Conservation Of Mass, Charge, And Energy
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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"
11. Conservation of Mass, Charge, and Energy
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Introduction: You Can't Get Something for Nothing...
The first ten chapters of this book have been mainly descriptive. They have portrayed the material universe as seen by an observer who has the ability to adjust his field of view to encompass entire galaxies or single atoms. At the lowest level we have seen how electrons can be arranged around nuclei in atoms, and how this arrangement limits the different kinds of atoms that can exist. At a slightly higher level of organization we have seen the way in which electrons hold groups of atoms together in molecules of definite size and shape, and how the properties of matter depend on molecular structure. This is the essence of chemistry: to explain matter in molecular terms. Like any other branch of science, chemistry eventually becomes trivial if it remains descriptive. The essence of science is control of matter by means of successful predictions of behavior; and predictions without measurement are hazy. Sooner or later we must adopt the viewpoint of William Thomson (Lord Kelvin), a pioneer in thermodynamics and electricity, who said in 1891: "When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind. It may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science."
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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"
11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
In the fusion reaction and should be understood as representing nuclei, with total mass numbers given by superscripts, and nuclear charge (equal to the number of protons) given by subscripts. An electron is represented as , with a zero mass number (not counted among the nucleons) and a -1 charge. In the fusion equation four hydrogen nuclei (not atoms) combine with two of the four electrons around the atoms to form a helium nucleus. The two remaining original electrons, not shown in the equation, associate with the nucleus to build a neutral helium atom. A proton in this representation is p, and a neutron is . The uranium fission reaction tells us that a uranium nucleus, when bombarded by a neutron, breaks down into barium and krypton nuclei with the release of three more neutrons. Therefore this is a chain reaction, with more neutrons released than were used up. These neutrons can bombard neighboring nuclei and produce even more fission. Mass number (superscripts) and charge (subscripts) are conserved in nuclear reactions of this type, just as the number of each kind of atom is conserved in a chemical reaction. You should verify that the sum of subscripts, and sum of superscripts, are constant on the two sides of the equation.
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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"
11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
In this region, beyond atomic number 26, energy is released by fission rather than fusion. At the far left of the curve, hydrogen fusion in stars releases energy:
and at the far right, uranium fission in atomic reactors also releases energy:
(This is only one of many ways in which the
235
U nucleus can break down.)
The maximum stability of the iron nucleus is the reason why the elementbuilding process by successive fusion reactions, outlined in Chapter 8, stops at iron. Beyond iron the fusion process is energy-requiring instead of energyyielding. The heavier elements are built up by more indirect processes involving neutron capture. In spite of the fact that mass is not conserved in nuclear reactions, conservation principles do apply to the total number of heavy particles (protons and neutrons) and total charge. This is implicit in the equations for the two previous nuclear reactions.
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The mass loss or binding energy per nuclear particle (protons and neutrons) rises rapidly to a maximum at iron, then falls. Iron is the most stable nucleus of all. The mass losses or binding energies per nucleon are plotted above for all nuclei from helium through uranium. After some initial minor irregularities in the first- and second-row elements, the values settle down to a smooth curve, which rises to a maximum at iron, then begins a long descending slope through uranium and beyond.
This curve gives us information that was used in the discussion of stellar synthesis of elements in Chapter 8. Iron is the most stable nucleus of all. For elements with smaller atomic numbers than iron, fusion of nuclei to produce heavier elements releases energy, because the products are lighter and more stable on a per-nucleon basis than the reactants. In contrast, beyond iron, fusion absorbs energy because the products are heavier on a pernucleon basis than the reactants.
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Mass and Energy: Nuclear Reactions
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Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"
11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
Example. What is the mass loss per nucleon for the component protons, neutrons, and electrons?
atom, compared with its
Solution. The atom contains 26 protons, 26 electrons, and 30 neutrons, so the mass calculation is performed as opposite: Notice that the mass loss per nucleon, and hence the binding energy per nucleon, is greater for iron than for helium. This means that the iron nucleus is more stable relative to protons and neutrons than the helium nucleus is. If some combination of helium nuclei could be induced to produce an iron nucleus, energy would be given off, which would correspond to the increased stability of the product nucleus per nuclear particle.
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Mass and Energy: Nuclear Reactions -1
(Compare this energy with the 83 kcal mole required to break carbon-carbon bonds in chemical reactions.) Protons and neutrons collectively are known as nuclear particles or nucleons. The number of nucleons in the nucleus is the sum of the number of protons and neutrons and, as was mentioned in Chapter 2, is known as the mass number of the nucleus. Every atomic nucleus is lighter than the sum of the masses of the nucleons from which it is built, and this mass loss corresponds to the binding energy of the nucleus. The relative stability of two nuclei with different numbers of nucleons can be assessed by comparing their mass loss per nucleon. Example: What is the mass loss per nucleon for the helium atom? Solution: The total mass loss is 0.0305 amu, and since the nucleus has four nucleons, the mass loss per nucleon is 0.00763 amu.
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11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
(We must include electrons in this calculation because 4.0026 amu is the mass of the helium-4 atom, not the nucleus.) This missing mass corresponds to 0.0305 x 931.4 MeV = 28.4 MeV of energy. If we could put together a helium atom directly from two neutrons, two protons, and two electrons, then 28.4 MeV of energy would be given off for every atom formed:
Compared to common chemical reactions, this is an enormous quantity of energy. Since 1 electron volt per atom is equivalent to 23.06 kcal per mole,
= 655,000,000 kcal mole
-1
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11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
With nuclear reactions, the energies involved are so great that the changes in mass become easily measurable. One no longer can assume that mass and energy are conserved separately, but must take into account their interconversion via Einstein's relationship, E = mc . If mass is in grams and the velocity of light is expressed as c = 3 x 10 cm sec , then the energy is in units of g cm sec , or ergs. A useful conversion is from mass in amu to energy in million electron volts (MeV): 1 amu = 931.4 MeV What holds a nucleus together? If we attempt to bring two protons and two neutrons together to form a helium nucleus, we might reasonably expect the positively charged protons to repel one another violently. Then what keeps them together in the nucleus? The answer, as we mentioned in Chapter 2, is that a helium atom is lighter than the sum of two protons, two neutrons, and two electrons. Some of the mass of the separated particles is converted into energy and dissipated when the nucleus is formed. Before the helium nucleus can be torn apart into its component particles, this dissipated energy must be restored and turned back into mass. Unless this energy is provided, the nucleus cannot be taken apart. This energy is termed the binding energy of the helium nucleus.
