times~$ $\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}~=~$ $3.934\times10^{7}\text{mol H}_{2}$
We require less than the amount of H2 present, so it is the excess reactant.
When the reaction ends, 3.934 x 107 mol of H2 will have reacted with 1.967 x 107 mol O2 and there will be
(5.271 x 107 mol - 3.934 x 107 mol) = 1.337 x 107 mol H2 left over. Oxygen is therefore the limiting reagent.
b. Since the hydrogen doesn't all react, we need to calculate the amount of water produced from the amount of oxygen consumed, by using the stoichiometric ratio:
$\text{n}_{\text{H}_{2}\text{O}}~=~ \text{n}_{\text{O}_{2}}~~\times~~$ $~~\frac{\text{2 mol H}_{2}\text{O}}{\text{1 mol O}_{2}}~~$ $~=~3.934\times10^{7}\text{mol H}_{2}\text{O}$
The mass of water is then calculated by using the molar mass:
$3.934\times10^{7}\text{mol H}_{2}\text{O}~\times~$ $\frac{\text{18.01 g}}{\text{1 mol H}_{2}\text{O}}$ $~=~7.08\times10^{8}\text{g H}_{2}\text{O}$
c. The excess hydrogen is not a mistake. The reaction is so exothermic that it ejects some of the unreacted hydrogen. This doesn't matter, since any ejected mass contributes to the backward momentum of fuel, and the forward momentum of the rocket. Since the mass of hydrogen is small, its velocity is large, and it can contribute to a large forward velocity of the rocket.
These calculations can be organized as a table, with entries below the respective reactants and products in the chemical equation. We can calculate (hypothetically) how much of each reactant would be required if the other were completely consumed to demonstrate which is in excess, and which is limiting. We use the amount of limiting reagent to calculate the amount of product formed.
Calculations of Solutions to Example 1
2 H2 + O2 → H2O
m (g) 1.063 x 108 6.293 x 108
M (g/mol) 2.016 31.999 18.015
n (mol) 5.271 x 108 1.967 x 108 --
if all H2 reacts -5.271 x 107 -2.636 x 107 +5.271 x 107
if all O2 reacts -3.934 x 107 -1.967 x 107 +3.934 x 107
Actual Reaction
Amounts
-3.933 x 107 -1.967 x 107 +3.933 x 107
Actual Reaction
Masses
-7.930 x 107 -6.293 x 108 +7.086 x 108
In the end, 1.063 x 108 - 7.930 x 107 = 2.70 x 107 g of H2 will remain, along with 7.08 x 108 g of water, for a total of 7.35 x 108 g. The mass of the reactants was also 1.063 x 108 + 6.29 x 108 = 7.35 x 108 g.
General Strategy for Limiting Reactant Problems
From this example you can begin to see what needs to be done to determine which of two reagents, X or Y, is limiting. We must compare the stoichiometric ratio S(X/Y) with the actual ratio of amounts of X and Y which were initially mixed together. In Example 1 this ratio of initial amounts
$\frac{n_{\text{H}_{2}}\text{(initial)}}{n_{\text{O}_{2}}\text{(initial)}}$ $~=~\frac{5.271\times10^{7}\text{mol H}_{2}}{1.967 \times 10^{7}\text{ mol O}_{2}}$ $~=~\frac{\text{2.68 mol H}_{2}}{\text{1 mol O}_{\text{2}}}$
was less than the stoichiometric ratio $\text{S}\left( \frac{\text{H}_{2}}{\text{O}_{2}} \right)=\frac{\text{2 mol H}_{2}}{\text{1 mol O}_{2}}$ This indicated that there was not enough O2 to react with all the hydrogen, and oxygen was the limiting reagent. The corresponding general rule, for any reagents X and Y, is \begin{align} & \text{If}~ \frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is less than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then X is limiting}\text{.} \ & \ & \text{If}~\frac{n_{\text{X}}\text{(initial)}}{n_{\text{Y}}\text{(initial)}}~\text{is greater than S}\left( \frac{\text{X}}{\text{Y}} \right)\text{, then Y is limiting}\text{.} \ \end{align}
Of course, when the amounts of X and Y are in exactly the stoichiometric ratio, both reagents will be completely consumed at the same time, and neither is in excess.
As you can see from the example, in a case where there is a limiting reagent, the initial amount of the limiting reagent must be used to calculate the amount of product formed. Using the initial amount of a reagent present in excess would be incorrect, because such a reagent is not entirely consumed.
APCP Solid Propellants
Solid Rocket Boosters (SRBs) are the type implicated in the infamous Challenger Disaster[10], where a poor seal allowed escape of hot gases. Each SRB motor contains a propellant mixture (weighting about 590,000kg) consisting of ammonium perchlorate (oxidizer, 69.6% by weight), and aluminum (fuel,16%). A catalyst (iron oxide, 0.4%), a binder (which also acts as secondary fuel, 12.04%), and a curing agent (1.96%).[11][12] This propellant is commonly referred to as Ammonium Perchlorate Composite Propellant (APCP).
a. An amateur rocketeer wants to make 100g of propellant. He mixes the propellant with 69.6 g of ammonium perchlorate (the required 69.6%), but he has no binder, so uses 16.0 g of Al plus another 12.0 g Al in place of the binder (which is also a fuel), or 28% aluminum in the mixture. Is this ratio correct? b. If not, which is the limiting reactant? c. Calculate the mass of Al2O3 produced by the first reaction below.
6 NH4ClO4 + 10 Al → 5 Al2O3 + 6 HCl + 3 N2 + 9 H2O or 6 NH4ClO4 + 10 Al → 4 Al2O3 + 2 AlCl3 + 3 N2 + 12 H2O
Solution
The stoichiometric ratio connecting Al and NH4ClO4 is $\text{S}\left( \frac{\text{Al}}{\text{NH}_{4}\text{ClO}_{3}} \right)$ $~=~\frac{\text{10 mol Al}}{\text{6 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}$ $~=~\frac{\text{1.67 mol Al}}{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}$ The initial amounts of Al and NH4ClO4 are calculated using appropriate molar masses nAl(initial) $~=~28.0\text{g}~\times~ \frac{\text{1 mol Al}}{\text{26.982 g}}$ $~=~\text{1.04 mol Al}$
$\text{n}_{\text{NH}_{\text{4}}\text{ClO}_{\text{4}}\text{(initial)}}$ $~=~\text{69.6 g}~\times~ \frac{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}{\text{117.489 g}}$ $~=~\text{0.592 mol NH}_{\text{4}}\text{ClO}_{\text{4}}$ Their ratio is $\frac{n_{\text{Al}\text{(initial)}}}{n_{\text{NH}_{\text{4}}\text{ClO}_{\text{4}}\text{(initial)}}}$ $~=~\frac{\text{1.04 mol Al}}{\text{0.592 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}$ $~=~\frac{\text{1.76 mol Al}}{\text{1 mol NH}_{\text{4}}\text{ClO}_{\text{4}}}$ Since this ratio is larger than the stoichiometric ratio, you have more than enough Al to react with all the NH4ClO4. NH4ClO4 is the limiting reagent, so the ratio isn't stoichiometric. b) The amount of product formed in a reaction may be calculated via an appropriate stoichiometric ratio from the amount of a reactant which was consumed. Some of the excess reactant Al will be left over, but all the initial amount of NH4ClO4 will be consumed. Therefore we use nNH4 ClO4 (initial) to calculate how much Al2O3 can be obtained $n_{\text{NH}_{4}\text{ClO}_{4}}~~$ $\xrightarrow{S\text{(Al}_{2}\text{O}_{3}\text{/NH}_{4}\text{ClO}_{4}\text{)}}$ $~~n_{\text{Al}_{2}\text{O}_{3}}$ $\xrightarrow{M_{\text{Al}_{2}\text{O}_{3}}}$ $~~\text{m}_{\text{Al}_{2}\text{O}_{3}}$ $m_{\text{Al}_{2}\text{O}_{3}}$ $~=~\text{0.592 mol NH}_{4}\text{ClO}_{4}~~$ $\times~~\frac{\text{5 mol Al}_{2}\text{O}_{3}}{\text{6 mol NH}_{4}\text{ClO}_{4}}~~$ $\times ~~\frac{\text{101.961 g}}{\text{mol Al}_{2}\text{O}_{3}}$ $~=~\text{50.3 g Al}_{2}\text{O}_{3}$
You may want to verify the rest of the values in the table:
6 NH4ClO4 + 10 Al → 5 Al2O3 + 6 HCl + 3 N2 + + 9 H2O
m (g) 69.6 28.0
M (g/mol) 117.49 26.982 101.961 36.461 28.013 18.015
n (mol) 0.592 1.04 -- -- -- --
if all NH4ClO4 reacts -0.592 -0.987
if all Al reacts -1.04 -0.622
Actual Reaction
Amounts
-0.592 -0.987 +0.493 +0.592 +0.296 +0.887
Actual Reaction
Masses
-69.6 -26.63 +50.30 +21.59 +8.29 +15.99 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.03%3A_The_Limiting_Reagent/3.3.09%3A_Physics-_Rocket_Propellants.txt |
Not all chemical reactions are as simple as the ones we have considered, so far. Quite often a mixture of two or more products containing the same element is formed. For example, when octane (or gasoline in general) burns in an excess of air, the reaction is
$2 \text{C}_{8} \text{H}_{18} + 25 \text{O}_{2} \rightarrow 16 \text{CO}_{2} + 18 \text{H}_{2} \text{O}\label{1}$
If oxygen is the limiting reagent, however, the reaction does not necessarily stop short of consuming all the octane available. Instead, some carbon monoxide (CO) forms:
$2 \text{C}_{8} \text{H}_{18} + {24} \text{O}_{2} \rightarrow 14 \text{CO}_{2} + {\text{2CO}}+ 18 \text{H}_{2} \text{O} \nonumber$
Burning gasoline in an automobile engine, where the supply of oxygen is not always as great as that demanded by the stoichiometric ratio, often produces carbon monoxide, a poisonous substance and a major source of air pollution.
In other cases, even though none of the reactants is completely consumed, no further increase in the amounts of the products occurs. We say that such a reaction does not go to completion. When a mixture of products is produced or a reaction does not go to completion, the effectiveness of the reaction is usually evaluated in terms of percent yield of the desired product. A theoretical yield is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage:
$\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent} \nonumber$
The video below (modeled after the octane example given earlier in the chapter) demonstrates visually what the percent yield is, first showing the theoretical yield, then showing the actual yield (where the reaction doesn't go to completion) and finally comparing the actual yield to the theoretical yield to find the percent yield.
Example $1$ : Percent Yield
When 100.0 g N2 gas and 25.0 g H2 gas are mixed at 350°C and a high pressure, they react to form 28.96 g NH3 (ammonia) gas. Calculate the percent yield.
Solution:
We must calculate the theoretical yield of NH3, and to do this, we must first discover whether N2 or H2 is the limiting reagent. For the balanced equation
$\ce{N2 + 3H2 -> 2NH3} \nonumber$
the stoichiometric ratio of the reactants is
$\text{S}\left( \frac{\ce{H2}}{\ce{N2}} \right)= \frac{\text{3 mol H}_2}{\text{1 mol N}_2} \nonumber$
Now, the initial amounts of the two reagents are
and
\begin{align} & n_{\text{H}_{\text{2}}}\text{(initial)}&=\text{25}\text{.0 g H}_{\text{2}}\times \frac{\text{1 mol H}_{\text{2}}}{\text{2}\text{.016 g H}_{\text{2}}}=\text{12}\text{.4 mol H}_{\text{2}} \ & \ & n_{\text{N}_{\text{2}}}\text{(initial)}&=\text{100}\text{.0 g N}_{\text{2}}\times \frac{\text{1 mol N}_{\text{2}}}{\text{28}\text{.02 g N}_{\text{2}}}=\text{3}\text{.569 mol N}_{\text{2}} \ \end{align}
The ratio of initial amounts is thus
$\frac{n_{\text{H}_{\text{2}}}\text{(initial)}}{n_{\text{N}_{\text{2}}}\text{(initial)}}=\frac{\text{12}\text{.4 mol H}_{\text{2}}}{\text{3}\text{.569 mol N}_{\text{2}}}=\frac{\text{3}\text{.47 mol H}_{\text{2}}}{\text{1 mol N}_{\text{2}}} \nonumber$
Since this ratio is greater than $\text{S}\left( \frac{\text{H}_{\text{2}}}{\text{N}_{\text{2}}} \right)$, there is an excess of H2. N2 is the limiting reagent. Accordingly we must use 3.569 mol N2 (rather than 12.4 mol H2) to calculate the theoretical yield of NH3. We then have
$n_{\text{NH}_{\text{3}}}\text{(theoretical)}=\text{3}\text{.569 mol N}_{\text{2}}\times \frac{\text{2 mol NH}_{\text{3}}}{\text{1 mol N}_{\text{2}}} =\text{7}\text{.138 mol NH}_{\text{3}} \nonumber$
so that
$m_{\text{NH}_{\text{3}}}\text{(theoretical)}=\text{7}\text{.138 mol NH}_{\text{3}}\times \frac{\text{17}\text{.03 g NH}_{\text{3}}}{\text{1 mol NH}_{\text{3}}} =\text{121}\text{.6 g NH}_{\text{3}} \nonumber$
The percent yield is then
\begin{align} \text{Percent yield} & =\frac{\text{actual yield}}{\text{theoretical yield}} \times \text{100 percent } \ { } \ &=\frac{\text{28.96 g}}{\text{121.6 g}}\times \text{100 percent} \ { } \ &=\text{23.81 percent} \end{align}
Combination of nitrogen and hydrogen to form ammonia is a classic example of a reaction which does not go to completion. Commercial production of ammonia is accomplished using this reaction in what is called the Haber process. Even at the rather unusual temperatures and pressures used for this industrial synthesis, only about one-quarter of the reactants can be converted to the desired product. This is unfortunate because nearly all nitrogen fertilizers are derived from ammonia and the world has come to rely on them in order to produce enough food for its rapidly increasing population. Ammonia ranks third [after sulfuric acid (H2SO4) and oxygen (O2)] in the list of most-produced chemicals, worldwide. It might rank even higher if the reaction by which it is made went to completion. Certainly ammonia and the food it helps to grow would be less expensive and would require much less energy to produce if this were the case.
3.04: Percent Yield
Biological oils are different from petroleum oils ("regular" diesel fuel) in molecular structure and properties.
Some cities are converting their bus lines to biodiesel
Common petroleum diesel fuel is a mixture of simple hydrocarbons, with the average chemical formula C12H23 (shown below), but components may range approximately from C10H20 (dodecane) to C15H28(pentadecane).[1] By contrast, biological oils are "triglycerides" (classified as "esters") like the glyceryl trilinoleate shown below:
C12H23, dodecane
A triglyceride: glyceryl trilinoleate
Because of their large size and consequent large intermolecular attractions, the viscosity of biological oils is generally too high for use in conventional diesel engines. Biological oils also burn a little less readily, and with a sootier flame than petroleum diesel. Biological oils can be used in conventional diesel engines if they are preheated to reduce their viscosity, but this requires an auxiliary electrical heater until the engine warms up. For these resons, bilogical oils require processing for use as biodiesel.
A biological oil is an ester, which is a type of organic compound having the atom linkage shown below.
The ester linkage. R and R' represent chains of carbon and hydrogen atoms. R = -CH3 and R' = -C18H35O2 for the methyl stearate in Example 1
glycerol
The ester linkage in biological oils is created when a glycerol molecule reacts with organic acids. The glycerol molecule has a chain of 3 carbon atoms, each with an -OH (alcohol) group on it. The figure below shows how an organic alcohol reacts with a organic acid. Organic chemists abbreviate molecular structures--the "zig-zag" lines in the figure represent carbon chains with a C atom at each "zig" or "zag". Each carbon has 4 bonds, and if fewer than 4 are shown, it's assumed that they go to H atoms. So the alcohol is C2H5OH (ethanol), and the acid is acetic acid (or ethanoic acid, CH3COOH) in the Figure:
Organic Acid + Alcohol → (with sulfuric acid catalyst) an ester and water
Since glycerol has 3 -OH groups, 3 long chain organic "fatty acids" attach to make the bulky "triglyceride".
Stearin, or glyceryl tristearate
But just as easily as esters can be made from alcohols and acids, they can switch alcohols or acids. In the presence of a strong base catalyst, like NaOH, a triglyceride can react with 3 small alcohol molecules, like methanol (CH3OH), which replace the glycerol "backbone", making 3 separate esters of lower molecular weight
Quite often a mixture of two or more products is formed. For example, when a vegetable oil reacts with methanol, only one or two of the acids may be displaced from the glycerine, producing only 1 or 2 FAMEs.
C3H5(C18H35O2)3 + NaOH + 2 CH3OH → C3H5(C18H35O2)2(OH) + 2 C17H35COOCH3
C3H5(C18H35O2)3 + NaOH + 1 CH3OH → C3H5(C18H35O2)(OH)2 + 1 C17H35COOCH3
C3H5(C18H35O2)3 + NaOH + 3 CH3OH → C3H5(OH)3 + 3 C17H35COOCH3
Usually, a large excess of methanol and sodium hydroxide are added, so that the reaction produces the maximum amount of FAME.
But in the case of a transesterification, even though none of the reactants is completely consumed, no further increase in the amounts of the products occurs. We say that such a reaction does not go to completion. When a mixture of products is produced or a reaction does not go to completion, the effectiveness of the reaction is usually evaluated in terms of percent yield of the desired product. A theoretical yield is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage:
$\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent}$
EXAMPLE 1 When 100.0 g C3H5(C18H35O2)3 gas and 15.0 g CH3OH are mixed at 55°C with NaOH catalyst, they react to form 90.96 g C17H35COOCH3 methyl stearate biodiesel. Calculate the percent yield.
Solution
We must calculate the theoretical yield of NH3, and to do this, we must first discover whether N2 or H2 is the limiting reagent. For the balanced equation
C3H5(C18H35O2)3 + NaOH + 3 CH3OH → C3H5(OH)3 + 3 C17H35COOCH3
stearin + sodium hydroxide + 3 CH3OH → glycerol + 3 methyl stearate
The stoichiometric ratio of the reactants is
$\text{S}\left( \frac{\text{stearin}}{\text{CH}_{\text{3}}\text{OH}} \right)=\frac{\text{1 mol stearin}}{\text{3 mol CH}_{\text{3}}\text{OH}}$ Now, the initial amounts of the two reagents are and \begin{align} & n_{\text{stearin}}\text{(initial)}=\text{100}\text{.0 g stearin}\times \frac{\text{1 mol stearin}}{\text{891}\text{.5 stearin}}=\text{0}\text{.1122 mol stearin} \ & \ & n_{\text{CH}_{\text{3}}\text{OH}}\text{(initial)}=\text{15}\text{.0 g CH}_{\text{3}}\text{OH}\times \frac{\text{1 mol CH}_{\text{3}}\text{OH}}{\text{32}\text{.04 g CH}_{\text{3}}\text{OH}}=\text{0}\text{.4682 mol CH}_{\text{3}}\text{OH} \ \end{align} The ratio of initial amounts is thus $\frac{n_{\text{stearin}}\text{(initial)}}{n_{\text{CH}_{\text{3}}\text{OH}}\text{(initial}}~=~ \frac{\text{0}\text{.1122 mol stearin}}{\text{0}\text{.4682 mol CH}_{\text{3}}\text{OH}}~=~\frac{\text{0}\text{.240 mol stearin}}{\text{1 mol CH}_{\text{3}}\text{OH}}$ Since this ratio is less than $\text{S}\left( \frac{\text{stearin}}{\text{CH}_{\text{3}}\text{OH}} \right)~=~0.33$, there is an excess of CH3OH. Stearin is the limiting reagent. Accordingly we must use 0.1122 mol stearin and 0.3366 mol CH3OH (rather than 0.4682 mol CH3OH) to calculate the theoretical yield of C17H35COOCH3 (methyl stearate). We then have $n_{\text{methyl stearate}}\text{(theoretical)}=\text{0}\text{.1122 mol stearin}\times \frac{\text{3 mol methyl stearate}}{\text{1 mol stearin}}=\text{0}\text{.3365 mol methyl stearate}$ so that $\text{m}_{\text{methyl stearate}}\text{(theoretical)}=\text{0}\text{.3365 mol methyl stearate}\times \frac{\text{298}\text{.51 g methyl stearate}}{\text{1 mol methyl stearate}}=\text{100}\text{.5 g methyl stearate}$ We can organize these calculations in a table:
C3H5(C18H35O2)3(s) + 3 CH3OH (l) → 1 C3H5(OH)3 (l) + 3 C18H35O2)3CH3(s)
m, g 100.0 g 15.00 g 90.96 g
M, g/mol 891.5 32.04 298.5 92.1
n present, mol 0.1122 mol 0.4682 mol
n actual, mol 0.1122 0.3366 0.1122 0.3366
m actual, mass 100.0 10.78 10.33 100.47
The percent yield is then $\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent }=\frac{\text{90}\text{.96 g}}{\text{100}\text{.5 g}}\times \text{100 percent}=\text{90}\text{.55 percent}$
Transesterification is a classic example of a reaction which does not go to completion. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.04%3A_Percent_Yield/3.4.01%3A_Environment-_Synthesis_of_Biodiesel_Fuel.txt |
Dietary intake of vegetable oils and hydrogenated vegetable oils has significant health effects. Not only do they have about twice as many calories per gram as sugars and proteins, but they have long term effects on circulatory system health. Crisco®, with saturated oils, may not be as healthy as olive oil, with more unsaturated oils.
Crisco contains hydrogenated vegetable oil[1]
olive oil is 55-83% oleic acid[2]
To understand these effects, we need to look at the structure of triglycerides. The triglyceride [1] is an important part of the blood test done with an annual physical exam.
Triglycerides
Vegetable oils are all triglycerides, which contain a glycerol () three carbon "backbone" with 3 long chain "fatty acids" attached through ester linkages, as in the figure below.
The long chain fatty acids may be saturated with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are unsaturated and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is polyunsaturated, with multiple double bonds (it is linolenic acid). Various cooking oils have |known concentrations of saturated and unsaturated fatty acids.
Saturated vs. Unsaturated vs. Trans Fatty Acids
Generally, triglycerides with more unsaturated fatty acid substituents are more healthful, but food companies hydrogenate them to make them solid saturated fats (like margarine or Crisco), and to reduce the tendency to spoil. Unsaturated fats have kinks in their molecular structures that reduce the tendency for them to cause arthereosclerosis (clogged arteries), pretty much for the same reason that kinks reduce the tendency to pack efficiently and form solids. Saturated fats have more linear fatty acid chains that pack well and solidify easily. Compare the Jmol models of saturated palmitic acid and unsaturated oleic acid below. Partial hydrogenation of polyunsaturated fats also produces trans-fatty acids, which have structures like saturated fats and consequently are as unhealty (see Elaidic Acid Below):
Trans- fatty acids have the hydrogens on opposite sides of the C=C double bond, like this while cis-fatty acids have the hydrogen atoms on the same side, like this .
Composition of Crisco, "partially hydrogenated" vegetable oil
As of 2010 Crisco is made of soybean oil, fully hydrogenated cottonseed oil, and partially hydrogenated soybean and cottonseed oils. According to the product information label, one 12 g serving of Crisco contains 3 g of saturated fat, 0g of trans fat, 6 g of polyunsaturated fat, and 2.5 g of monounsaturated fat.[3] Notice that the fat masses don't add up[4] because the weights of glycerol are not included in the separately listed components. Trans fatty acids are now recognized as a major dietary risk factor for cardiovascular diseases, and the US FDA has revised food labeling requirements to include trans fats.[5]
The composition of the soybean oil in Crisco is shown below.[6]
Name Fatty Acid Structure Formula Percent
Palmitic Acid C16H32O2 10
Stearic Acid C18H36O2 4
Oleic Acid C18H34O2 23
Linoleic Acid C18H32O2 51
Linolenic Acid C18H30O2 7-10
Other 2-5
Percent yield of hydrogenation products
Quite often a mixture of two or more products is formed in a chemical reaction. For example, when a vegetable oil like palm oil is hydrogenated, we might want to make just mono-unsaturated products. But the many triglycerides it contains with varied fatty acid chains. No single process could work for all of them. Suppose we start with just one possible palm oil molecule, a glycerol with 2 linolenic acid, and 1 linoleic acid substituents (we'll abbreviate it GLLL). The desired product might be the oil with three oleic acid substituents (we'll abbreviate it GOOO, which also might be a good description of it) so the equation is:
(C18H29O2)CH2CH(C18H29O2)CH2-(C18H31O2) + 5 H2 → (C18H33O2)CH2CH(C18H33O2)CH2-(C18H33O2)
"GLLL" + 5 H2 → "GOOO"
A large excess of hydrogen is usually present under pressure, with a palladium or "Raney Nickel" catalyst[7]. A large number of products is obtained, including completely saturated fats like Stearin (glyceryl tristearate), and trans fats. The products are usually analyzed by converting the oils to simpler (methyl) esters and running a gas chromatogam.
The effectiveness of the reaction is usually evaluated in terms of percent yield of the desired product. A theoretical yield is calculated by assuming that all the limiting reagent is converted to product. The experimentally determined mass of product is then compared to the theoretical yield and expressed as a percentage:
$\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent}$
Example $1$
Suppose a |hydrogenation of 100.0 g of (C18H29O2)CH2CH(C18H29O2)CH2-(C18H31O2), abbreviated "GLLL" (M = 875.4 g/mol) is carried out with 2.000 g H2, sealed in a high pressure steel reaction vessel with a catalyst at 55°C. The products include 90.96 g(C18H33O2)CH2CH(C18H33O2)CH2-(C18H33O2), abbreviated "GOOO" (M = 885.5 g/mol). Calculate the percent yield.
Solution
We must calculate the theoretical yield of (C18H33O2)CH2CH(C18H33O2)CH2-(C18H33O2), and to do this, we must first discover whether (C18H29O2)CH2CH(C18H29O2)CH2-(C18H31O2) or H2 is the limiting reagent. For the balanced equation above,
The stoichiometric ratio of the reactants is:
$\text{S}\left( \frac{\text{GLLL}}{\text{H}_{\text{2}}} \right)=\frac{\text{1 mol GLLL}}{\text{5 mol H}_{\text{2}}} \nonumber$
Now, the initial amounts of the two reagents are and:
\begin{align*} & n_{\text{GLLL}}\text{(initial)}=\text{100}\text{.0 g GLLL}\times \frac{\text{1 mol stearin}}{\text{875}\text{.4 g GLLL}}=\text{0}\text{.1142 mol GLLL} \ & \ & n_{\text{H}_2}\text{(initial)}=\text{2}\text{.000 g H}_2\times \frac{\text{1 mol H}_2}{\text{2}\text{.016 g H}_2}=\text{0}\text{.9921 mol H}_2 \ \end{align*}
The ratio of initial amounts is thus:
$\frac{n_{\text{GLLL}}\text{(initial)}}{n_{\text{H}_2}\text{(initial}} ~=~ \frac{\text{0}\text{.1142 mol stearin}}{\text{0}\text{.9921 mol H}_2} ~=~\frac{\text{0}\text{0.1151 mol stearin}}{\text{1 mol H}_2} \nonumber$
Since this ratio is less than $\text{S}\left( \frac{\text{GLLL}}{\text{H}_2} \right)~=~0.20$, there is an excess of H2. GLLL is the limiting reagent. Accordingly we must use 0.1142 mol GLLL and 0.5712 mol H2 (rather than 0.9921 mol H2) to calculate the theoretical yield of (C18H33O2)CH2CH(C18H33O2)CH2-(C18H33O2), or "GOOO". We then have
$n_{\text{GOOO}}\text{(theoretical)}=\text{0}\text{.1142 mol GLLL}\times \frac{\text{1 mol GOOO}}{\text{1 mol GLLL}}=\text{0}\text{.1142 mol GOOO} \nonumber$
so that:
$\text{m}_{\text{GOOO}}\text{(theoretical)}=\text{0}\text{.1142 mol GOOO}\times \frac{\text{885}\text{.5 g GOOO}}{\text{1 mol GOOO}}=\text{101}\text{.2 g GOOO} \nonumber$
We can organize these calculations in a table:
(C18H29O2)CH2CH(C18H29O2)CH2-(C18H31O2)
"GLLL"
+ 5 H2 → (C18H33O2)CH2CH(C18H33O2)CH2- (C18H33O2)
"GOOO"
m, g 100.0 g 2.000 g 90.96 g
M, g/mol 875.4 2.016 885.5
n present, mol 0.1142 mol 0.9921 mol
n actual, mol 0.1142 0.5712 0.1142
m actual, mass 100.0 1.1515 101.2
The percent yield is then:
$\text{Percent yield}=\frac{\text{actual yield}}{\text{theoretical yield}}\times \text{100 percent }=\frac{\text{90}\text{.96 g}}{\text{101}\text{.2 g}}\times \text{100 percent}=\text{89}\text{.9 percent} \nonumber$ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.04%3A_Percent_Yield/3.4.02%3A_Foods-_Vegetable_Oil_Hydrogenation_Trans_Fats_and_Percent_Yield.txt |
Up to this point we have obtained all stoichiometric ratios from the coefficients of balanced chemical equations. Chemical formulas also indicate relative amounts of substance, however, and stoichiometric ratios may be derived from them, too. For example, the formula HgBr2 tells us that no matter how large a sample of mercuric bromide we have, there will always be 2 mol of bromine atoms for each mole of mercury atoms. That is, from the formula HgBr2 and the submicroscopic image below, we have the stoichiometric ratio
$\text{S}\left( \frac{\text{Br}}{\text{Hg}} \right)=\frac{\text{2 mol Br}}{\text{1 mol Hg}}\label{1}$
We could also determine that for HgBr2
$\text{S}\left( \frac{\text{Hg}}{\text{HgBr}_{\text{2}}} \right)=\frac{\text{1 mol Hg}}{\text{1 mol HgBr}_{\text{2}}} \nonumber$
$\text{S}\left( \frac{\text{Br}}{\text{HgBr}_{\text{2}}} \right)=\text{ }\frac{\text{2 mol Br}}{\text{1 mol HgBr}_{\text{2}}} \nonumber$
(The reciprocals of these stoichiometric ratios are also valid for HgBr2.)
Stoichiometric ratios derived from formulas instead of equations are involved in the most common procedure for determining the empirical formulas of compounds which contain only C, H, and O. The process is depicted in the image below for reference. A weighed quantity of the substance to be analyzed is placed in a combustion train and heated in a stream of dry O2. All the H in the compound is converted to H2O(g) which is trapped selectively in a previously weighed absorption tube. All the C is converted to CO2(g) and this is absorbed selectively in a second tube. The increase of mass of each tube tells, respectively, how much H2O and CO2 were produced by combustion of the sample.
Example $1$ : Empirical Formula
A 6.49-mg sample of ascorbic acid (vitamin C) was burned in a combustion train. 9.74 mg CO2 and 2.64 mg H2O were formed. Determine the empirical formula of ascorbic acid.