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Conservation of Electrons
Exercise. For practice, balance the following reactions by both the oxidation-number and half-reaction methods:
Although half-reactions have been introduced here merely as a means of obtaining a balanced overall reaction, they can have physical meaning of their own. If the iron oxidation and manganese reduction can be carried out in separate containers, and if these containers can be given suitable electrical connections, then we can make use of the energy released by this reaction as the electrons flow through a wire from the iron container to the manganese. This is the principle of the electrolytic cell or battery. An ordinary flashlight battery uses manganese reduction, and oxidation of the zinc battery casing. We will return to electrolytic cells and energy production in Chapter 17.
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Conservation of Electrons
The iron half-reaction is simple:
The final step is to add these two individually balanced half-reactions and obtain an overall reaction in which electrons do not appear explicitly. To accomplish this, we must add one unit of the manganese reaction to five units of the iron reaction. The result is the same as before:
This overall equation is now balanced with respect to charge and number of atoms because the half-reactions were balanced, and with respect to oxidation number because the proper multiples of the half-reactions were chosen to make the electrons cancel.
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Conservation of Electrons
There is another useful method for balancing redox equations: the half-reaction method. In this method each substance that changes its oxidation number is balanced separately in a half-reaction that includes electrons explicitly. The two balanced half-reactions then are combined in such a way as to cancel electrons from the final expression. Example. Balance the iron and permanganate reaction by the half-reaction method. Solution. The unbalanced permanganate half-reaction is
Five electrons were needed because manganese goes from +7 to +2 oxidation state. Water molecules are added on the right to balance the oxygen atoms in the permanganate, and protons are added to the left:
This half-reaction is now balanced with respect to both number of atoms and overall charge, +2 on each side.
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Conservation of Electrons
It is obvious that five iron atoms are required for every manganese atom, in order that the changes in oxidation numbers cancel:
Eight hydrogen ions are needed to use up the four oxygen atoms on the left, leading to four water molecules on the right:
In choosing a 1-to-5 ratio of Mn-to-Fe, we ensured that oxidation number was conserved. By adding the hydrogen ions we obtained an equation that balanced the number of each atom. As a final check, the net charge on each side of the equation can be tested, and found to be the same: +17. This oxidation-reduction equation now is balanced with respect to electrons, atoms, and charge.
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Conservation of Electrons
The corresponding equation for acidic solutions can be obtained by adding enough hydrogen ions to each side to eliminate the hydroxide ions:
The tetrahedral permanganate ion Example. Potassium permanganate, , is a common inorganic oxidizing agent, which becomes reduced to manganous ion, Mn , in acidic solution. Write a balanced equation for the reaction in which permanganate oxidizes ferrous iron, 2+
3+
Fe , to ferric ion, Fe . Solution.
In the permanganate ion manganese has an oxidation number of +7 which decreases to +2 in the manganous ion, a change of -5. Iron increases in oxidation number by one:
Formula MnO4-
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Conservation of Electrons
For electrons to balance properly, three moles of H must be oxidised for every mole of NO3- reduced: 3 (+1)
+
(-3)
=
The planar, delocallised nitrate ion
0
and the reaction can be written
In any reaction in aqueous solution, one is at liberty to assume as many H O, +
-
+
2
H , or OH as are required to balance the reaction, using H O and H under 2 acidic conditions, or and OH if the solution is basic. The redox (oxidation reduction) part of the balancing has been done, and what remains is only an accounting for O and H atoms. One possible answer is
This has accounted for the three hydrogen atoms and the negative charge. The
Formula NO3-
-
equation is correct for basic solutions where OH ions are present.
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Conservation of Electrons
How can we write a properly balanced equation for the oxidation of foods with NO as the waste product?
It might be tempting to balance by inspection along the following lines, making the number of each kind of atom the same on left and right:
But this leaves an electron on the left side unaccounted for on the right side. The approach that is most nearly foolproof and at the same time shows you what is happening is the oxidation-number method. In nitrates, N has an oxidation number of +5 ( verify this for yourself ). In NO the oxidation number of N is +2, a decrease of three. Hydrogen, as before, goes from 0 to +1, an increase of one:
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Conservation of Electrons
These oxidation numbers (ON) of H and O in H O arise because O is more 2
electronegative than H, so both the electrons in each O H bond are assigned to O. If you are unsure of this process, look back at Chapter 6. This is such a simple chemical reaction that it can be balanced by inspection - by making sure that the same number of atoms of H and O are on each side of the equation. The balanced equation is
We also could have balanced the equation by seeing to it that the net change in oxidation number of all substances was zero. If the oxidation number of one oxygen atom decreases by two, then two hydrogen atoms each must increase by one. In physical terms, if one oxygen atom pulls two electrons toward itself, then two hydrogen atoms are required to donate one electron each. In terms of changes in oxidation number,
Changes in ON:
2H 2(+1)
O (-2)
+
= 0
This was a trivial example, but the following example is not quite so trivial. If oxygen is in short supply, some bacteria can respire using nitrates as sources of oxidizing power instead of O . Rather than reducing oxygen to water, these 2 bacteria reduce nitrates to NO , NO, or N . 2
2
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11. Conservation of Mass, Charge, and Energy
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Conservation of Electrons
Mass is not the only property that is conserved in chemical reactions. In Chapter 6 we saw that, since oxidation and reduction represent only the moving of electrons away from or toward atoms, whenever something is oxidized something else must be reduced. Moving an electron away from one atom in a chemical reaction means moving it toward another one. Thus we can say that, in any chemical reaction in which oxidation and reduction take place, the net change in oxidation numbers of all of the atoms taking part is zero. Total oxidation number is conserved. This is merely an indirect way of saying that electrons are neither created nor destroyed during the reaction. As an example, the combustion of foods during respiration in all oxygen-using forms of life requires the oxidation of carbon and hydrogen compounds. The hydrogen atoms in these compounds are assigned oxidation number zero, because each shares electrons equally with the atoms to which they are bonded. These zero-oxidation-state hydrogen atoms often are represented symbolically in brackets as , without reference to the particular source compound. The energy-yielding oxidation reaction taking place during respiration then can be written in unbalanced form as...