Solution
We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric ratios
$\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}} \nonumber$
$\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}} \nonumber$
Thus
\begin{align*} n_{\text{C}}&=\text{9}\text{.74}\times \text{10}^{\text{-3}}\text{g CO}_{\text{2}}\times {\frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}} \times {\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}} = \text{2.21}\times \text{10}^{-4}\text{ mol C}\ { } \ n_{\text{H}} &=\text{2}\text{.64}\times \text{10}^{\text{-3}}\text{g H}_{\text{2}}\text{O}\times {\frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}} \times {\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}} = \text{2.93}\times \text{10}^{\text{-4}}\text{ mol H} \end{align*} \nonumber
The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample
\begin{align*} m_{\text{C}}&=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{2}\text{.65}\times \text{10}^{\text{-3}}\text{g C}=\text{2}\text{.65 mg C} \ & \ m_{\text{H}}&=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{2}\text{.95}\times \text{10}^{\text{-4}}\text{g H}=\text{0}\text{0.295 mg H} \ \end{align*} \nonumber
Thus we have
$6.49 \text{mg sample} – 2.65 \text{mg C} – 0.295 \text{mg H} = 3.54 \text{mg O} \nonumber$
and
$n_{\text{O}}=\text{3}\text{.54}\times \text{10}^{\text{-3}}\text{ g O}\times {\frac{\text{1 mol O}}{\text{16.00 g O}}} =\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O} \nonumber$
The ratios of the amounts of the elements in ascorbic acid are therefore
${\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1}\text{.33 mol H}}{\text{1 mol C}}=\frac{\text{1}\tfrac{1}{3}\text{mol H}}{\text{1 mol C}}=\frac{\text{4 mol H}}{\text{3 mol C}}} \nonumber$
${\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1 mol O}}{\text{1 mol C}}=\frac{\text{3 mol O}}{\text{3 mol C}}} \nonumber$
Since nC:nH:nO is 3 mol C:4 mol H:3 mol O, the empirical formula is C3H4O3.
A drawing of a molecule of ascorbic acid is shown here. You can determine by counting the atoms that the molecular formula is C6H8O6—exactly double the empirical formula. It is also evident that there is more to know about a molecule than just how many atoms of each kind are present. In ascorbic acid, as in other molecules, the way the atoms are connected together and their arrangement in three-dimensional space are quite important. A picture showing which atoms are connected to which is called a structural formula. Empirical formulas may be obtained from percent composition or combustion-train experiments, and, if the molecular weight is known, molecular formulas may be determined from the same data. More complicated experiments are required to find structural formulas. In Example 2 we obtained the mass of O by subtracting the masses of C and H from the total mass of sample. This assumed that only C, H, and O were present. Sometimes such an assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that it contained sulfur was missed. This mistake was not discovered for some time because the atomic weight of sulfur is almost exactly twice that of oxygen. Two atoms of oxygen were substituted in place of one sulfur atom in the formula.
A 3D representation of L-Ascorbic Acid
3.05: Analysis of Compounds
The caloric value for fats is about 9 Cal/g, while for carbohydrates (sugars or starches) and proteins, it's about 4 Cal/g [1], so a teaspoon of sugar is only about 20 Calories, but a teaspoon of oil is about 45 calories. Our body stores fats as a long term, high energy per gram energy source, while sugars can be metabolized quickly, but don't give as much energy per gram. The energy is released when each is metabolized, giving the same amount of energy as combustion in air.
Fats and Oils
Vegetable fats and oils are all triglycerides, which contain a glycerol () three carbon "backbone" with 3 long chain "fatty acids" attached through ester linkages, as in the figure below. The actual shape is shown in the Jmol model, which can be rotated with the mouse. Triglycerides are called "fats" when they're solids or semisolids, and "oils" when they're liquids.
A triglyceride, overall unsaturated, with the glycerol "backbone" on the left, and saturated palmitic acid, monounsaturated oleic acid, and polyunsaturated alpha-linolenic acid. Carbon atoms are at each bend in the structure, and hydrogen atoms are omitted.
The long chain fatty acids may be saturated with hydrogen atoms, in which case they have all single bonds like the top fatty acid in the Figure (which is palmitic acid). If they have fewer hydrogen atoms, they are unsaturated and have double bonds like the middle fatty acid in the Figure (which is oleic acid). The bottom fatty acid is polyunsaturated, with multiple double bonds (it is linolenic acid). Various cooking oils have |known concentrations of saturated and unsaturated fatty acids.
Carbohydrates
Carbohydrates are made up of simple sugar units. The sucrose (ordinary table sugar) molecule shown below is made of a glucose and fructose "monosaccharides".
Sucrose,C12H22O11 (C atoms are at each bend, H atoms are not shown)
A key to why fats have more than twice the caloric value of sugars comes from the combustion reactions:
A typical sugar combustion reaction is
C12H22O11 + 6 O2 → 6 CO2 + 6 H2O
It requires 6 mol O2 for every 6 mol of C, a 1:1 ratio. Since the molar mass of sucrose is 180 g/mol, about 0.033 mol O2 is required per gram.
While a typical fat combustion might be
C56H108O6 + 80 O2 → 56 CO2 + 54 H2O
It requires 80 mol O2 for every 56 mol of C, a 1.42:1 ratio. Or, since the molar mass of the fat shown is 878 g/mol, about 0.091 mol O2 is required per gram. The fat requires much more oxygen to burn, and consequently produces more energy. We see why by looking at the C:O mole ratios in fats and sugars:
In the fat above,
$\text{S}\left( \frac{\text{C}}{\text{O}} \right)=\frac{\text{56 mol C}}{\text{6 mol O}}$ (a 9.3:1 ratio). Or for the sugar,
$\text{S}\left( \frac{\text{C}}{\text{O}} \right)=\frac{\text{6 mol C}}{\text{6 mol O}}$ (a 1:1 ratio).
Since the sugar is already more oxygenated, it produces less energy when it is burned.
Looking at it another way, the carbon, which reacts with oxygen and releases energy in combustion, is a bigger part of the fat:
In fat: $S\left( \frac{\text{C}}{\text{C}_{56}\text{H}_{108}\text{O}_{6}} \right)=\frac{\text{56 mol C}}{\text{1 mol C}_{56}\text{H}_{108}\text{O}_{6}}$ or 56 mol C/877.5 g C56H108O6 = 0.064 mol C/g fat.
In Sugar: $S\left( \frac{\text{C}}{\text{C}_{6}\text{H}_{12}\text{O}_{6}} \right)=\frac{\text{6 mol C}}{\text{1 mol C}_{6}\text{H}_{12}\text{O}_{6}}$ or 6 mol C/180 g C6H12O6 = 0.033 mol C/g fat.
We'll see below that these ratios actually allow us to determine the chemical formula for a fat, sugar, or any other compound.
Stoichiometric ratios derived from formulas instead of equations are involved in the most common procedure for determining the empirical formulas of compounds which contain only C, H, and O. A weighed quantity of the substance to be analyzed is placed in a combustion train and heated in a stream of dry O2. All the H in the compound is converted to H2O(g) which is trapped selectively in a previously weighed absorption tube. All the C is converted to CO2(g) and this is absorbed selectively in a second tube. The increase of mass of each tube tells, respectively, how much H2O and CO2 were produced by combustion of the sample
EXAMPLE 1 A 1.000 g sample of a fat was burned in a combustion train, producing 2.784 g of CO2 and 1.140 g of H2O. Determine the empirical formula of the fat.
Solution
We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric ratios
$\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}$ $\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}$ Thus
$n_{\text{C}}=\text{2.784 g CO}_{\text{2}}\times \frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}\times \frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}=\text{0.06326 mol C}$
$n_{\text{H}}=\text{ 1.140 g H}_{\text{2}}\text{O}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}\times \frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{0.1265 mol H}$ The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample
\begin{align} & m_{\text{C}}=\text{0.06326 mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{0.7598 g C} \ & \ & m_{\text{H}}=\text{0.1265 mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{0.1275g H} \ \end{align}
Thus we have 1.000 g sample – 0.7598 g C – 0.1275 g H = 0.1128 g O and $n_{\text{O}}=\text{0.1128 g O}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g O}}=\text{0.007050 mol O}$ The ratios of the amounts of the elements in ascorbic acid are therefore $\frac{n_{\text{H}}}{n_{\text{O}}}=\frac{\text{0.1265 mol H}}{\text{0.00705 mol C}}=\frac{\text{17.94 mol H}}{\text{1 mol O}}$
$\frac{n_{\text{C}}}{n_{\text{O}}}=\frac{\text{20.06326 mol O}}{\text{0.00705 mol O}}=\frac{\text{8.97 mol C}}{\text{1 mol C}}$
Since nC:nH:nO is 9 mol C : 18 mol H : 1 mol O, the empirical formula is C9H18O1.
Because most fats have 3 fatty acids bonded to glycerol with 2 oxygen atoms in each "ester" bond, the molecular formula is probably C54H108O6. This molecule would contain the 3 carbon glycerol backbone, the 3 fatty acid chains would share the remaining 51 carbon atoms, and would be of average length 51/3 = 17 carbon atoms.
EXAMPLE 2 A 6.49-mg sample of ascorbic acid (vitamin C) was burned in a combustion train. 9.74 mg CO2 and 2.64 mg H2O were formed. Determine the empirical formula of ascorbic acid.
Solution
We need to know the amount of C, the amount of H, and the amount of O in the sample. The ratio of these gives the subscripts in the formula. The first two may be obtained from the masses of CO2 and H2O using the molar masses and the stoichiometric ratios
$\text{S}\left( \frac{\text{C}}{\text{CO}_{\text{2}}} \right)=\frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}$ $\text{S}\left( \frac{\text{H}}{\text{H}_{\text{2}}\text{O}} \right)=\frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}$ Thus
$n_{\text{C}}=\text{9}\text{.74}\times \text{10}^{\text{-3}}\text{g CO}_{\text{2}}\times \frac{\text{1 mol CO}_{\text{2}}}{\text{44}\text{.01 g CO}_{\text{2}}}\times \frac{\text{1 mol C}}{\text{1 mol CO}_{\text{2}}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}$
$n_{\text{H}}=\text{2}\text{.64}\times \text{10}^{\text{-3}}\text{g H}_{\text{2}}\text{O}\times \frac{\text{1 mol H}_{\text{2}}\text{O}}{\text{18}\text{.02 g H}_{\text{2}}\text{O}}\times \frac{\text{2 mol H}}{\text{1 mol H}_{\text{2}}\text{O}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}$ The compound may also have contained oxygen. To see if it does, calculate the masses of C and H and subtract from the total mass of sample
\begin{align} & m_{\text{C}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}\times \frac{\text{12}\text{.01 g C}}{\text{1 mol C}}=\text{2}\text{.65}\times \text{10}^{\text{-3}}\text{g C}=\text{2}\text{.65 mg C} \ & \ & m_{\text{C}}=\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}\times \frac{\text{1}\text{.008 g H}}{\text{1 mol H}}=\text{2}\text{.95}\times \text{10}^{\text{-4}}\text{g H}=\text{0}\text{.295 mg H} \ \end{align}
Thus we have 6.49 mg sample – 2.65 mg C – 0.295 mg H = 3.54 mg O and $n_{\text{O}}=\text{3}\text{.54}\times \text{10}^{\text{-3}}\text{ g O}\times \frac{\text{1 mol O}}{\text{16}\text{.00 g O}}=\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}$ The ratios of the amounts of the elements in ascorbic acid are therefore $\frac{n_{\text{H}}}{n_{\text{C}}}=\frac{\text{2}\text{.93}\times \text{10}^{\text{-4}}\text{ mol H}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1}\text{.33 mol H}}{\text{1 mol C}}=\frac{\text{1}\tfrac{1}{3}\text{mol H}}{\text{1 mol C}}=\frac{\text{4 mol H}}{\text{3 mol C}}$
$\frac{n_{\text{O}}}{n_{\text{C}}}=\frac{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol O}}{\text{2}\text{.21}\times \text{10}^{\text{-4}}\text{ mol C}}=\frac{\text{1 mol O}}{\text{1 mol C}}=\frac{\text{3 mol O}}{\text{3 mol C}}$
Since nC:nH:nO is 3 mol C:4 mol H:3 mol O, the empirical formula is C3H4O3.
A drawing of a molecule of ascorbic acid is shown here. You can determine by counting the atoms that the molecular formula is C6H8O6—exactly double the empirical formula. It is also evident that there is more to know about a molecule than just how many atoms of each kind are present. In ascorbic acid, as in other molecules, the way the atoms are connected together and their arrangement in three-dimensional space are quite important. A picture showing which atoms are connected to which is called a structural formula. Empirical formulas may be obtained from percent composition or combustion-train experiments, and, if the molecular weight is known, molecular formulas may be determined from the same data. More complicated experiments are required to find structural formulas. In Example 2 we obtained the mass of O by subtracting the masses of C and H from the total mass of sample. This assumed that only C, H, and O were present. Sometimes such an assumption may be incorrect. When penicillin was first isolated and analyzed, the fact that it contained sulfur was missed. This mistake was not discovered for some time because the atomic weight of sulfur is almost exactly twice that of oxygen. Two atoms of oxygen were substituted in place of one sulfur atom in the formula.
A 3D representation of L-Ascorbic Acid <chemeddl-jmol2>ascorbic|size=300</chemeddl-jmol2> | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.05%3A_Analysis_of_Compounds/3.5.01%3A_Foods-_Burning_or_Metabolizing_Fats_and_Sugars.txt |
When a chemical reaction occurs, there is usually a change in temperature of the chemicals themselves and of the beaker or flask in which the reaction is carried out. If the temperature increases, the reaction is exothermic—energy is given off as heat when the container and its contents cool back to room temperature. (Heat is energy transferred from one place to another solely because of a difference in temperature.) An endothermic reaction produces a decrease in temperature. In this case heat is absorbed from the surroundings to return the reaction products to room temperature. Thermochemistry, a word derived from the Greek thermé, “heat,” is the measurement and study of energy transferred as heat when chemical reactions take place. It is extremely important in a technological world where a great deal of work is accomplished by transforming and harnessing heat given off during combustion of coal, oil, and natural gas.
If your workstation is authorized to view JCE Software, you will see a video below which shows an example of an endothermic reaction. Ammonium thiocyanate is mixed with barium hydroxide, and the reaction takes in enough heat to freeze water.
Find others videos by searching YouTube for "endothermic reaction".
3.07: Energy
Energy is usually defined as the capability for doing work. For example, a billiard ball can collide with a second ball, changing the direction or speed of motion of the latter. In such a process the motion of the first ball would also be altered. We would say that one billiard ball did work on (transferred energy to) the other.
Kinetic Energy
Image source: Smart Learning for All
Energy due to motion is called kinetic energy and is represented by Ek. For an object moving in a straight line, the kinetic energy is one-half the product of the mass and the square of the speed:
$E_{k} = \frac{1}{2} mu^{2} \label{1}$
where
• m = mass of the object
• u = speed of object
If the two billiard balls mentioned above were studied in outer space, where friction due to their collisions with air molecules or the surface of a pool table would be negligible, careful measurements would reveal that their total kinetic energy would be the same before and after they collided. This is an example of the law of conservation of energy, which states that energy cannot be created or destroyed under the usual conditions of everyday life. Whenever there appears to be a decrease in energy somewhere, there is a corresponding increase somewhere else.
Example $1$ : Kinetic Energy
Calculate the kinetic energy of a Volkswagen Beetle of mass 844 kg (1860 lb) which is moving at 13.4 m s–1 (30 miles per hour).
Solution:
$\large E_{k} = \frac{1}{2} m u^{2} = \frac{1}{2} \times 844 \text{ kg} \times ( 13.4 \text{ m} \text{ s}^{-1} )^{2} = 7.58 \times 10^{4} \text{ kg}\text{ m}^{2} \text{ s}^{-2}$
In other words the units for energy are derived from the SI base units kilogram for mass, meter for length, and second for time. A quantity of heat or any other form of energy may be expressed in kilogram meter squared per second squared. In honor of Joule’s pioneering work this derived unit 1 kg m2 s–2 called the joule, abbreviated J. The Volkswagen in question could do nearly 76 000 J of work on anything it happened to run into.
Potential Energy
Image source: Smart Learning for All
Potential Energy is energy that is stored by rising in height, or by other means. It frequently comes from separating things that attract, like rising birds are being separated from the Earth that attracts them, or by pulling magnets apart, or pulling an electrostatically charged balloon from an oppositely charged object to which it has clung. Potential Energy is abbreviated EP and gravitational potential energy is calculated as follows:
$\large E_{P} = mgh \tag{2}$
where
• m = mass of the object in kg
• g = gravitational constant, 9.8 m s2
• h = height in m
Notice that EP has the same units, kg m2 s–2 or Joule as kinetic energy.
Example $2$: Kinetic Energy Application
How high would the VW weighing 844 kg and moving at 30 mph need to rise (vertically) on a hill to come to a complete stop, if none of the stopping power came from friction?
Solution:
The car's kinetic energy is 7.58 × 104 kg m2 s–2(from EXAMPLE $1$ ), so all of this would have to be converted to EP. Then we could calculate the vertical height:
$\large E_{P} = mgh = 7.58 \times 10^{4} \text{ kg} \text{ m}^{2} \text{ s}^{-2} = 844 \text{ kg} \times 9.8 \text{m} \text {s}^{-2} \times h$
$\large h = 9.2 \text{ m}$
Even when there is a great deal of friction, the law of conservation of energy still applies. If you put a milkshake on a mixer and leave it there for 10 min, you will have a warm, rather unappetizing drink. The whirling mixer blades do work on (transfer energy to) the milkshake, raising its temperature. The same effect could be produced by heating the milkshake, a fact which suggests that heating also involves a transfer of energy. The first careful experiments to determine how much work was equivalent to a given quantity of heat were done by the English physicist James Joule (1818 to 1889) in the 1840s. In an experiment very similar to our milkshake example, Joule connected falling weights through a pulley system to a paddle wheel immersed in an insulated container of water. This allowed him to compare the temperature rise which resulted from the work done by the weights with that which resulted from heating. Units with which to measure energy may be derived from the SI base units of Table 1 from The International System of Units (SI)(opens in new window) by using Eq. $\ref{1}$.
Another unit of energy still widely used by chemists is the calorie. The calorie used to be defined as the energy needed to raise the temperature of one gram of water from 14.5°C to 15.5°C but now it is defined as exactly 4.184 J. | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.06%3A_Thermochemistry.txt |
Energy changes which accompany chemical reactions are almost always expressed by thermochemical equations, such as
$\text{C} H_{4} (g) + 2 \text{O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) + 2 \text{H}_{2} \text{O} (l) \text{ (25°C, 1 atm pressure)} \ \Delta H_{m} = –890 \text{kJ} \label{1}$
which is displayed on the atomic level below. To get an idea of what this reaction looks like on the macroscopic level, check out the flames on the far right.
Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs as written, and also enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. $\ref{1}$, indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. $\ref{1}$, the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of ΔHm.
It is important to notice that ΔHm is the energy for the reaction as written. In the case of Equation $\ref{1}$, that represents the formation of 1 mol of carbon dioxide and 2 mol of water. The quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. $\ref{1}$ tells us that 890.4 kJ of heat energy is given off for every mole of CH4 which is consumed. Alternatively, it tells us that 890.4 kJ is released for every 2 moles of H2O produced. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed or released when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed or released and n is the amount of substance involved, then
$\large \Delta H_{\text{m}}=\frac{q}{n} \nonumber$
Example $1$ : Heat Energy
How much heat energy is obtained when 1 kg of ethane gas, C2H6, is burned in oxygen according to the equation:
$2 \text{C}_{2} \text{H}_{6} (g) + 7 \text{O}_{2} (g) \rightarrow 4 \text{C} \text{O}_{2} (g) + 6 \text{H}_{2} \text{O} (l) \nonumber$
with $\Delta H_{m} = –3120 \text{ kJ}$.
Solution
The mass of C2H6 is easily converted to the amount of C2H6 from which the heat energy q is easily calculated by means of Eq. (2). The value of ΔHm is –3120 kJ per per 2 mol C2H6. The road map is
$\large m_{\text{C}_{\text{2}}\text{H}_{\text{6}}}\text{ }\xrightarrow{M}\text{ }n_{\text{C}_{\text{2}}\text{H}_{\text{6}}}\text{ }\xrightarrow{\Delta H_{m}}\text{ }q \nonumber$
so that
\begin{align*} q &= 1 \times 10^3 \text{ g }\ce{C2H6} \times \frac{\text{1 mol }\ce{C2H6}}{\text{30.07 g }\ce{C2H6}} \times \frac{-3120\text{ kJ}}{\text{2 mol }\ce{C2H6}} \ &= -\text{51 879 kJ} = -\text{51.88 MJ} \end{align*} \nonumber
By convention, a negative value of q corresponds to a release of heat energy by the matter involved in the reaction.
The quantity ΔHm is referred to as an enthalpy change for the reaction. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy.
It is important to realize that the value of ΔHm given in thermochemical equations like $\ref{1}$ or $\ref{3}$ depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. $\ref{1}$, the value of ΔHm would be different from -890.4 kJ. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified they usually refer to 25°C and to normal atmospheric pressure.
Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change. For example,
$\text{H}_{2} \text{O} (l) \rightarrow \text{H}_{2} \text{O} (g) \ \Delta \text{H}_{m} = 44 \text{ kJ} \nonumber$
In the image above, the flames input energy into the water, giving it the energy necessary to transition to the gas phase. Since flames provide the energy for the phase transition, this is an endothermic reaction (energy is absorbed).
tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. $\text{H}_{2} \text{O} (g) \rightarrow \text{H}_{2} \text{O} (l) \ \Delta \text{H}_{m} = –44 \text{kJ} \nonumber$
It's counterintuitive, but the common summer occurrence seen above is actually exothermic. Since the reaction isn't highly exothermic (like the combustion of CH4), we find it hard to associate with a release of energy. Thermodynamics allows us to better understand on a micro level energy changes like this one.
To see why this must be true, suppose that ΔHm [Eq. (4a)] = 44 kJ while ΔHm [Eq. (4b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is,
$\Delta \text{H}_{m} \text{forward} = –\Delta \text{H}_{m} \text{reverse} \nonumber$
3.08: Thermochemical Equations
As we've seen, the diet of Eagles (along with all other animals including us) includes certain masses of food. For eagles, it's 250-550 g/day. We also saw that this provides the energy for all the day's activities, and food that's left after energy production goes to weight gain. How is the mass of food related to the energy produced? The first step in answering this question is as simple one, and involves writing an overall "thermochemical equation" for the metabolism of sugar, which turns out to be the same as the equation for the combustion of sugar. A thorough answer to this question requires us to consider other factors, which we'll take up later.
Thermochemical equations are used to relate energy changes to the chemical reactions that produce them. For example, we've already seen in Metabolism of Dietary Sugar that sugar is metabolized according to the equation[1]
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) (25o, 1 Atm) ΔHm = –2808 kJ (1) Here the sign of ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and the value enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. (1), indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (1), the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of ΔHm.
It is important to notice that ΔHm is the energy change for the equation as written. This is necessary because the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (1) tells us that 2805 kJ of heat energy is given off for every mole of C6H12O6 which is consumed. Alternatively, it tells us that 2808 kJ is released for every 6 mole of H2O produced, i.e., 468 kJ is produced for every mol H2O. ΔHm for Equation (1) also tells us that 2808 kJ of heat is released when 6 mol of carbon dioxide is produced, or 6 mol of oxygen is consumed. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed and n is the amount of substance involved, then
$\Delta H_{\text{m}}=\frac{q}{n}$ (2)
EXAMPLE 1 How much heat energy is obtained if we assume that the eagle's diet of 250-550 g includes 350 g of glucose, C6H12O6, which is burned in oxygen according to the equation:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
ΔHm = –2808 kJ (3)
Solution
The mass of C6H12O6 is easily converted to the amount of C6H12O6 from which the heat energy q is easily calculated by means of Eq. (2). The value of ΔHm is –2805 kJ per mole of C6H12O6,
$m_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{M}\text{ }n_{\text{C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\xrightarrow{\Delta H_{m}}\text{ }q$
so that
$q=\text{350}\text{ g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}\text{ }\times \text{ }\frac{\text{1 mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}{\text{180}\text{.16 g C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}\text{ }\times \text{ }\frac{-\text{2808 kJ}}{\text{mol C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{\text{6}}}=-\text{5 455 kJ}$ Note: By convention a negative value of q corresponds to a release of heat energy by the matter involved in the reaction. The quantity ΔHm is the enthalpy change for the reaction equation as written. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Enthalpy For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy.
It is important to realize that the value of ΔHm given in thermochemical equations like (1) or (3) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. (1), the value of ΔHm would be different from -2808 kJ. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure.
Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change. For example,
H2O(l) → H2O(g) ΔHm = 44 kJ (4a) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H2O(g) → H2O(l) ΔHm = –44 kJ (4b) To see why this must be true, suppose that ΔHm [Eq. (4a)] = 44 kJ while ΔHm [Eq. (4b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is, ΔHm forward = –ΔHm reverse
References
1. Atkins, P. Physical Chemistry, 6th Ed., W.H. Freeman &Co. New York, 1998, p. 69 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.08%3A_Thermochemical_Equations/3.8.01%3A_Biology-_Weight_of_Food_and_Energy_Production.txt |
You may have noticed "E85" gasoline selling for $2.39/gallon for FlexFuel vehicles while regular gas was selling for$2.79. Is E85 use advantageous if you have a FlexFuel vehicle capable of using either E85 (85% ethanol) or regular gas? Note that many states sell regular gas that is E10 (10% ethanol) for standard (non-FlexFuel) vehicles.
Ethanol for fuel is not a new idea
85% Ethanol at a modern pump
The model T was first designed to run on ethanol[1]
There are really (at least) two questions here:
1. Is it a good deal for the consumer (is energy from ethanol economical)?
2. Is it a good deal for the environment (is the Energy Balance favorable)?
The Energy Balance
The energy balance is the ratio of the energy produced by 1 kg of the fuel (i.e. ethanol), to the energy necessary to produce it (cultivation of plants, fermentation, transportation, irrigation, etc.). The energy balance for ethanol in the US is only 1.3 to 1.6 (we get 1.3-1.6 J out for 1 J energy input), while in Brazil it's 8.3 to 10.2. [2][3]
The energy balance for oil is about 5: Today, about 5 barrels of oil extracted for every 1 barrel of oil is consumed in the process (a century ago, when oil was more plentiful the ratio was 50:1).[4]
The reason for the poor energy balance for fuel ethanol in the US is that we use corn. We extract the cornstarch, and then hydrolyze it at high temperatures (over 90 oC and ferment the resulting sugars to give the ethanol (simultaneous saccharification and fermentation, SSF). More energy efficient processes are being developed. [5] Starch is a polymer of glucose composed of amylose (shown below) and amylopectin, a similar polymer with more branching. The chemical reactions that occur are first, hydrolysis of starches to maltose (C12H22O11):
[C6H10O5]n + n/2 H2O → n/2 C12H22O11 (1)
Amylose Maltose
The complex sugar maltose is further hydrolyzed (sometimes in the same step) to glucose (C6H12O6):
C12H22O11 + H2O → 2 C6H12O6 (2)
Finally, during ethanol fermentation, glucose is decomposed into ethanol (C2H5OH) and carbon dioxide.
C6H12O6 → 2 C2H5OH + 2 CO2 + heat (3)
Switchgrass requires similar processing, but has the potential for a higher energy balance. In Brazil, sugarcane is used in ethanol production, eliminating the high energy cost reaction (1), because sugarcane produces sucrose, a sugar similar to maltose. In all cases, a large energy expenditure is necessary to separate the ethanol from the water in which the reactions take place, and other impurities.
Sources of Ethanol
↑ Sugarcane
← Switchgrass
Corn
But even in Brazil, rising sugar costs led to a rebound in reliance on petroleum, and E100 capable vehicles declined sharply in the late 1980s[6]. Nonetheless, the existence of ethanol pumps at a large percentage of gas stations has helped sustain the use of other blends. Clearly, the economics and history of fuel use should be studied carefully when investing in future fuel development.
The Cost Advantage
So which is a better deal: filling your 10-gallon tank with E85 costing $2.69/gallon or with gasoline costing$2.99/gallon?[7].
Energy changes which accompany chemical reactions are almost always expressed by thermochemical equations. For example, during combustion ethanol reacts with oxygen to produce carbon dioxide, water vapor, and heat according to the thermochemical equation:
C2H5OH + 3 O2 → 2 CO2 + 3 H2O (l) (4) ΔHm = -1367 kJ[8]
The quantity ΔHm is the enthalpy change for the reaction equation as written. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy. Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. (3), indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (4), stronger bonds have formed, leading to a decrease in potential energy, and it is this decrease which is indicated by a negative value of ΔHm.
It is important to notice that the quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (4) tells us that 1367 kJ of heat energy is given off for every mole of C2H5OH which is consumed. Alternatively, it tells us that 1367 kJ is released for every 3 moles of H2O produced, or every 2 mol of carbon dioxide produced, or every 3 mol of oxygen consumed. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed and n is the amount of substance involved, then
$\Delta H_{\text{m}}=\frac{q}{n}$ (5) Equation (4) represents the standard heat of combustion, which is related to the "higher heating value" of a fuel.[9] Because liquid water is produced in Equation (4) and the heat (1367 kJ/mol) is given for a theoretical reaction occurring at 25 oC, it is only an approximation of the heat produced in an actual combustion, where the reaction takes place at a high temperature and produces water vapor. A better estimate may be the Lower Heating Value (LHV) which is adjusted by adding the heat of vaporization of water, and heat required to raise the temperature of reactants to the combustion temperature and products to 150 oC (an arbitrarily chosen standard).[10][11] We'll use the LHV for ethanol, -1330 kJ/mol,[12] and abbreviate ethanol (C2H5OH) as EtOH to calculate the heat in 1 gallon of ethanol:
$V_{\text{EtOH}}~\xrightarrow{M}~m_{\text{EtOH}}\text{ }\xrightarrow{M}\text{ }n_{\text{EtOH}}~\xrightarrow{\Delta H_{m}}~q$
so that
$q=\text {1 gallon EtOH} ~\times~\frac{\text{3.79 L}}{\text{1 gallon}} ~\times~\frac{\text{1000 mL}}{\text{1 L}} ~\times~\frac{\text{0.789 g}}{\text{mL}} ~\times~\frac{\text{1 mol EtOH}}{\text{46.07 g EtOH}} ~\times~\frac{-\text{1330 kJ}}{\text{mol EtOH}}$
$=-86~330\text{ kJ}=-\text{86}\text{.33 MJ (estimate)}$
We can calculate the energy value for octane, which represents gasoline fairly well, in a similar way. The thermochemical equation for the combustion of octane (C8H18) is:
C8H18 + 25/2 O2 → 8 CO2 + 9 H2O (l) (6) ΔH ~ -5430 kJ/mol[13]
Again, we'll use the LHV for octane (-5064 kJ/mol[14]) in our calculation:
$q=\text {1 gallon octane} ~\times~\frac{\text{3.79 L}}{\text{1 gallon}} ~\times~\frac{\text{1000 mL}}{\text{1 L}} ~\times~\frac{\text{0.737 g}}{\text{mL}} ~\times~\frac{\text{1 mol octane}}{\text{114.23 g octane}} ~\times~\frac{\text{-5 064 kJ}}{\text{mol octane}}$
$=-123~800\text{ kJ}=-\text{123}\text{.80 MJ (estimate)}$
So 1 gallon of gasoline has (123,800 / 86,330) or 1.4 times as much heating value, but costs only $2.79 /$2.39 or 1.2 times as much. It's the better buy. If the Energy Balance may only be about 1.3 in the US, it actually is a net loss to burn ethanol, at least at the current price and availability of petroleum. As the Energy Balance and availability of petroleum decreases, we had better develop more energy efficient means of ethanol production.
Indeed, based on EPA tests for all 2006 E85 models, the average fuel economy for E85 vehicles was 25.56% lower than unleaded gasoline.[15][16]
The Lower Heat Value (LHV)
It is important to realize that the value of ΔHm given in thermochemical equations like (4) or (6) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a gas instead of a liquid in the reaction in Eq. (4), the value of ΔHm would be different from -1367 kJ. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure.
Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change. For example,
H2O(l) → H2O(g) ΔHm = 44 kJ (7a) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H2O(g) → H2O(l) ΔHm = –44 kJ (7b) To see why this must be true, suppose that ΔHm [Eq. (7a)] = 44 kJ while ΔHm [Eq. (7b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is,
ΔHm forward = –ΔHm reverse
Since Reaction (4) produces 3 mol of H2O(l), it would produce 3 mol x 44 kJ/mol = 132 kJ less energy if the water were produced as the vapor, because the heat released in condensation to the liquid would not be included. The enthalpy change would then be -1367 + 132 kJ or -1235 kJ. This is still different from the LHV because of the heat required to change the temperature of the reactants and products from the standard temperature (25 oC) to the combustion temperature.
FlexFuel Facts
In 2007, only 3.3% of American cars were FlexFuel and only 1% of the filling stations provided FlexFuel, but the percentage has been increasing rapidly. World ethanol production for transport fuel tripled between 2000 and 2007 from 17 billion to more than 52 billion litres, with 89% produced in Brazil and the US.[17] The price differential shows large changes (it was about 30% in 2007).[18] The National Alcohol Program in Brazil, a world leader in using ethanol fuel, mandated decreased reliance on petroleum after the first oil crisis in 1973, and by 1979 several automakers provided cars that ran on pure ethanol (E100). After reaching more than 4 million cars and light trucks running on pure ethanol by the late 1980s,[3] the use of E100-only vehicles sharply declined after increases in sugar prices produced shortages of ethanol fuel. The emphasis has been on E85 (85% ethanol) vehicles in the US, because engines won't start reliably on E100 below about 60 oF. Even E85 cannot be used below about 30 oF, and in northern states E70 is delivered by E85 pumps without changing the label.[19] | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.08%3A_Thermochemical_Equations/3.8.02%3A_Environment-_Gas.txt |
Energy from Fats & Sugars
Earlier we discussed the nature of fats and mentioned that fats typically provide 9 Cal/g of food energy, while sugars provide about 4 Cal/g. So in order to store the energy in 10 lb of fat, your body would need to store 22.5 lb of carbohydrates or sugars; but it's more extreme than that. Because sugars carry about their own weight of associated water in the body, 67.5 lb (31 kg) of hydrated glycogen has the energy equivalent of 10 lb (5 kg) of fat![1]
The food energy in various food types is given approximately in the following table [2], and you can find the fat content (as well as all other nutritional information) about nearly all foods in the United States Department of Agriculture's Bulletin #8 which has a searchable USDA Nutrient database. In the database, fats are list under "lipids"/"Fatty Acids" and then under "saturated" and "18:0", indicating the number of carbon atoms (18) in the fatty acid, and the number of double bonds (0) (see Example 1).
Food component Energy Density
kJ/g kcal/g
Fat 37 9
Ethanol (drinking alcohol) 29 7
Proteins 17 4
Carbohydrates 17 4
Sorbitol, sugar alcohol sweeteners) 10 2.4
dietary Fiber 8 2
These caloric values are measured in a "bomb calorimeter" like the one in the Figure.
A bomb Calorimeter
A 1-2 g sample of food is sealed in a heavy walled steel cylinder (about 4" in diameter and 7" high), shown in the center of the Figure, which is then filled with pure oxygen at 30-40 atmospheres pressure, and immersed in a few liters of water. The sample is ignited electrically, and the heat released is determined by measuring the temperature increase of water that surrounds the "bomb".
Let's investigate the basis for these caloric values in terms of the chemical reactions that provide the energy. Energy changes which accompany chemical reactions are almost always expressed by thermochemical equations. The combustion of stearic acid, which is the main component of saturated fats, is written:
C18H36O2 (s) + 24 O2(g) → 18 CO2(g) + 18 H2O(l) (25°C, 1 atm pressure)
ΔHm = –11 407 kJ[3] (1)
Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. (1), indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (1), the C, H, and O atoms have collectively lost energy and it is this loss which is indicated by a negative value of ΔHm.
It is important to notice that ΔHm is the enthalpy for the reaction as written. The quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (1) tells us that 890.4 kJ of heat energy is given off for every mole of C18H36O2 which is consumed. Alternatively, it tells us that 11407 kJ is released for every 18 mole of H2O produced, or for every 18 mol of carbon dioxide produced, or 24 mol of oxygen consumed. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed and n is the amount of substance involved, then
$\Delta H_{\text{m}}=\frac{q}{n}$ (2)
EXAMPLE 1
Stearic Acid, 18:0
In the molecular model, each bend in the structure is occupied by a carbon atom, and each carbon atom has 4 bonds; missing bonds are to hydrogen atoms, which are not shown.
a. How much heat energy is obtained when 1 g of C18H36O2, is burned in oxygen according to the equation above? The molar mass of steric acid is 284.48 g/mol.
b. What is the caloric value of 1 g of stearic acid, given that ΔHm = –11407 kJ for equation (1)?
Solution
a. The mass of C18H36O2 is easily converted to the amount of C18H36O2 from which the heat energy q is easily calculated by means of Eq. (2). The value of ΔHm is –11407 kJ per mole of C18H36O2. The road map is
$m_{\text{C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{\text{2}}}\text{ }\xrightarrow{M}\text{ }n_{\text{C}_{\text{18}}\text{H}_{\text{36}}\text{O}_2}\text{ }\xrightarrow{\Delta H_{m}}\text{ }q$ so that $q=\text{1.0 g C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{2}\text{ }\times \text{ }\frac{\text{1 mol C}_{\text{18}}\text{H}_{\text{36}}\text{O}_2}{\text{284}\text{.48 g C}_{\text{18}}\text{H}_{\text{36}}\text{O}_{2}}\text{ }\times \text{ }\frac{-\text{11407 kJ}}{\text{1 mol C}_{\text{182}}\text{H}_{\text{36}}\text{O}_{2}}=\text{-40.09 kJ}$
b. $\text{-40.09 kJ}~\times~\frac{\text{1 kcal}}{\text{4.184 kJ}}~\times~\frac{\text{1 Cal}}{\text{1 kcal}}~=~-\text{9.84 Cal}$
Note: By convention a negative value of q corresponds to a release of heat energy by the matter involved in the reaction.
This is close to the estimated 9 Cal/g for fats. We saw earlier that most fats are triglycerides, that is, they would have 3 fatty acid substituents (like stearic acid) attached to a glycerol "backbone" in a fat like glyceryl tristearate ("stearin") (C57H110O6, M = 891.48). Stearin has a heat of combustion of -35 663 kJ/mol, so 1 g produces (35 663 kJ/mol) / (891.48 g/mol) x (1 Cal / 4.184 kJ) = 9.57 Cal. When energy is required by our body, triglycerides are converted free fatty acids, and transported by serum albumin in the blood to cells where energy is required. Serum albumin is necessary because the solubility of fatty acids is low in water-based blood.[4]
In comparison, sucrose (C12H22O11) has a molar mass of 342.3 g/mol and a heat of combustion of -5645 kJ/mol, so it produces 16.49 kJ/g or 3.94 Cal/g, very close to the estimated value, by the combustion:
C12H22O11 (s) + 12 O2(g) → 12 CO2(g) + 11 H2O(l) (25°C, 1 atm pressure)
ΔHm = –5 645 kJ mol–1[5]
The quantity ΔHm is the enthalpy change for the reaction as written. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy.
It is important to realize that the value of ΔHm given in thermochemical equations like (1) depends on the physical state of both the reactants and the products. Thus, if water were obtained as a liquid instead of a gas in the reaction in Eq. (1), the value of ΔHm would be different from -890.4 kJ. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure.
Two more characteristics of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change. For example,
H2O(l) → H2O(g) ΔHm = 44 kJ (3a) tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat. H2O(g) → H2O(l) ΔHm = –44 kJ (3b) To see why this must be true, suppose that ΔHm [Eq. (4a)] = 44 kJ while ΔHm [Eq. (4b)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is, ΔHm forward = –ΔHm reverse | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.08%3A_Thermochemical_Equations/3.8.03%3A_Foods-_Energy_from_Fats_and_Sugars.txt |
The Earth's mantle is composed largely of ultramafic rock which has high Fe or Mg content, and low silicon content. This rock may be converted to "serpentinites" by a process (logically) called serpentinization", which is an exothermic process, releasing a lot of heat energy-- about 660,000,000 joules of heat per cubic meter of rock according to the NOAA (National Oceanic and Atmospheric Administration). This is enough energy to raise the temperature of the rock by 260°C (550°F)[1]. The "Lost City", located 20 km west of the Mid-Atlantic Ridge, is a hydrothermal vent field characterized by carbonate edifices that tower 60 m above the ocean floor and extreme conditions found nowhere else in the marine environment[2]
Serpentinites at Lost City. Photo taken with the robotic vehicle Hercules. The wreckfish is ~ 1 m in length[3]
Serpentine itself is often chrysotile, Mg3Si2O5(OH)4. The name "serpentine" derives from the Latin serpentinus ("serpent rock") because the mineral is often greenish with a smooth to scaly surface[4].
Serpentine
Polished Serpentine Sample
Serpentine marbles are used in architecture and jewelry, but other serpentines include asbestos (which is a lung cancer risk when mined) and another 20 varieties of hydrous magnesium/iron phyllosilicates. Serpentinites often are toxic to plants because they may contain significant levels of nickel, chromium, and cobalt. They are often mixed, and thus treated collectively as a group called "serpentinites."
In the case of the "Lost City" of hydrothermal vents in the mid-Atlantic, serpentinization is a greater source of energy than the radioactivity of the Earth's core . Radioactivity normally accounts for about 80% of the internal heat of the Earth [5]. But Thermochemical reactions like serpentinization, which produce or consume significant amounts of heat, are an inextricable part of all geological processes.
Thermochemical Equations
Since serpentinites are mixtures, several equations can be used to describe their reactions. While natural minerals may have indefinite compositions like olivine, (Fe,Mg)2SiO4, we always look at equations for specific reactions in order to associate definite energies with them. These are called thermochemical equations. For example, serpentinization may involve:
Fayalite + water → magnetite + aqueous silica + hydrogen
3 Fe2SiO4 + 2 H2O → 2 Fe3O4 + 3 SiO2 + 3 H2 ΔHm = 40.1 kJ[6] (25°C, 1 atm pressure) (1)
Or: Forsterite + aqueous silica → serpentine (crysotile)
3 Mg2SiO4 + SiO2 + 4 H2O → 2 Mg3Si2O5(OH)4 ΔHm = -179.7 kJ[7] (25°C, 1 atm pressure) (2)
Or: Forsterite + water → serpentine (chrysotile) + brucite
3 Mg2SiO4 +3 H2O → 2 Mg3Si2O5(OH)4 + Mg(OH)2 ΔHm = -2267.2 kJ[8] (25°C, 1 atm pressure) (3)
Here the ΔHm (delta H subscript m) tells us whether heat energy is released or absorbed when the reaction occurs and also enables us to find the actual quantity of energy involved. By convention, if ΔHm is positive, as in Equation (1), heat is absorbed by the reaction; i.e., it is endothermic. More commonly, ΔHm is negative as in Eq. (2), indicating that heat energy is released rather than absorbed by the reaction, and that the reaction is exothermic. This convention as to whether ΔHm is positive or negative looks at the heat change in terms of the matter actually involved in the reaction rather than its surroundings. In the reaction in Eq. (2), the strength of the bonds increases as products are formed, so the products are lower in potential energy, and the lost energy is indicated by a negative value of ΔHm.
The values are calculated, as we'll see later, from standard tabulated values found in databases developed especially for geologists[9] [10] [11] [12]
It is important to notice that ΔHm is the energy for the reaction as written. The quantity of heat released or absorbed by a reaction is proportional to the amount of each substance consumed or produced by the reaction. Thus Eq. (2) tells us that 179.7 kJ of heat energy is given off for every mole of SiO2 which is consumed, or for every 3 mol of Mg2SiO4 consumed. Alternatively, it tells us that 179.7 kJ is released for every 2 moles of Mg3Si2O5(OH)4 produced. Seen in this way, ΔHm is a conversion factor enabling us to calculate the heat absorbed when a given amount of substance is consumed or produced. If q is the quantity of heat absorbed and n is the amount of substance involved, then:
$\Delta H_{\text{m}}=\frac{q}{n}$ (4)
EXAMPLE 1
How much heat energy is obtained when 1 kg of the serpentine chrysotile, Mg3Si2O5(OH)4, is formed according to Equation (2)?
Solution
The mass of Mg3Si2O5(OH)4 is easily converted to the amount of Mg3Si2O5(OH)4 from which the heat energy q is easily calculated by means of Eq. (4). The value of ΔHm is –179.7 kJ per 2 moles of Mg3Si2O5(OH)4. The road map is:
$m_{\text{Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{OH}_{4}}~$ $\xrightarrow{M}~$ $n_{\text{Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{OH}_{4}}~$ $\xrightarrow{\Delta H_{m}}~q$ so that $q=\text{1 }\times \text{ 10}^{\text{3}}\text{ g Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}~$ $\times~\frac{\text{1 mol Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}}{\text{277.112 g} \text{ Si}_{2}\text{O}_{5}\text{(OH)}_{4}}~$ $\times ~\frac{-179.7 kJ}{\text{2 mol Mg}_{2}\text{Si}_{2}\text{O}_{5}\text{(OH)}_{4}}$ = q = -324 kJ Note: By convention a negative value of q corresponds to a release of heat energy by the matter involved in the reaction. The quantity ΔHm is the enthalpy change as the reaction proceeds as written. In this context the symbol Δ (delta) signifies change in” while H is the symbol for the quantity being changed, namely the enthalpy. We will deal with the enthalpy in some detail in Chap. 15. For the moment we can think of it as a property of matter which increases when matter absorbs energy and decreases when matter releases energy.
It is important to realize that the value of ΔHm given in thermochemical equations like (1), (2) or (3) depends on the physical state of both the reactants and the products. Thus, if water were present as a gas instead of a liquid in the reaction in Eq. (1), the value of ΔHm would be different from 40.1 kJ. These reactions may occur under conditions where water may be supercritical (above 374°C) and yet a different value would be obtained. It is also necessary to specify both the temperature and pressure since the value of ΔHm depends very slightly on these variables. If these are not specified [as in Eq. (3)] they usually refer to 25°C and to normal atmospheric pressure. Since geochemical processes like those above normally occur at hundreds of atmospheres (hundreds of bars) pressure and elevated temperatures, geologists adjust the standard enthalpies to give values appropriate for the conditions. Although the adjustments are not difficult, computer programs exist to do the work.
Forward and Reverse Thermochemical Equations
Another characteristic of thermochemical equations arise from the law of conservation of energy. The first is that writing an equation in the reverse direction changes the sign of the enthalpy change.
Calcite/Aragonite
For example[13], the conversion of the two forms of calcium carbonate
CaCO3 ⇔ CaCO3 ΔHm = -0.17 kJ (5)
calcite aragonite
Therefore, the forward direction (calcite to aragonite) is exothermic, releasing heat as a more stable crystal lattice forms. Logically, in the reverse direction, disrupting the stable lattice of aragonite must require energy, so the conversion of aragonite to calcite endothermic:
CaCO3 ⇔ CaCO3 ΔHm = +0.17 kJ (6)
aragonite calcite
Water/Ice
Melting or freezing water can release or absorb significant amounts of heat:
H2O(l) → H2O(g) ΔHm = 44 kJ (7)
tells us that when a mole of liquid water vaporizes, 44 kJ of heat is absorbed. This corresponds to the fact that heat is absorbed from your skin when perspiration evaporates, and you cool off. Condensation of 1 mol of water vapor, on the other hand, gives off exactly the same quantity of heat.