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Conservation of Mass in Chemical Reactions
Note that we proceeded in five steps: 1. Write an unbalanced expression with the correct reactants and products. 2. Balance the equation properly, and obtain the ratio of number of moles of the reactants and products of interest. 3. Calculate molecular weights of the reactants and products of interest. 4. Calculate the number of moles of reactant used, and use the mole ratio from step 2 to find the number of moles of product. 5. Use the molecular weight of the product to obtain the weight in grams. Example. Glucose is a sugar with the chemical formula , and is a common energy source for living organisms. How many moles of oxygen are required to burn a mole of glucose, and how many grams of are needed for a kilogram of glucose? In passing, observe that burning a kilogram of glucose requires only 1066 g of oxygen, whereas burning the same weight of propane requires 3627 g of oxygen. This is because glucose already is partially oxidized. You should not be surprised later when we calculate that the combustion of glucose produces only half as much heat per gram as combustion of propane. Glucose is a poorer fuel than propane on a weight basis. In Chapter 21 we will return to the question as to why plants selected glucose for energy storage (in plant starch), whereas animals developed fats as a kind of "solid propane."
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Conservation of Mass in Chemical Reactions
A single molecule of propane The balanced equation indicates that each mole of propane gas burned requires five moles of oxygen. The problem as stated involved 1000g of propane, and the number of moles is
Five times this many moles of oxygen are needed:
The quantity of oxygen in grams then is -1
113.4 moles O x 32.00 g mole = 3629 g O 2
2
The entire calculation could have been set up in one step:
Formula C3H8
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Conservation of Mass in Chemical Reactions
Solution. The unbalanced reaction, showing only the reactants and products, is
since we know that the combustion products are carbon dioxide and water. The balancing process is given in the illustration opposite: In brief, if three molecules are formed from the three carbon atoms in one propane molecule, then six oxygen atoms or three molecules will be required. In addition, the eight hydrogen atoms in propane will lead to four water molecules, thereby requiring two more molecules of . The balanced equation is
The molecular weights of reactants and products are : 3 x 12.01 g + 8 x 1.01 g = 44.11 g : 2 x 16.00 g = 32.00 g : 1 x 12.01 g + 2 x 16.00 g = 44.01 g : 2 x 1.01 g + 1 x 16.00 g = 18.02 g From the balanced equation and these molecular weights we can verify that mass is conserved during the reaction:
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Conservation of Mass in Chemical Reactions
states that water can be made from hydrogen and oxygen molecules. In addition, it says that two moles of hydrogen and one mole of oxygen are required to produce two moles of water. As an expression of the conservation of atoms, it indicates that for every four atoms of hydrogen and two atoms of oxygen (in and molecules), only two molecules of water can be obtained, which contain the same total of four H and two O atoms. From the balanced equation, one can obtain information about the relative amounts of reactants and products involved. The molecular weights of , and are 2.02 g, 32.00 g, and 18.02 g, respectively. Hence 2 x 2.02 g of hydrogen react with 1 x 32.00 g of oxygen to form 2 x 18.02 g of water: 2(2.02) g + 32.00 g = 2(18.02) g 36.04 g = 36.04 g The total weight of reactants before the reaction is the same as the weight of the products after the reaction; mass has been conserved. Example. Propane, , is a low-pollution fuel gas that can be burned in existing automobile engines with only minor engine adjustments. How many moles of are required to burn one mole of propane, and how many grams of are used with a kilogram of propane?
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Conservation of Mass in Chemical Reactions
In chemical processes, the most important property to be conserved is the number of atoms of each kind that are present. Unlike nuclear processes, chemical reactions do not create or destroy atoms, or change one kind of atom into another. They only reshuffle the atoms that were originally present into different molecular combinations. What we would like to be able to do is to count each kind of atom before and after a reaction and make sure that none has been gained or lost. Counting atoms directly is not practical, but because mass-energy conversion is negligible in chemical reactions, conservation of the number of atoms effectively means the conservation of mass. From the discussion of moles and Avogadro's number in Chapter 2, we know that the mass of a substance divided by its atomic or molecular weight is the quantity of the substance in moles, and that one mole of any chemical substance contains the same number of particles. That number is Avogadro's number, N, which is 6.022 x 10 particles per mole. Hence by weighing moles of a substance, we effectively can count atoms with a laboratory balance. The symbols in a properly balanced chemical equation tell much more than just the identity of reactant and product molecules. The equation
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Weight and Mass
Strictly speaking, mass is related to the inertia of a moving object and the effort needed to stop it. Even though two astronauts weigh only one sixth as much on the moon as on Earth, if they begin fighting the blows land just as hard and hurt just as much because the mass behind the blows is the same as on Earth. In contrast, weight is the force with which a planet pulls on a given mass. If m is the mass in grams and g is the gravitational acceleration in centimeters per second per second, the weight of the object, w, in grams cm sec , or dynes, is given by w = m x g. We ordinarily do not make the distinction between weight and mass. When we say that an object "weighs one gram," what we really mean is that it "weighs what a one-gram mass would weigh on Earth." The gravitational constant, g, is one sixth as large on the moon, so the same mass will have only one sixth the weight there that it has on Earth. When we say that a one-gram object "weighs only one sixth of a gram on the moon," we mean that it is pulled toward the moon with the same force that the Earth would exert on a onesixth-gram mass. As long as we are making only Earth-bound comparisons, no confusion need arise between weight and mass. We will use the terms "atomic weights" and "atomic masses" interchangeably.