H2O(g) → H2O(l) ΔHm = –44 kJ (8)
To see why this must be true, suppose that ΔHm [Eq. (7)] = 44 kJ mol while ΔHm [Eq. (8)] = –50.0 kJ. If we took 1 mol of liquid water and allowed it to evaporate, 44 kJ would be absorbed. We could then condense the water vapor, and 50.0 kJ would be given off. We could again have 1 mol of liquid water at 25°C but we would also have 6 kJ of heat which had been created from nowhere! This would violate the law of conservation of energy. The only way the problem can he avoided is for ΔHm of the reverse reaction to be equal in magnitude but opposite in sign from ΔHm of the forward reaction. That is, ΔHm forward = –ΔHm reverse | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.08%3A_Thermochemical_Equations/3.8.04%3A_Geology-_Heat_Engine_at_Lost_City.txt |
Perhaps the most useful feature of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is reaction of 1 mol C(s) and 0.5 mol O2(g) to form 1 mol CO(g):
$\text{C} (s) + \frac{1}{2} \text{O}_{2} (g) \rightarrow \text{C} \text{O} (g) ~~~~~~~~~~~~~~~~~~~~ \Delta H_{m} = –110.5 \text{kJ} = \Delta H_{1} \nonumber$
(Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) In step 2 the mole of CO reacts with an additional 0.5 mol O2 yielding 1 mol CO2: $\text{C} \text{O} (g) + \frac{1}{2} \text{O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) ~~~~~~~~~~~~~~~~~~~~~~~~~ \Delta H_{m} = –283.0 \text{ kJ} = \Delta H_{2} \nonumber$
The net result of this two-step process is production of 1 mol CO2 from the original 1 mol C and 1 mol O2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2.
Using the molecular representations of each reaction pictured above, confirm this conclusion, using the molecules to visually represent what is occurring.
On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction
$\text{C} (s) + \small\frac{1}{2}\normalsize\text{O}_2 (g) \qquad\quad\quad \space\space\space\rightarrow \cancel{\text{CO} (g)} \qquad \qquad \nonumber$
$\underline{\qquad\quad\space \small\frac{1}{2} \normalsize\text{O}_2 (g) + \cancel{\text{CO} (g)} \rightarrow \qquad \qquad \space\space \text{CO}_2 (g)} \nonumber$
$\text{C}(s) + \space\space\space \text{O}_2 (g) + \cancel{\text{CO} (g)} \rightarrow \cancel{\text{CO} (g)} + \text{CO}_2 (g) \nonumber$
Experimentally it is found that the enthalpy change for the net reaction is the sum of the enthalpy changes for steps 1 and 2: $\Delta H_{net} = –110.5 \text{ kJ} + (–283.0 \text{ kJ} ) = –393.5 \text{ kJ} = \Delta H_{1} + \Delta H_{2} \nonumber$ That is, the thermochemical equation $\text{C} (s) + \text{ O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) \ \Delta H_{m} = –393.5 \text{ kJ} \nonumber$
Notice how the equation above represents the reaction symbolically, while the 3D molecules show the microscopic view, and the final images show this process as we see it, on the macroscopic level.
is the correct one for the overall reaction.
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Example $1$ : Enthalpy Change
Acetylene (C2H2) cannot be prepared directly from its elements according to the equation
$2 \text{C} (s) + \text{H}_{2} (g) \rightarrow \text{C}_{2} \text{H}_{2} (g) \tag{1}$ Calculate ΔHm for this reaction from the following thermochemical equations, all of which can be determined experimentally:
\begin{align} \text{C} (s) &+ \space\space\space\text{O}_{2} (g) \rightarrow \text{C} \text{O}_{2} (g) &\Delta H_{m} &= –393.5 \text{kJ} \tag{2a} \ \text{H}_{2} (g) &+ \tfrac{1}{2} \text{O}_{2} (g) \rightarrow \text{H}_{2} \text{O} (l) &\Delta H_{m} &= –285.8 \text{kJ} \tag{2b} \ \text{C}_{2} \text{H}_{2} (g) &+ \tfrac{\text{5}}{\text{2}} \text{O}_{2} (g) \rightarrow 2 \text{C} \text{O}_{2} (g) + \text{H}_{2} \text{O} (l) &\Delta H_{m} &= –1299.8 \text{kJ} \tag{2c} \end{align}
Solution: We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):
a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2a) by 2.
b) Since Eq. (1) has 1 mol H2 on the left, we leave Eq. (2b) unchanged.
c) Since Eq. (1) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of Eq. (2c) we write Eq. (2c) in reverse.
We then have
\begin{align} \text{2 C}(s) &+ 2\space\ce{O2}(g) \space\space ~\to~ \ce{2CO2}(g) &\Delta H_{m}&=\text{ 2 (}-\text{393.5) kJ} \ \ce{H2}(g) &+ \tfrac{1}{2}\space\ce{O2}(g) \space ~\to~ \ce{H2O}(l) &\Delta H_{m} &= -\text{285.8 kJ} \ \underline{ \ce{2 CO2}(g)} &\underline{\text{ +} \space\space\space\ce{H2O}(l) ~\to~ \ce{C2H2}(g) + \tfrac{5}{2}\ce{O2}(g)} &\Delta H_{m}&=-\text{(}-\text{1299.8 kJ)} \ \ce{2 C}(s) + \ce{H2}(g) &+ 2\tfrac{1}{2}\ce{O2}(g) ~\to~ \ce{C2H2}(g) +\tfrac{5}{2}\ce{O2}(g) \end{align} \nonumber
\begin{align} & \Delta H_{m} = (-787.0 -285.8 + 1299.8) \text{ kJ} \ & = 227.0 \text{ kJ}\end{align} \nonumber
Cancelling 5/2 O2 on each side, the desired result is
$2 \text{C} (s) + \text{H}_{2} (g) \rightarrow \text{C}_{2} \text{H}_{2} (g) ~~~~~~~~~~~~~~~~~~~~~ \Delta H_{m} = 227.0 \text{ kJ} \nonumber$
3.09: Hess' Law
Pasteur showed that sugar fermentation in the presence of oxygen (aerobic fermentation) leads to a maximum rate of yeast growth, but minimum alcohol production. Excluding air (so the process continues anaerobically) slows yeast growth, but increases alcohol production. Anaerobic processes usually produce less energy than aerobic ones. We notice the same thing in our muscles when all the blood oxygen is used up, and lactic acid is produced (see below).
Fermentation product
Let's see how thermochemical equations help explain the "Pasteur Effect" and energy associated with lactic acid buildup. In case you're wondering, the heat energy values below can be obtained from the National Institute of Standars and Technology website NIST, from the home page use a name search for "ethanol".
Perhaps the most useful feature of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions. Consider, for example, the following two-step sequence. Step 1 is anaerobic fermentation of glucose, C6H12O6, to make 2 mol of ethanol, C2H5OH and 2 mol CO2(g):
$\ce{C6H12O6(l) → 2 C2H5OH(l) + 2 CO2(g) ΔHm = –74.4 kJ = ΔH1} \nonumber$
Note that a small amount of energy is produced anaerobically. \
If oxygen becomes available, the C2H5OH reacts with 6 mol O2 yielding 4 mol CO2:
$\ce{2 C2H5OH + 6 O2(g) → 4 CO2(g) + 6H2O(l) ΔHm = –2734 kJ = ΔH2} \nonumber$
The net result of this two-step process is production of 6 mol CO2 from the original 1 mol C6H12O6 and 6 mol O2. All the ethanol produced in step 1 is used up in step 2. The overall effect is the same as the aerobic fermentation of glucose:
$\ce{C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O (25o, 1 Atm) ΔHm = –2808 kJ} \nonumber$
Now we see an explanation for the "Pasteur Effect". If yeast grows in air, it can produce 2808 kJ/mol sugar, just like we do. That's energy that can be used to synthesize compounds and grow (we could use it to move around, but yeast can't do that!). If yeast ferments the sugar anaerobically, it can only produce 74 kJ/mol sugar, so growth is retarded, but it produces a lot of alcohol! The difference is the energy that comes from the aerobic metabolism of alcohol, producing 2734 kJ (for 2 mol ethanol).
On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The ethanol produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction
$\ce{C6H12O6(l) → 2 C2H5OH(l) + 2 CO2(g) ΔHm = –74.4 kJ = ΔH1} \nonumber$
2 C2H5OH + 6 O2(g) → 4 CO2(g) + 6H2O ΔHm = –2734 kJ = ΔH2
$^{\underline{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }}$
$\ce{C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O (25o, 1 Atm) ΔHm = –2808 kJ} \nonumber$
Experimentally it is found that the enthalpy change for the net reaction is the sum of the enthalpy changes for steps 1 and 2:
ΔHnet = –74.4 kJ + (–2734 kJ) + –2808 kJ = ΔH1 + ΔH2} \nonumber \]
That is, the thermochemical equation
$\ce{C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O ΔHm = –2808 kJ} \nonumber$
is the correct one for the overall reaction. Note that this is the same equation, and same heat of reaction, that we used [Weight of Food and Energy Production |before].
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Example $1$:
During exercise, glucose is first metabolized according to equation (1) above because there is plenty of oxygenated blood around muscle tissue. But as the oxygen is depleted, glucose is metabolized anaerobically to lactate ion to produce energy (we'll simplify by showing it as solid lactic acid):
Lactic acid
When blood once more supplies oxygen to the muscle, the lactic acid is metabolized according to the equation below:
$\ce{C3H6O3(s) + 3 O2(g) → 3 CO2(g) + 3 H2O (l)ΔHm = –1344 kJ (1)} \nonumber$
Use ΔHm for this reaction,and for the aerobic metabolism of glucose:
$\ce{C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O ΔHm = –2808 kJ (2) } \nonumber$
to calculate ΔHm for the reaction,
$\ce{2 C3H6O3(s) → C6H12O6(s) ΔHm = ? kJ } \label{3}$
Solution
We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):
a) Since Eq. (3) has 2 mol C3H6O3(s) on the left, we multiply Eq. (1) by 2. We also double the heat energy produced.
b) Since Eq. (3) no H2O or CO2, we need to cancel these molecules. Since Eq. (1) has 6 mol CO2 and 6 mol H2O on the right, whereas there also CO2 and H2O on the right of Eq. (2), we write Eq. (2) in reverse so they'll cancel. We also change the sign on the heat energy, indicating that it is absorbed rather than released.
c) Reversing (2) also puts C6H12O6(s) on the right in (2'), where it appears in (3), so we can combine the equations, cancelling molecules that appear on both sides.
We then have
a. $\ce{2 C3H6O3(s) + 6 O2(g) → 6 CO2(g) + 6 H2O (la) ΔHm = 2 x –1344 kJ = -2688 kJ (1')} \nonumber$
b. $\ce{6 CO2(g) + 6 H2O → C6H12O6(s) + 6 O2(g) ΔHm = +2808 kJ (2')} \nonumber$
c. $\ce{2 C3H6O3(s)→ C6H12O6(s) ΔHm = +2808 kJ - 2688kJ = +120 kJ (3)} \nonumber$
The result (3) is interesting; in reverse, it's C6H12O6(s) → 2 C3H6O3(s) ΔHm = -120 kJ (3)
which is the amount of energy from anaerobic conversion of glucose to lactic acid. We can see that it isn't much, but the reaction proceeds rapidly again and again, producing significant amounts of energy. The alternative would be no energy at all in the absence of oxygen.
Since lactic acid is normally metabolized as soon as oxygen is again available, it isn't the cause of "day after" muscle ache, as people erroneously say. It's actually the cause of burning muscles that you feel during exercise. But even then, it's an indirect effect; the lactic acid itself doesn't cause the burning, but causes formation of a flood of ATP, which hydrolyzes to give the acid which causes the |pain | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.09%3A_Hess%27_Law/3.9.01%3A_Biology-_Anaerobic_Fermentation_in_Beer_and_Lactic_Acid_in_Muscles.txt |
We noted previously that chemist's "standard enthalpy (heat) of combustion" may be a misleading measure of a fuel's heating value in real conditions. Standard enthalpies are precisely defined as the heat energy absorbed or released when a process occurs at 25 oC to give substances in their most stable state at that temperature. For example, for propane, water is produced as a liquid:
C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (l) ΔHm = –2219.2 kJ[1] (1)
Most of the plume above smokestacks is condensed water droplets, which release heat of condensation to the atmosphere
But boilers are almost always operated with flue temperatures near the combustion temperature, and maintained at over 100 oC to prevent water from condensing to liquid. Since water is produced as a gas, the standard thermodynamic equation (1), does not apply. If it were formed, liquid water would dissolve acidic flue gases like HCl and SO2 to make acidic solutions which corrode the system. These gases are normally removed by limestone "scrubbers",[2] leaving just the water vapor. The white plumes over smokestacks are water droplets forming by condensation of water vapor. They often disappear as the droplets evaporate.
The precision of our definition of standard enthalpy guarantees that our calculations of heat energy will be sound, but it requires that we pay attention to each energy cost and source. We'll see that inattention to such details has led to misinformation and confusion. Since Higher Heating Value (HHV) may be used by some air quality management authorities,[3] while Lower Heating Values are used by many engineers and European power facilities, it is important to know precisely what the terms mean and which is being used. The USDA[4] reports the following heating costs apparently, but not explicitly, stated to be LHVs:
USDA[4] reports the following heating costs
Fuel
Gross
heating value
Efficiency
(%)
Net
heating valueb
Fuel required
for 1 million Btu of usable heat
Average
cost/unit
Total
annual fuel costa
Natural gas 1.03 million Btu/1000 ft3 80 0.82 million Btu/1000 ft3 1,220 ft3 $7/1000 ft3$854
Propane 91,200 Btu/gal 79 72,000 Btu/gal 13.86 gal $1.25/gal$1,730
Fuel oil #2 138,800 Btu/gal 83 115,000 Btu/gal 8.68 gal $1.40/gal$1,220
Seasoned firewood 20 million Btu/cord 77 15.4 million Btu/cord 0.065 cord $115/cord$747
Electricity 3,413 Btu/kWh 98 3,340 Btu/kWh 299 kWh $0.08/kWh$2,390
Premium wood pellets 16.4 million Btu/ton 83 13.6 million Btu/ton 0.073 ton $120/ton$882
a Based on 100 million Btu of energy for the heating season, a typical value for an average sized house.
There are many variables that affect total heating costs (furnace efficiencies, local energy costs, etc.), but one variable that can potentially be understood is the heat available from fuel combustion. Without precise definitions, even this may be obscured. To illustrate this point, let's examine what are known as the Higher Heating Value (HHV)[5] and Lower Heating Value (LHV)[6] are calculated for propane.
The HHV
The standard heat of combustion of propane is
C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (l) ΔHm = -2219.2 kJ[7] = HHV = ΔH1 (1)
This is the process which yields the HHV, because the fuel is burned and the combustion products are cooled to 25 oC, removing all heat resulting from the condensation of water in the process.
The LHV
But we want the enthalpy for the reaction:
C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (g) ΔHm = LHV = ΔH2 (2)
In this case, the water is not condensed, so some of the energy is not recovered, so the ΔHm is the Lower Heating Value. We can imagine reaction (2) occurring in two steps. First, reaction (1), then
H2O(l) → H2O(g) ΔHm = 44.0 kJ (3)
We notice that equation (1) produces 4 mol of H2O (l), so we multiply equation (3) by 4, so that 4 moles of H2O (l) are consumed in equation (3a):
4 H2O(l) → 4 H2O(g) ΔHm = 4 x 44.0 kJ = 176.0 kJ (3a)
If we add equations (1) and (3a) as below, canceling the 4 mol H2O (l) that are produced (appear on the right) with the 4 mol H2O (l) which are consumed (appear on the left), we get equation (2).