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Conservation Principles
However, in the much more powerful nuclear reactions, these principles of separate conservation of mass and energy must be combined into the conservation of the total of mass and energy. Mass can be converted into energy and energy into mass according to Einstein's relationship E = mc , in which E is energy, m is mass, and c is the velocity of light. In the last half of this chapter we will discuss nuclear reactions for which this mass-energy conversion is important. The conversion of mass into energy is central to both nuclear fission and nuclear fusion, on this planet and in the sun. In principle, if a reaction gives off energy, the products formed must have lower energy and be lighter than the reactants. But a release of 100 kcal by a typical chemical reaction corresponds (via the Einstein relationship) to a mass loss of only 5 x 10 amu per molecule, or one hundred thousandth the mass of an electron. This amounts to only 5 x 10 gram per mole, which is far less than we can measure. This is why we can say that, for chemical reactions, mass and energy are conserved independently. Many properties other than mass are not conserved in chemical reactions: volume, density, shape, thermal conductivity, hardness, color, and others. It was Antoine Lavoisier, the brilliant French chemist who revolutionized chemistry before he went to the guillotine in 1794, who realized that mass was more fundamental than any of these properties. When you ask "How much?" in chemistry, you basically are asking "What mass?"
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Conservation Principles
A third, rather subtle quantity also is conserved during propane combustion. This quantity, the oxidation number, is a measure of the location of the electrons. Carbon and hydrogen atoms are oxidized, because they begin by sharing electrons equally with neighboring atoms, but end by forming C O and H O bonds in which oxygen exerts the greater pull on the electrons. Conversely, oxygen atoms are reduced because they begin by sharing electrons equally in O=O molecules and end by monopolizing electrons in their bonds with C and H. The sum of changes in oxidation numbers of all the atoms in propane combustion is zero, because every atom that loses its grip on an electron must be matched by another atom that pulls the electron toward it. In this chapter we consider the conservation of mass and oxidation number; the following chapters are devoted to energy. If we look at mass and energy closely enough, the principles that they individually are conserved turn out to be only approximately true. Mass and energy actually are interconvertible, and are different manifestations of the same thing. We can uncouple them in thinking about chemical reactions only because the quantities of energy involved in chemical processes correspond to infinitesimal amounts of mass.
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Conservation Principles
Science is full of principles of conservation: conservation of mass, conservation of energy, conservation of charge, conservation of symmetry or "parity," and others. These principles all are statements that, when physical and chemical changes take place, certain properties do not change. Throughout the first part of this book we have been using a conservation principle, although we have not spelled it out explicitly: In chemical reactions matter is neither created nor destroyed, within the limits of our ability to measure mass. The amount of material that comes out of any chemical process is no more and no less than the amount that went in, although the appearance of the material may be changed greatly. In the illustration opposite, propane and oxygen gases react to produce another gas and a liquid. The substances produced look and behave differently, but the total number of atoms of each type is unchanged in the course of the reaction. Energy also is conserved in chemical reactions, within the limits of our ability to measure it. The amount of energy in the universe at the end of the propane reaction is the same as at the beginning. If a process gives off energy (the propane reaction does), then the product molecules must have less energy than the reactants, by the amount given off.
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Introduction: You Can't Get Something for Nothing...
The next seven chapters will be devoted to using numbers in chemistry and making it exact. This will lead to an understanding of why substances react, why they appear to react only so far and no farther, and why they do so rapidly or slowly. One of the practical triumphs of chemistry is the ability to control the rates and course of chemical processes, to produce useful substances and energy. The advantages in industrial synthesis are obvious; but the advantages in biosynthesis are no less important. These next seven chapters are an introduction to some aspects of physical chemistry, although quantum theory normally is included also as a part of physical chemistry. Chapters 18 through 21 introduce the subject of organic chemistry, and Chapters 22 through 26 bring us to biochemistry and the evolution of life. One should not pay too much attention to these categories, however, because the most active research today is being done in borderline areas that do not fall easily into any category. It is the overall view of the unity of chemistry that is important.
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11. Conservation of Mass, Charge, and Energy
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Introduction: You Can't Get Something for Nothing...
The next seven chapters will be devoted to using numbers in chemistry and making it exact. This will lead to an understanding of why substances react, why they appear to react only so far and no farther, and why they do so rapidly or slowly. One of the practical triumphs of chemistry is the ability to control the rates and course of chemical processes, to produce useful substances and energy. The advantages in industrial synthesis are obvious; but the advantages in biosynthesis are no less important. These next seven chapters are an introduction to some aspects of physical chemistry, although quantum theory normally is included also as a part of physical chemistry. Chapters 18 through 21 introduce the subject of organic chemistry, and Chapters 22 through 26 bring us to biochemistry and the evolution of life. One should not pay too much attention to these categories, however, because the most active research today is being done in borderline areas that do not fall easily into any category. It is the overall view of the unity of chemistry that is important.
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Introduction: You Can't Get Something for Nothing...
The first ten chapters of this book have been mainly descriptive. They have portrayed the material universe as seen by an observer who has the ability to adjust his field of view to encompass entire galaxies or single atoms. At the lowest level we have seen how electrons can be arranged around nuclei in atoms, and how this arrangement limits the different kinds of atoms that can exist. At a slightly higher level of organization we have seen the way in which electrons hold groups of atoms together in molecules of definite size and shape, and how the properties of matter depend on molecular structure. This is the essence of chemistry: to explain matter in molecular terms. Like any other branch of science, chemistry eventually becomes trivial if it remains descriptive. The essence of science is control of matter by means of successful predictions of behavior; and predictions without measurement are hazy. Sooner or later we must adopt the viewpoint of William Thomson (Lord Kelvin), a pioneer in thermodynamics and electricity, who said in 1891: "When you can measure what you are speaking about, and express it in numbers, you know something about it; but when you cannot measure it, when you cannot express it in numbers, your knowledge is of a meagre and unsatisfactory kind. It may be the beginning of knowledge, but you have scarcely, in your thoughts, advanced to the stage of science."
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11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
In the fusion reaction and should be understood as representing nuclei, with total mass numbers given by superscripts, and nuclear charge (equal to the number of protons) given by subscripts. An electron is represented as , with a zero mass number (not counted among the nucleons) and a -1 charge. In the fusion equation four hydrogen nuclei (not atoms) combine with two of the four electrons around the atoms to form a helium nucleus. The two remaining original electrons, not shown in the equation, associate with the nucleus to build a neutral helium atom. A proton in this representation is p, and a neutron is . The uranium fission reaction tells us that a uranium nucleus, when bombarded by a neutron, breaks down into barium and krypton nuclei with the release of three more neutrons. Therefore this is a chain reaction, with more neutrons released than were used up. These neutrons can bombard neighboring nuclei and produce even more fission. Mass number (superscripts) and charge (subscripts) are conserved in nuclear reactions of this type, just as the number of each kind of atom is conserved in a chemical reaction. You should verify that the sum of subscripts, and sum of superscripts, are constant on the two sides of the equation.
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11. Conservation of Mass, Charge, and Energy
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Mass and Energy: Nuclear Reactions
In this region, beyond atomic number 26, energy is released by fission rather than fusion. At the far left of the curve, hydrogen fusion in stars releases energy:
and at the far right, uranium fission in atomic reactors also releases energy:
(This is only one of many ways in which the
235
U nucleus can break down.)
The maximum stability of the iron nucleus is the reason why the elementbuilding process by successive fusion reactions, outlined in Chapter 8, stops at iron. Beyond iron the fusion process is energy-requiring instead of energyyielding. The heavier elements are built up by more indirect processes involving neutron capture. In spite of the fact that mass is not conserved in nuclear reactions, conservation principles do apply to the total number of heavy particles (protons and neutrons) and total charge. This is implicit in the equations for the two previous nuclear reactions.
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The mass loss or binding energy per nuclear particle (protons and neutrons) rises rapidly to a maximum at iron, then falls. Iron is the most stable nucleus of all. The mass losses or binding energies per nucleon are plotted above for all nuclei from helium through uranium. After some initial minor irregularities in the first- and second-row elements, the values settle down to a smooth curve, which rises to a maximum at iron, then begins a long descending slope through uranium and beyond.
This curve gives us information that was used in the discussion of stellar synthesis of elements in Chapter 8. Iron is the most stable nucleus of all. For elements with smaller atomic numbers than iron, fusion of nuclei to produce heavier elements releases energy, because the products are lighter and more stable on a per-nucleon basis than the reactants. In contrast, beyond iron, fusion absorbs energy because the products are heavier on a pernucleon basis than the reactants.
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Mass and Energy: Nuclear Reactions
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Mass and Energy: Nuclear Reactions
Example. What is the mass loss per nucleon for the component protons, neutrons, and electrons?
atom, compared with its
Solution. The atom contains 26 protons, 26 electrons, and 30 neutrons, so the mass calculation is performed as opposite: Notice that the mass loss per nucleon, and hence the binding energy per nucleon, is greater for iron than for helium. This means that the iron nucleus is more stable relative to protons and neutrons than the helium nucleus is. If some combination of helium nuclei could be induced to produce an iron nucleus, energy would be given off, which would correspond to the increased stability of the product nucleus per nuclear particle.
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Mass and Energy: Nuclear Reactions -1
(Compare this energy with the 83 kcal mole required to break carbon-carbon bonds in chemical reactions.) Protons and neutrons collectively are known as nuclear particles or nucleons. The number of nucleons in the nucleus is the sum of the number of protons and neutrons and, as was mentioned in Chapter 2, is known as the mass number of the nucleus. Every atomic nucleus is lighter than the sum of the masses of the nucleons from which it is built, and this mass loss corresponds to the binding energy of the nucleus. The relative stability of two nuclei with different numbers of nucleons can be assessed by comparing their mass loss per nucleon. Example: What is the mass loss per nucleon for the helium atom? Solution: The total mass loss is 0.0305 amu, and since the nucleus has four nucleons, the mass loss per nucleon is 0.00763 amu.
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Mass and Energy: Nuclear Reactions
(We must include electrons in this calculation because 4.0026 amu is the mass of the helium-4 atom, not the nucleus.) This missing mass corresponds to 0.0305 x 931.4 MeV = 28.4 MeV of energy. If we could put together a helium atom directly from two neutrons, two protons, and two electrons, then 28.4 MeV of energy would be given off for every atom formed:
Compared to common chemical reactions, this is an enormous quantity of energy. Since 1 electron volt per atom is equivalent to 23.06 kcal per mole,
= 655,000,000 kcal mole
-1
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Mass and Energy: Nuclear Reactions
With nuclear reactions, the energies involved are so great that the changes in mass become easily measurable. One no longer can assume that mass and energy are conserved separately, but must take into account their interconversion via Einstein's relationship, E = mc . If mass is in grams and the velocity of light is expressed as c = 3 x 10 cm sec , then the energy is in units of g cm sec , or ergs. A useful conversion is from mass in amu to energy in million electron volts (MeV): 1 amu = 931.4 MeV What holds a nucleus together? If we attempt to bring two protons and two neutrons together to form a helium nucleus, we might reasonably expect the positively charged protons to repel one another violently. Then what keeps them together in the nucleus? The answer, as we mentioned in Chapter 2, is that a helium atom is lighter than the sum of two protons, two neutrons, and two electrons. Some of the mass of the separated particles is converted into energy and dissipated when the nucleus is formed. Before the helium nucleus can be torn apart into its component particles, this dissipated energy must be restored and turned back into mass. Unless this energy is provided, the nucleus cannot be taken apart. This energy is termed the binding energy of the helium nucleus.