C3H8(g) + 5 O2(g) → 3 CO2(g) + 4 H2O (l) ΔHm = –2219.2 kJ (1)
4 H2O(l) → 4 H2O(g) ΔHm = 4 mol x 44.0 kJ mol–1 = 176.0 kJ (3a)
————————————————————————————————————————
C3H8(g) + 5 O2(g) → 3 CO2(g)+ 4 H2O (g) ΔHm = LHV = ΔH2=–2043.2 kJ (2)
Experimentally it is found that the enthalpy change for reaction (2) is the sum of the enthalpy changes for reactions (1) and (3a):
ΔH2 = –2219.2 kJ + (4 x 44.0 kJ) = –2043.2 kJ = ΔH1 + 4 x ΔH3
This value of ΔHm should be the Lower Heating Value. Let's see if it matches the USDA value of 91,200 BTU/gal in the table above, given that a gallon of propane is about 4.23 lb[8]
$\frac{\text{2043.2.2 kJ}}{mol} ~\times~\frac{\text{1 mol}}{\text{44.1 g}} ~\times~\frac{\text{1 BTU}}{\text{1.055 kJ}} ~\times~\frac{\text{453.6 g}}{\text{lb}} ~\times~\frac{\text{4.23 lb}}{\text{gal}} ~=~\text{84 300 BTU/gal}$
Our value is not close to the USDA LHV, but it matches the Oak Ridge National Laboratory (ORNL) value for the HHV of 84 250 BTU/gal[9].
The Higher Heating Value is the enthalpy change for reaction (1), which includes the heat released when 4 mol of gaseous water from the combustion cool to 25 oC, so its value is more negative than the LHV by four times the heat of condensation of water (–2043.2 + 4 x (–-44) = –2219.2 kJ):
H2O(g) → H2O(l) ΔHm = –44 kJ (4)
Repeating the calculation for the heat liberated in equation (1) in BTU/gal, we get 91 500 BTU/gal, which is the HHV reported by ORNL. The USDA table above apparently reports the HHV, perhaps unknowingly. The variety of values found on the web for both HHV and LHV attests to the fact that much confusion can result from not being careful with the meaning and application of a standard enthalpy.[10] Technically, the lower heating value of a fuel is defined as the amount of heat released in the combustion of the fuel to give products at 150°C.[11]
Hess' Law
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Example 1
Acetylene (C2H2) cannot be prepared directly from its elements according to the equation
2C(s) + H2(g) → C2H2(g) (1) Calculate ΔHm for this reaction from the following thermochemical equations, all of which can be determined experimentally: C(s) + O2(g) → CO2(g) ΔHm = –393.5 kJ (2a) H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ (2b) C2H2(g) + $\tfrac{\text{5}}{\text{2}}$O2(g) → 2CO2(g) + H2O(l) ΔHm = –1299.8 kJ (2c)
Solution
We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (1):
a) Since Eq. (1) has 2 mol C on the left, we multiply Eq. (2a) by 2.
b) Since Eq. (1) has 1 mol H2 on the left, we leave Eq. (2b) unchanged.
c) Since Eq. (1) has 1 mol C2H2 on the right, whereas there is 1 mol C2H2 on the left of Eq. (2c) we write Eq. (2c) in reverse.
We then have \begin{align} \text{2 C(s) + O}_{\text{2}}\text{(g)}&\to \text{CO}_{\text{2}}\text{(g)}~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{ 2 (}\text{-393.5 kJ)} \ \text{H}_{\text{2}}\text{(g) + }\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}&\to \text{H}_{\text{2}}\text{O}(l)~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{-285.8 kJ} \ \underline{\text{2 CO}_{\text{2}}\text{(g) + H}_{\text{2}}\text{O}(l)}&\underline{\to \text{ C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)}}~~\Delta H_{\text{m}}=-\text{(}-\text{1299}\text{.8 kJ)} \ \text{2 C(s) + H}_{\text{2}}\text{(g) + 2}\tfrac{1}{2}\text{O}_{\text{2}}\text{(g)}&\to \text{C}_{\text{2}}\text{H}_{\text{2}}\text{(g) +}\tfrac{5}{2}\text{O}_{\text{2}}\text{(g)} \ \Delta H_{\text{m}}=\text{(}-\text{787}\text{.0}&-\text{285}\text{.8 + 1299}\text{.8) kJ}=\text{227}\text{.0 kJ} \ \end{align} Thus the desired result is 2C(s) + H2(g) → C2H2(g) ΔHm = 227.0 kJ | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.09%3A_Hess%27_Law/3.9.02%3A_Environment-_Heating_Values_of_Various_Fuels.txt |
One of the most useful features of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions which have never been observed. We might want to see what ΔHm would be if we could carry out a reaction that has never been done, or it might be interesting for theoretical reasons.
For example, we have noted that the body would have to store up to 67.5 lb of sugar complexes for the energy equivalent of 10 lb of fat.[1]
In Example 1 below, we calculate the energy for the hypothetical reaction in which a fat is converted to sugar:
C18H36O2 (s) + 8 O2(g) → 3 C6H12O6 (s)
ΔHmfor this reaction is the extra energy our body can get from a fat.
A Simple Case
But first, a simpler example may help to make the method clear. Instead of the oxidation of a complicated fat molecule, we'll consider the simplest possible oxidation, a sequence in which carbon itself is oxidized. Step 1 is the oxidation of 1 mol C(s) and 0.5 mol O2(g) to form 1 mol CO(g):
C(s) + ½O2(g) → CO(g) ΔHm = –110.5 kJ = ΔH1 (Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.)
In step 2 the some of the mole of CO reacts with an additional 0.5 mol O2 yielding 1 mol CO2:
CO(g) + ½O2(g) → CO2(g) ΔHm = –283.0 kJ = ΔH2 The net result of this two-step process is production of 1 mol CO2 from the original 1 mol C and 1 mol O2 (0.5 mol in each step). All the CO produced in step 1 is used up in step 2.
On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The CO produced is canceled by the CO consumed since it is both a reactant and a product of the overall reaction
Experimentally it is found that the enthalpy change for the net reaction is the sum of the enthalpy changes for steps 1 and 2: ΔHnet = –110.5 kJ + (–283.0 kJ ) = –393.5 kJ = ΔH1 + ΔH2 That is, the thermochemical equation C(s) + O2(g) → CO2(g) ΔHm = –393.5 kJ is the correct one for the overall reaction.
Hess' Law
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Example 1: Fat vs. Sugar Metabolism
EXAMPLE 1
Although fat metabolism is a complicated process (called "beta oxidation") which yields the ATP that releases energy to muscle, we could imagine a reaction that helps us understand why fats store so much energy compared to sugar. We could imagin the combustion of steric acid to the sugar, glucose, according to the equation
(1) C18H36O2 (s) + 8 O2(g) → 3 C6H12O6 (s) ΔHm1
This would represent the "extra" energy that fats provide, over the energy that metabolism of a sugar like glucose provides.
Calculate ΔHm1 for this reaction from the following thermochemical equations, (which are heats of combustion that are easily determined experimentally):
(2) C18H36O2 (s) + 26 O2(g) → 18 CO2(g) + 18 H2O(l) (25°C, 1 atm pressure)
ΔHm2 = –11 407 kJ [2]
(3) C6H12O6 (s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l) (25°C, 1 atm pressure)
ΔHm3 = –2 800 kJ [3]
Solution
We see that reaction (3) has glucose (C6H12O62) on the left, but the target reaction (1) has it on the right. We'll need to reverse equation (3), and then combine it with equation (2) to get the target equation (1). If we reverse (3), we change the sign on ΔHm3:
(3a) 6 CO2(g) + 6 H2O(l)→ C6H12O6 (s) + 6 O2(g) (-) ΔHm3 = +2 800 kJ
But we also might notice that the target equation contains no CO2(g) or H2O(l), so we'll need to multiply equation (3) by 3, so that there will be an equal amount of CO2(g) or H2O(l) on the left and right, and they will cancel. Multiplying equation (3a) by 3:
(3b) 18 CO2(g) + 18 H2O(l)→ 3 C6H12O6 (s) + 18 O2(g) (-3)ΔHm2 = +8 400 kJ
When we combine this equation, and its associated ΔHm with Equation (2), we get the target reaction, (1):
$\text{C}_{18}\text{H}_{36}\text{O}_{2}\text{(s)} + \text{26 O}_{\text{2}}\text{(g)}\to$$\text{18 CO}_{\text{2}}\text{(g)}~+~\text{18 H}_{2}\text{O}~~~~~~~~~~~~~~~~\Delta H_{\text{m}}=\text{11 407 kJ }$
$\underline{\text{18 CO}_{\text{2}}\text{(g) + 18 H}_{\text{2}}\text{O}(l)}\underline{\to \text{3 C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{6}\text{(s) +}\text{18 O}_{\text{2}}\text{(g)}}~~~~~~~~\Delta H_{\text{m}}=-3\text{(}-\text{2 800 kJ)}$
$\text{C}_{18}\text{H}_{36}\text{O}_{2}\text{(s)} + \text{8 O}_{\text{2}}\text{(g)}\to \text{3 C}_{\text{6}}\text{H}_{\text{12}}\text{O}_{6}\text{(g)}$ $\Delta H_{\text{m}}=\text{(}\text{-11 407}+\text{8 400 kJ}\text{)}~=~\text{-3007 kJ}$
ΔHm1 = ΔHm2 + (-3)ΔHm3 = –11 407 + 8400 kJ
ΔHm1 = –3 007 kJ
So one mole (284.48 g) of stearic acid releases 3 007 kJ when it's oxidized to 3 mol of glucose. This is 10.57 kJ/g, or 2.5 Cal/g that we get from fat but not from sugar.
Additionally, for every gram of stearic acid, we get the energy from 1.900 g of glucose (see the stoichiometry summary table below), which provides 4 Cal per gram. This is 1.90 g x 4 Cal/g = 7.59 Cal, so the total energy from 1 g of fat is 2.5 + 7.6 = 10.1 Cal in this case (similar to the 9 Cal/g estimate for typical fats).
Solution to Example 1
C18H36O2 + 8 O2 → 3 C6H12O6
m (g) 1.00 0.900 1.90
M (g/mol) 284.4 32.0 180
n (mol) 0.00352 0.281 0.0106 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.09%3A_Hess%27_Law/3.9.03%3A_Foods-_Fat_vs._Sugar_Metabolism.txt |
Perhaps the most useful feature of thermochemical equations is that they can be combined to determine ΔHm values for other chemical reactions. For example, iron forms several oxides, including iron(II) oxide or wüstite (FeO), iron(III) oxide or hematite (Fe2O3), and finally, iron(II,III) oxide or magnetite (FeO·Fe2O3 or Fe3O4). These oxides form by thermochemical reactions which depend on, and influence, their environment by producing or absorbing heat. Hematite exists in several phases (denoted &alpha--hematite;, β, &gamma--maghemite; and ε), and they are all different from ordinary rust, which is also often given the formula Fe2O3 [1]. Fe2O3 is the chief iron ore used in production of iron metal. FeO is nonstoichiometric. Magnetite is the most magnetic of all the naturally occurring minerals on Earth. Naturally magnetized pieces of magnetite, called lodestone, will attract small pieces of iron. We'll see evidence below that Fe3O4 is not simply a mixture of FeO and Fe2O3.
Iron(II) oxide, wüstite
Iron(III) oxide, hematite
Iron(II,III) oxide, Magnetite
Consider, for example, the following two-step sequence. Step 1 is reaction of 2 mol Fe(s) and 1 mol O2(g) to form 2 mol FeO(s):
(1) 2 Fe(s) + 1 O2(g) → 2 FeO(s) ΔHm = -544 kJ = ΔH1 In step 2 the 2 moles of FeO react with an additional 0.5 mol O2 yielding 1 mol Fe2O3: (2) 2 FeO(s) + ½O2(g) → Fe2O3(s) ΔHm = –280.2 kJ = ΔH2 (Note that since the equation refers to moles, not molecules, fractional coefficients are permissible.) The net result of this two-step process is production of 1 mol Fe2O3 from the original 2 mol Fe and 1.5 mol O2 (1 mol in the first step and 0.5 mol in the second step). All the FeO produced in step 1 is used up in step 2.
On paper this net result can be obtained by adding the two chemical equations as though they were algebraic equations. The FeO produced is canceled by the FeO consumed since it is both a reactant and a product of the overall reaction
2 Fe(s) + 1 O2(g) → 2 FeO(s) ΔHm = –-544 kJ
½O2(g) + 2 FeO(s) → Fe2O3(s) ΔHm = –280.2 kJ
2 Fe(s) + 1.5 O2(g) → 1 Fe2O3(s) (3) ΔHm
Experimentally it is found that the enthalpy change for the net reaction is the sum of the enthalpy changes for steps 1 and 2: ΔHnet = –544 kJ + (–280.2 kJ ) = = –824 kJ = ΔH1 + ΔH2 That is, the thermochemical equation (3) 2 Fe(s) + 1.5 O2(g) → 1 Fe2O3(s) ΔHm = –824 kJ is the correct one for the overall reaction.
In the general case it is always true that whenever two or more chemical equations can be added algebraically to give a net reaction, their enthalpy changes may also be added to give the enthalpy change of the net reaction.
This principle is known as Hess' law. If it were not true, it would be possible to think up a series of reactions in which energy would be created but which would end up with exactly the same substances we started with. This would contradict the law of conservation of energy. Hess’ law enables us to obtain ΔHm values for reactions which cannot be carried out experimentally, as the next example shows.
Example 1
Magnetite has been very important in understanding the conditions under which rocks form and evolve. Magnetite reacts with oxygen to produce hematite, and the mineral pair forms a buffer that can control the activity of oxygen. One way magnetite is formed is decomposition of FeO.
FeO is thermodynamically unstable below 575 °C, disproportionating to metal and Fe3O4[2].
(4) 4FeO → Fe + Fe3O4
The direct reaction of iron with oxygen does not occur in nature, because iron does not occur in the elemental form in the presence of oxygen, but we know the enthalpy of reaction from laboratory studies:
(5) 3 Fe(s) + 2 O2(g) → Fe3O4 ΔHm = –1118.4 kJ
Calculate the enthalpy change for Reaction (4) from the enthalpies of other reactions given on this page.
Solution
We use the following strategy to manipulate the three experimental equations so that when added they yield Eq. (4):
a) Since the target reaction (4) has FeO on the left, but the reaction (1) above with ΔHm1 has FeO on the right, we can reverse it, changing the sign on ΔHm1:
(1b) 2 FeO(s) → 2 Fe(s) + 1 O2(g) ΔHm = +544 kJ = - ΔH1
But the target reaction requires 4 mol of FeO on the left, so we need to multiply this reaction, and its associated enthalpy change, by 2:
(1c) 4 FeO(s) → 4 Fe(s) + 2 O2(g) ΔHm = +1088 kJ = -2 x ΔH1
b) Since the target equation has 1 mole of Fe3O4 on the right, as does equation (5) above, we can combine equation (5) with (1c):
(1c) 4 FeO(s) → 4 Fe(s) + 2 O2(g) ΔHm = +1088 kJ = -2xΔH1
(5) 3 Fe(s) + 2 O2(g) → Fe3O4 ΔHm = –1118.4 kJ
Combining the equations and canceling 2O2 on the left and right, and canceling 3 Fe on the left, leaving 1 Fe on the right, we get equation (4):
(4) 4FeO → Fe + Fe3O4
The enthalpy change will be the sum of the enthalpy changes for (1c) and (5):
ΔHm = -2ΔHm1Hm5
ΔHm = +1088 kJ + (-1118.4) = -30.4 kJ
Example 2
Fe3O4 is not simply a mixture of FeO and Fe2O3, but a novel structure. Prove this by using thermochemical equations on this page to calculate the enthalpy for reaction (6) below. If the enthalpy change is zero, no significant chemical change occurs.