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Conservation of Electrons
Exercise. For practice, balance the following reactions by both the oxidation-number and half-reaction methods:
Although half-reactions have been introduced here merely as a means of obtaining a balanced overall reaction, they can have physical meaning of their own. If the iron oxidation and manganese reduction can be carried out in separate containers, and if these containers can be given suitable electrical connections, then we can make use of the energy released by this reaction as the electrons flow through a wire from the iron container to the manganese. This is the principle of the electrolytic cell or battery. An ordinary flashlight battery uses manganese reduction, and oxidation of the zinc battery casing. We will return to electrolytic cells and energy production in Chapter 17.
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Conservation of Electrons
The iron half-reaction is simple:
The final step is to add these two individually balanced half-reactions and obtain an overall reaction in which electrons do not appear explicitly. To accomplish this, we must add one unit of the manganese reaction to five units of the iron reaction. The result is the same as before:
This overall equation is now balanced with respect to charge and number of atoms because the half-reactions were balanced, and with respect to oxidation number because the proper multiples of the half-reactions were chosen to make the electrons cancel.
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Conservation of Electrons
There is another useful method for balancing redox equations: the half-reaction method. In this method each substance that changes its oxidation number is balanced separately in a half-reaction that includes electrons explicitly. The two balanced half-reactions then are combined in such a way as to cancel electrons from the final expression. Example. Balance the iron and permanganate reaction by the half-reaction method. Solution. The unbalanced permanganate half-reaction is
Five electrons were needed because manganese goes from +7 to +2 oxidation state. Water molecules are added on the right to balance the oxygen atoms in the permanganate, and protons are added to the left:
This half-reaction is now balanced with respect to both number of atoms and overall charge, +2 on each side.
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Conservation of Electrons
It is obvious that five iron atoms are required for every manganese atom, in order that the changes in oxidation numbers cancel:
Eight hydrogen ions are needed to use up the four oxygen atoms on the left, leading to four water molecules on the right:
In choosing a 1-to-5 ratio of Mn-to-Fe, we ensured that oxidation number was conserved. By adding the hydrogen ions we obtained an equation that balanced the number of each atom. As a final check, the net charge on each side of the equation can be tested, and found to be the same: +17. This oxidation-reduction equation now is balanced with respect to electrons, atoms, and charge.
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Conservation of Electrons
The corresponding equation for acidic solutions can be obtained by adding enough hydrogen ions to each side to eliminate the hydroxide ions:
The tetrahedral permanganate ion Example. Potassium permanganate, , is a common inorganic oxidizing agent, which becomes reduced to manganous ion, Mn , in acidic solution. Write a balanced equation for the reaction in which permanganate oxidizes ferrous iron, 2+
3+
Fe , to ferric ion, Fe . Solution.
In the permanganate ion manganese has an oxidation number of +7 which decreases to +2 in the manganous ion, a change of -5. Iron increases in oxidation number by one:
Formula MnO4-
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Conservation of Electrons
For electrons to balance properly, three moles of H must be oxidised for every mole of NO3- reduced: 3 (+1)
+
(-3)
=
The planar, delocallised nitrate ion
0
and the reaction can be written
In any reaction in aqueous solution, one is at liberty to assume as many H O, +
-
+
2
H , or OH as are required to balance the reaction, using H O and H under 2 acidic conditions, or and OH if the solution is basic. The redox (oxidation reduction) part of the balancing has been done, and what remains is only an accounting for O and H atoms. One possible answer is
This has accounted for the three hydrogen atoms and the negative charge. The
Formula NO3-
-
equation is correct for basic solutions where OH ions are present.
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Conservation of Electrons
How can we write a properly balanced equation for the oxidation of foods with NO as the waste product?
It might be tempting to balance by inspection along the following lines, making the number of each kind of atom the same on left and right:
But this leaves an electron on the left side unaccounted for on the right side. The approach that is most nearly foolproof and at the same time shows you what is happening is the oxidation-number method. In nitrates, N has an oxidation number of +5 ( verify this for yourself ). In NO the oxidation number of N is +2, a decrease of three. Hydrogen, as before, goes from 0 to +1, an increase of one:
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Conservation of Electrons
These oxidation numbers (ON) of H and O in H O arise because O is more 2
electronegative than H, so both the electrons in each O H bond are assigned to O. If you are unsure of this process, look back at Chapter 6. This is such a simple chemical reaction that it can be balanced by inspection - by making sure that the same number of atoms of H and O are on each side of the equation. The balanced equation is
We also could have balanced the equation by seeing to it that the net change in oxidation number of all substances was zero. If the oxidation number of one oxygen atom decreases by two, then two hydrogen atoms each must increase by one. In physical terms, if one oxygen atom pulls two electrons toward itself, then two hydrogen atoms are required to donate one electron each. In terms of changes in oxidation number,
Changes in ON:
2H 2(+1)
O (-2)
+
= 0
This was a trivial example, but the following example is not quite so trivial. If oxygen is in short supply, some bacteria can respire using nitrates as sources of oxidizing power instead of O . Rather than reducing oxygen to water, these 2 bacteria reduce nitrates to NO , NO, or N . 2
2
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Conservation of Electrons
Mass is not the only property that is conserved in chemical reactions. In Chapter 6 we saw that, since oxidation and reduction represent only the moving of electrons away from or toward atoms, whenever something is oxidized something else must be reduced. Moving an electron away from one atom in a chemical reaction means moving it toward another one. Thus we can say that, in any chemical reaction in which oxidation and reduction take place, the net change in oxidation numbers of all of the atoms taking part is zero. Total oxidation number is conserved. This is merely an indirect way of saying that electrons are neither created nor destroyed during the reaction. As an example, the combustion of foods during respiration in all oxygen-using forms of life requires the oxidation of carbon and hydrogen compounds. The hydrogen atoms in these compounds are assigned oxidation number zero, because each shares electrons equally with the atoms to which they are bonded. These zero-oxidation-state hydrogen atoms often are represented symbolically in brackets as , without reference to the particular source compound. The energy-yielding oxidation reaction taking place during respiration then can be written in unbalanced form as...