(6) FeO(s) + Fe2O3 → Fe3O4(s) ΔHm
Solution: It appears that we could start with (5) which has Fe3O4 on the right, like the target equation:
(5) 3 Fe(s) + 2 O2(g) → Fe3O4 ΔHm = –1118.4 kJ
We can introduce the Fe2O3 needed on the left of the target equation by using the reverse of Equation (2), changing the sign on ΔHm :
(2b) Fe2O3 → 2 FeO(s) + ½O2(g) (s) ΔHm = -(–280.2) kJ mol–1 = - ΔH2
This will introduce 2 FeO on the right, and we want 3 FeO on the left in the target equation. There are also 3 Fe on the left of Equation (3) that need to be canceled. We can accomplish both by adding the reverse of Equation (1):
(1b) 2 FeO(s) → 2 Fe(s) + 1 O2(g) ΔHm = -(-544) kJ mol–1 = ΔH1
Since the target equation has 1 FeO on the left, we need to multiply (1b) by 3/2 or 1.5:
(1c) 3 FeO(s) → 3 Fe(s) + 1.5 O2(g) ΔHm = -3/2 x (-544) kJ
Combining (5), (2b), and (c) we get the target equation, and the ΔH is calculated by combining the corresponding ΔH1 values:
(6) FeO(s) + Fe2O3 → Fe3O4(s) ΔHm
ΔH1 = -1118 kJ + -(-280.2 kJ) + (-3/2)x(-544 kJ) = -22 kJ
Since this is a significantly exothermic change, it appears that a chemical change occurs when FeO and Fe2O3 combine to make Fe3O4. Significant enthalpy changes occur when solutions are prepared (the dangerous heating observed when water is added to sulfuric acid is a prime example), but these always indicate that bonds have been broken or formed.
3.9.05: Lecture Demonstration- Carbide Cannon
Enthalpy of reactions:
Assuming you want about a 1 kJ explosion, how much CaC2 would you add to a carbide cannon?
(1) 2 C2H2 + 5 O2 → 4 CO2 + 5 H2O ΔH ?
given
(2) C + O2 → CO2 ΔHf = -393.5 kJ/mol
(3) H2 + 1/2 O2 → H2O (g) Δ Hf = -241.82 kJ/mol
(4) 2 C + H2 → C2H2(g) ΔHf = +226.7
Calculated enthalpy change for combustion of 2 mol of acetylene, ΔH = -2511 kJ
This is 48.2 kJ/g
for 1 kJ, we need 1 kJ/48.2 kJ/g = 0.02 g of acetylene
CaC2 + 2 H2O → HCCH + Ca(OH)
this requires 0.0008 mol of CaC2, which is 0.05 g.
Do it with a Bangsite Cannon! | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.09%3A_Hess%27_Law/3.9.04%3A_Geology-_Iron_and_its_Ores.txt |
By now chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess' law makes it possible to list a single value, the standard enthalpy of formation ΔHf, for each compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).
For example, if we know that $\Delta H_{f} [ \text{H}_{2} \text{O} (l) ] = –285.8 \text{ kJ } \text{mol}^{-1} \nonumber$, we can immediately write the thermochemical equation
$\text{H}_{2} (g) + \frac{1}{2} \text{O}_{2} (g) \rightarrow \text{H}_{2} \text{O} (l) ~~~~~~~~~~~~ \Delta H_{m} = –285.8 \text{ kJ} \text{mol}^{-1} \label{1}$
The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed. Equation $\ref{1}$ must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½. Note the 3 different views of this reaction, the equation providing a symbolic view, the 3D molecules providing a microscopic view, and the photos providing a macroscopic view of the reaction as we would see it with our own 2 eyes.
In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine ΔHf. Quite often, however, elements do not react directly with each other to form the desired compound, and ΔHf must be calculated by combining the enthalpy changes for other reactions. A case in point is the gas acetylene, C2H2. In Example 1 from the Hess' law section we showed that the thermochemical equation
$2 \text{C} (s) + \text{H}_{2} (g) \rightarrow \text{C}_{2} \text{H}_{2} (g) ~~~~~~~~~ \Delta H_{m} = 227.0 \text{ kJ } \text{mol}^{-1} \nonumber$
was valid. Since it involves 1 mol C2H2 and the elements are in their most stable forms, we can say that ΔHf[C2H2(g)] = 227.0 kJ mol–1.
One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. If we form mercury from its elements, for example, we are talking about the reaction
$\text{Hg} (l) \rightarrow \text{Hg} (l) \nonumber$
Since the mercury is unchanged, there can be no enthalpy change, and $\Delta H_{f} = 0 \text{ kJ} \text{ mol}^{-1} \nonumber$.
Table $1$: 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
Standard enthalpies of formation for some common compounds are given in Table $1$. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.
Example $1$: Change in Enthalpy
Use standard enthalpies of formation to calculate ΔHm for the reaction
$2 \text{CO} (g) + \text{O}_{2} (g) \rightarrow 2 \text{C} \text{O}_{2} (g) \nonumber$
Solution:
We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step CO (carbon monoxide) is decomposed to its elements:
$2 \text{CO} (g) \rightarrow 2 \text{C} (s) + \text{O}_{2} (g) ~~~~~~~~~~ \Delta H_{m} = \Delta H_{1}\label{2}$
Since this is the reverse of formation of 2 mol CO from its elements, the enthalpy change is
$\Delta H_{1} = 2 \times { – \Delta H_{f} [ \text{CO} (g) ] } = 2 \times [ – ( – 110.5 \text{ kJ} \text{ mol}^{-1} ) ] = +221.0 \text{ kJ} \text{ mol}^{-1} \nonumber$
In the second step the elements are combined to give 2 mol CO2(carbon dioxide):
$2 \text{C} (s) + 2 \text{O}_{2} (g) \rightarrow 2 \text{C} \text{O}_{2} (g) ~~~~~~~~~ \Delta H_{m} = \Delta H_{2} \label{3}$
In this case
$\Delta H_{2} = 2 \times \Delta H_{f} [ \text{C} \text{O}_{2} (g)] = 2 \times (– 393.5 \text{ kJ } \text{mol}^{-1}) = – 787.0 \text{ kJ } \text{mol}^{-1} \nonumber$
You can easily verify that the sum of Equations $\ref{2}$ and $\ref{3}$ is
$2 \text{CO} (g) + 2 \text{O}_{2} (g) \rightarrow 2 \text{C} \text{O}_{2} (g) ~~~~~~~~~~~ \Delta H_{m} = \Delta H_{net} \nonumber$
Therefore
$\Delta H_{net} = \Delta H_{1} + \Delta H_{2} = 221.0 \text{ kJ } \text{mol}^{-1} – 787.0 \text{mol}^{-1} = – 566.0 \text{mol}^{-1} \nonumber$
Note carefully how Example $1$ was solved. In step 1 the reactant compound CO(g) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH1 was opposite in sign from ΔHf. Step 1 also involved 2 mol CO(g) and so the enthalpy change had to be doubled. In step 2 we had the hypothetical formation of the product CO2(g) from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation
$\Delta H_{m} = \sum \Delta H_{f} \text{ (products)} – \sum \Delta H_{f} \text{ (reactants)} \label{4}$
The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation for which ΔHm is being calculated.
Example $2$: Enthalpy Change
Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction
$4 \text{N} \text{H}_{3} (g) + 5 \text{O}_{2} (g) \rightarrow 6 \text{H}_{2} \text{O} (g) + 4 \text{NO} (g) \nonumber$
Solution
Using Equation $\ref{4}$, we have
\begin{align} \Delta H_{m} & = \sum \Delta H_{f} \text{ (products)} – \sum \Delta H_{f} \text{ (reactants)} \ & = [ 6 \Delta H_{f} ( \text{H}_{2} \text{O} ) + 4 \Delta H_{f} \text{(NO)]} – [4 \Delta H_{f} \text{ (N} \text{H}_{3} ) + 5 \Delta H_{f} \text{ (O}_{2} )] \ & = 6 ( – 241.8) \text{ kJ} \text{ mol}^{-1} + 4 ( 90.3 ) \text{ kJ} \text{ mol}^{-1} – 4(–46.1 \text{ kJ} \text{ mol}^{-1} ) – 5 \times 0 \ & = –1450.8 \text{ kJ} \text{ mol}^{-1} + 361.2 \text{ kJ} \text{ mol}^{-1} + 184.4 \text{ kJ} \text{ mol}^{-1} \ & = –905.2 \text{ kJ} \text{ mol}^{-1} \end{align} \nonumber
Note that we were careful to use ΔHf [H2O(g)] not ΔHf [H2O(l)]. Even though water vapor is not the most stable form of water at 25°C, we can still use its ΔHf value. Also the standard enthalpy of formation of the element O2(g) is zero by definition. Obviously it would be a waste of space to include it in the table above.
3.10: Standard Enthalpies of Formation
We have looked at sugar as a source of [Weight of Food and Energy Production | energy production] in our bodies in the last few sections, but we also know that the mechanism by which chemical energy is delivered to muscles isn't the same as the mechanism for combustion of a sugar:
C6H12O6(s) + 6 O2(g) → 6 CO2(g) + 6 H2O(l)
(25o, 1 Atm) ΔHm = –2808 kJ mol–1 (1)
In our bodies, adenosine Triphosphate, ATP, and adenosine diphosphate, ADP, are central intermediaries in energy delivery.
ATP
ADP
The energy of sugar metabolism is stored in the form of accumulated ATP, and when energy is needed by muscle, it is delivered by the reaction [1]
ATP4- + H2O → ADP3- + HPO42- + H+ ΔHm = ~ –22 kJ mol–1[2]
which is shown in an animation here.
But it just doesn't make sense that this reaction should release energy, because it involves breaking a bond to remove HPO42-, and breaking a bond should require energy. Since so many biologists misinterpret this reaction, we'll try to explain where the heat energy that drives our muscles (and growth) comes from more clearly, in terms of standard enthalpies of formation, below.
It's clear that there is almost an infinite number of chemical reactions whose heat energy is important to know, and chemists have measured the enthalpy changes for so many reactions that it would take several large volumes to list all the thermochemical equations. Fortunately Hess' law makes it possible to list a single value, the standard enthalpy of formation ΔHf, for each compound. The standard enthalpy of formation is the enthalpy change when 1 mol of a pure substance is formed from its elements. Each element must be in the physical and chemical form which is most stable at normal atmospheric pressure and a specified temperature (usually 25°C).
For example, if we know that ΔHf[H2O(l)] = –285.8 kJ mol–1, we can immediately write the thermochemical equation
H2(g) + ½O2(g) → H2O(l) ΔHm = –285.8 kJ mol–1 (2) The elements H and O appear as diatomic molecules and in gaseous form because these are their most stable chemical and physical states. Note also that 285.8 kJ are given off per mole of H2O(l) formed, because stronger bonds are formed (2 H-O bonds), releasing a lot of energy, while weaker bonds (O-O and H-H) are broken, requiring less energy. Equation (1) must specify formation of 1 mol H2O(l), and so the coefficient of O2 must be ½.
The heat of formation of Cl atoms makes it clear that bond breaking is endothermic:
Cl2 → 2 Cl ΔHm = +121.68 kJ mol–1 (3)
In some cases, such as that of water, the elements will react directly to form a compound, and measurement of the heat absorbed serves to determine ΔHf. Quite often, however, elements do not react directly with each other to form the desired compound, and ΔHf must be calculated by combining the enthalpy changes for other reactions. This is the case for ATP
Some Standard Enthalpies of Formation at 25°C.
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
One further point arises from the definition of ΔHf. The standard enthalpy of formation for an element in its most stable state must be zero. If we form oxygen from its elements, for example, we are talking about the reaction
O2(g) → O2(g)
Since the oxygen is unchanged, there can be no enthalpy change, and ΔHf = 0 kJ mol–1.
Standard enthalpies of formation for some common compounds are given in the table above. These values may be used to calculate ΔHm for any chemical reaction so long as all the compounds involved appear in the tables. To see how and why this may be done, consider the following example.
EXAMPLE 1 Use standard enthalpies of formation below to estimate [3] ΔHm for the reaction
P2O74- + H2O → 2 HPO42- (1)
Note that this is like the decomposition (or "hydrolysis") of ATP to give ADP + HPO42-, but simpler. It still releases energy, even though a bond is ostensibly broken between ATP and HPO42-, requiring energy. What isn't seen is very important: The heats of formation of ions include "solvation energies", or in this case, energies released when water molecules bond to the HPO42-, releasing significant amounts of energy. The energy released by "hydration" is greater than the energy required to break the P-O-P bond in ATP.
chemical species ΔHf, kJ mol-1
ATP4- -2982
H2O (l) -287
ADP3- -2000
HPO42- -1299
P2O74- -2286
H+ O
Solution
We can imagine that the reaction takes place in two steps, each of which involves only a standard enthalpy of formation. In the first step the reactants are broken into their elements, and in the second step the elements are recombined to give the products.
First, P2O74- ("pyrophosphate ion") is decomposed to its elements:
P2O74-(aq) → 2P(s) + 7/2 O2(g) ΔHm = ΔH2 (2)
Since this is the reverse of formation of 1 mol P2O74- from its elements, the enthalpy change is ΔH2 = {–ΔHf [P2O74-(aq)]} = [– (–2286 kJ mol–1)] = +2286 kJ mol–1. It's positive (endothermic) because bond breaking takes energy.
Next, we break water into its elements:
H2O (l) → H2 + 1/2O2 (3)
Again ΔH3 = -(-287) = +287 kJ mol–1.
In the second step the elements are combined to give 2 mol HPO42-("inorganic monophosphate ion" or "hydrogen phosphate ion"): 2P(s) + H2 + 2O2(g) → 2 HPO42-(aq) ΔHm = ΔH4 (4) In this case ΔH4 = 2 × ΔHf [HPO42-(aq)] = 2 × (– 1299 kJ mol–1) = – -2598 kJ mol–1 You can easily verify that the sum of Eqs. (2) and (3) is P2O74- + H2O → 2 HPO42- ΔHm = ΔHnet Therefore ΔHnet = ΔH2 + ΔH3 + ΔH4= 287 kJ mol–1 +2286 kJ mol–1 – 2598 kJ mol–1 = – 25 kJ mol–1 Note carefully how Example 2 was solved. In step 1 the reactant compound P2O74- (aq) was hypothetically decomposed to its elements. This equation was the reverse of formation of the compound, and so ΔH2 was opposite in sign from ΔHf. In step 2 we had the hypothetical decomposition of the other reactant, water, again ΔH2 was opposite in sign from ΔHf. Finally, 2 moles of the product HPO42-(aq) were formed from its elements. Since 2 mol were obtained, the enthalpy change was doubled but its sign remained the same.
Any chemical reaction can be approached similarly. To calculate ΔHm we add all the ΔHf values for the products, multiplying each by the appropriate coefficient, as in step 2 above. Since the signs of ΔHf for the reactants had to be reversed in step 1, we subtract them, again multiplying by appropriate coefficients. This can he summarized by the equation
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants) (4) The symbol Σ means “the sum of.” Since ΔHf values are given per mole of compound, you must be sure to multiply each ΔHf by an appropriate coefficient derived from the equation for which ΔHm is being calculated.
EXAMPLE 2 Use the table of standard enthalpies of formation at 25°C to calculate ΔHm for the reaction
ATP4- + H2O → ADP3- + HPO42- + H+
Once more, remember that the energy released comes from the fact that bonding between water molecules and the HPO42- that is released releases more energy than it takes to break the P-O-P bond in ATP to give ADP + HPO42-.
Solution Using Eq. (4), we have
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants) = [ΔHf(ADP3-) + ΔHf (HPO42-)] – [ΔHf ((H2O)+ ΔHf (ATP4-)]
= (–2000) kJ mol–1 + (-1299) kJ mol–1 – (–2982 kJ mol–1) – (–287 kJ mol–1) = -30 kJ mol–1
Note that we were careful to use ΔHf [H2O(l)] not ΔHf [H2O(g)].
Even though water vapor is not the most stable form of water at 25°C, we can still use its ΔHf value to do an interesting calculation: Find the heat energy required to vaporize 1 mole of water (we know that should be positive. It takes energy to boil water because we're breaking bonds of attraction between water molecules. H2O(l) → H2O(g)] ΔHf = ?
ΔHm = ∑ ΔHf (products) – ∑ ΔHf (reactants)
= - 241.8 -(- 285.8) = - 241.8 + 285.8 = + 44 kJ mol–1 | textbooks/chem/General_Chemistry/ChemPRIME_(Moore_et_al.)/03%3A_Using_Chemical_Equations_in_Calculations/3.10%3A_Standard_Enthalpies_of_Formation/3.10.01%3A_Biology-_Muscle_Energy_from_ATP.txt |
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