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Conservation of Mass in Chemical Reactions
Note that we proceeded in five steps: 1. Write an unbalanced expression with the correct reactants and products. 2. Balance the equation properly, and obtain the ratio of number of moles of the reactants and products of interest. 3. Calculate molecular weights of the reactants and products of interest. 4. Calculate the number of moles of reactant used, and use the mole ratio from step 2 to find the number of moles of product. 5. Use the molecular weight of the product to obtain the weight in grams. Example. Glucose is a sugar with the chemical formula , and is a common energy source for living organisms. How many moles of oxygen are required to burn a mole of glucose, and how many grams of are needed for a kilogram of glucose? In passing, observe that burning a kilogram of glucose requires only 1066 g of oxygen, whereas burning the same weight of propane requires 3627 g of oxygen. This is because glucose already is partially oxidized. You should not be surprised later when we calculate that the combustion of glucose produces only half as much heat per gram as combustion of propane. Glucose is a poorer fuel than propane on a weight basis. In Chapter 21 we will return to the question as to why plants selected glucose for energy storage (in plant starch), whereas animals developed fats as a kind of "solid propane."
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Conservation of Mass in Chemical Reactions
A single molecule of propane The balanced equation indicates that each mole of propane gas burned requires five moles of oxygen. The problem as stated involved 1000g of propane, and the number of moles is
Five times this many moles of oxygen are needed:
The quantity of oxygen in grams then is -1
113.4 moles O x 32.00 g mole = 3629 g O 2
2
The entire calculation could have been set up in one step:
Formula C3H8
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Conservation of Mass in Chemical Reactions
Solution. The unbalanced reaction, showing only the reactants and products, is
since we know that the combustion products are carbon dioxide and water. The balancing process is given in the illustration opposite: In brief, if three molecules are formed from the three carbon atoms in one propane molecule, then six oxygen atoms or three molecules will be required. In addition, the eight hydrogen atoms in propane will lead to four water molecules, thereby requiring two more molecules of . The balanced equation is
The molecular weights of reactants and products are : 3 x 12.01 g + 8 x 1.01 g = 44.11 g : 2 x 16.00 g = 32.00 g : 1 x 12.01 g + 2 x 16.00 g = 44.01 g : 2 x 1.01 g + 1 x 16.00 g = 18.02 g From the balanced equation and these molecular weights we can verify that mass is conserved during the reaction:
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Conservation of Mass in Chemical Reactions
states that water can be made from hydrogen and oxygen molecules. In addition, it says that two moles of hydrogen and one mole of oxygen are required to produce two moles of water. As an expression of the conservation of atoms, it indicates that for every four atoms of hydrogen and two atoms of oxygen (in and molecules), only two molecules of water can be obtained, which contain the same total of four H and two O atoms. From the balanced equation, one can obtain information about the relative amounts of reactants and products involved. The molecular weights of , and are 2.02 g, 32.00 g, and 18.02 g, respectively. Hence 2 x 2.02 g of hydrogen react with 1 x 32.00 g of oxygen to form 2 x 18.02 g of water: 2(2.02) g + 32.00 g = 2(18.02) g 36.04 g = 36.04 g The total weight of reactants before the reaction is the same as the weight of the products after the reaction; mass has been conserved. Example. Propane, , is a low-pollution fuel gas that can be burned in existing automobile engines with only minor engine adjustments. How many moles of are required to burn one mole of propane, and how many grams of are used with a kilogram of propane?
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Conservation of Mass in Chemical Reactions
In chemical processes, the most important property to be conserved is the number of atoms of each kind that are present. Unlike nuclear processes, chemical reactions do not create or destroy atoms, or change one kind of atom into another. They only reshuffle the atoms that were originally present into different molecular combinations. What we would like to be able to do is to count each kind of atom before and after a reaction and make sure that none has been gained or lost. Counting atoms directly is not practical, but because mass-energy conversion is negligible in chemical reactions, conservation of the number of atoms effectively means the conservation of mass. From the discussion of moles and Avogadro's number in Chapter 2, we know that the mass of a substance divided by its atomic or molecular weight is the quantity of the substance in moles, and that one mole of any chemical substance contains the same number of particles. That number is Avogadro's number, N, which is 6.022 x 10 particles per mole. Hence by weighing moles of a substance, we effectively can count atoms with a laboratory balance. The symbols in a properly balanced chemical equation tell much more than just the identity of reactant and product molecules. The equation
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Weight and Mass
Strictly speaking, mass is related to the inertia of a moving object and the effort needed to stop it. Even though two astronauts weigh only one sixth as much on the moon as on Earth, if they begin fighting the blows land just as hard and hurt just as much because the mass behind the blows is the same as on Earth. In contrast, weight is the force with which a planet pulls on a given mass. If m is the mass in grams and g is the gravitational acceleration in centimeters per second per second, the weight of the object, w, in grams cm sec , or dynes, is given by w = m x g. We ordinarily do not make the distinction between weight and mass. When we say that an object "weighs one gram," what we really mean is that it "weighs what a one-gram mass would weigh on Earth." The gravitational constant, g, is one sixth as large on the moon, so the same mass will have only one sixth the weight there that it has on Earth. When we say that a one-gram object "weighs only one sixth of a gram on the moon," we mean that it is pulled toward the moon with the same force that the Earth would exert on a onesixth-gram mass. As long as we are making only Earth-bound comparisons, no confusion need arise between weight and mass. We will use the terms "atomic weights" and "atomic masses" interchangeably.
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Conservation Principles
However, in the much more powerful nuclear reactions, these principles of separate conservation of mass and energy must be combined into the conservation of the total of mass and energy. Mass can be converted into energy and energy into mass according to Einstein's relationship E = mc , in which E is energy, m is mass, and c is the velocity of light. In the last half of this chapter we will discuss nuclear reactions for which this mass-energy conversion is important. The conversion of mass into energy is central to both nuclear fission and nuclear fusion, on this planet and in the sun. In principle, if a reaction gives off energy, the products formed must have lower energy and be lighter than the reactants. But a release of 100 kcal by a typical chemical reaction corresponds (via the Einstein relationship) to a mass loss of only 5 x 10 amu per molecule, or one hundred thousandth the mass of an electron. This amounts to only 5 x 10 gram per mole, which is far less than we can measure. This is why we can say that, for chemical reactions, mass and energy are conserved independently. Many properties other than mass are not conserved in chemical reactions: volume, density, shape, thermal conductivity, hardness, color, and others. It was Antoine Lavoisier, the brilliant French chemist who revolutionized chemistry before he went to the guillotine in 1794, who realized that mass was more fundamental than any of these properties. When you ask "How much?" in chemistry, you basically are asking "What mass?"
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Conservation Principles
A third, rather subtle quantity also is conserved during propane combustion. This quantity, the oxidation number, is a measure of the location of the electrons. Carbon and hydrogen atoms are oxidized, because they begin by sharing electrons equally with neighboring atoms, but end by forming C O and H O bonds in which oxygen exerts the greater pull on the electrons. Conversely, oxygen atoms are reduced because they begin by sharing electrons equally in O=O molecules and end by monopolizing electrons in their bonds with C and H. The sum of changes in oxidation numbers of all the atoms in propane combustion is zero, because every atom that loses its grip on an electron must be matched by another atom that pulls the electron toward it. In this chapter we consider the conservation of mass and oxidation number; the following chapters are devoted to energy. If we look at mass and energy closely enough, the principles that they individually are conserved turn out to be only approximately true. Mass and energy actually are interconvertible, and are different manifestations of the same thing. We can uncouple them in thinking about chemical reactions only because the quantities of energy involved in chemical processes correspond to infinitesimal amounts of mass.
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Conservation Principles
Science is full of principles of conservation: conservation of mass, conservation of energy, conservation of charge, conservation of symmetry or "parity," and others. These principles all are statements that, when physical and chemical changes take place, certain properties do not change. Throughout the first part of this book we have been using a conservation principle, although we have not spelled it out explicitly: In chemical reactions matter is neither created nor destroyed, within the limits of our ability to measure mass. The amount of material that comes out of any chemical process is no more and no less than the amount that went in, although the appearance of the material may be changed greatly. In the illustration opposite, propane and oxygen gases react to produce another gas and a liquid. The substances produced look and behave differently, but the total number of atoms of each type is unchanged in the course of the reaction. Energy also is conserved in chemical reactions, within the limits of our ability to measure it. The amount of energy in the universe at the end of the propane reaction is the same as at the beginning. If a process gives off energy (the propane reaction does), then the product molecules must have less energy than the reactants, by the amount given off.
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Introduction: You Can't Get Something for Nothing...
The next seven chapters will be devoted to using numbers in chemistry and making it exact. This will lead to an understanding of why substances react, why they appear to react only so far and no farther, and why they do so rapidly or slowly. One of the practical triumphs of chemistry is the ability to control the rates and course of chemical processes, to produce useful substances and energy. The advantages in industrial synthesis are obvious; but the advantages in biosynthesis are no less important. These next seven chapters are an introduction to some aspects of physical chemistry, although quantum theory normally is included also as a part of physical chemistry. Chapters 18 through 21 introduce the subject of organic chemistry, and Chapters 22 through 26 bring us to biochemistry and the evolution of life. One should not pay too much attention to these categories, however, because the most active research today is being done in borderline areas that do not fall easily into any category. It is the overall view of the unity of chemistry that is important.
Page 2 of 37
http://neon.chem.ox.ac.uk/vrchemistry/Conservation/page02.htm
2006/12/10
Foundations to Chemistry - adapted from "Chemistry, Matter and the Universe"
11. Conservation of Mass, Charge, and Energy
Page 1 of 1
-- Jump to --
Introduction: You Can't Get Something for Nothing...
The next seven chapters will be devoted to using numbers in chemistry and making it exact. This will lead to an understanding of why substances react, why they appear to react only so far and no farther, and why they do so rapidly or slowly. One of the practical triumphs of chemistry is the ability to control the rates and course of chemical processes, to produce useful substances and energy. The advantages in industrial synthesis are obvious; but the advantages in biosynthesis are no less important. These next seven chapters are an introduction to some aspects of physical chemistry, although quantum theory normally is included also as a part of physical chemistry. Chapters 18 through 21 introduce the subject of organic chemistry, and Chapters 22 through 26 bring us to biochemistry and the evolution of life. One should not pay too much attention to these categories, however, because the most active research today is being done in borderline areas that do not fall easily into any category. It is the overall view of the unity of chemistry that is important.
Page 2 of 37
http://neon.chem.ox.ac.uk/vrchemistry/Conservation/page02.htm
2006/12/10
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