Heat, Energy, And Chemical Bonds

Introduction, page 1

12. Heat, Energy, and Chemical Bonds

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Introduction

In Chapter 11 we saw two examples of principles of conservation, involving matter and charge, in chemical reactions. Neither the total mass nor the net oxidation number of reactants and products can change during a chemical transformation. Mass and oxidation number are more fundamental properties of substances than are volume, color, texture, density, or electrical conductivity. A third important property that is conserved is energy. Combustions and many other reactions give off energy in the form of heat, and this makes these processes useful to us. But if we draw an imaginary box around the reacting substances, large enough to contain the substances and everything else they interact with, then the total energy within that box will not change during the reaction. (This is one form of the first law of thermodynamics.) If the reaction gives off heat energy, then the products must have less energy than the reactants, because this difference is the only source of the heat. Conversely, if the products have more energy than the reactants, this extra energy must be supplied to the reaction from outside.

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Introduction, page 2

12. Heat, Energy, and Chemical Bonds

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Introduction

Energy can be thought of as the capacity to do work. Potential energy is energy that a motionless object has by virtue of its position. A rock poised at a great height has gravitational potential energy because it can do work while falling, if harnessed to a suitable apparatus. Galileo's cannonball had considerable potential energy before he dropped it (see cartoon). Two separated positive and negative charges have electrostatic potential energy, for they can do electrical work as they come together again. A bar magnet turned crosswise to a magnetic field has magnetic potential energy. Work is required to turn the magnet out of alignment with the field, and this work can be regained when the magnet aligns itself with the field once more. Kinetic energy is the energy that an object has because of its motion. Work is required initially to get the object going, and this work can be obtained again when the projectile collides with a target and comes to rest. If the moving object has mass m, and is moving with velocity v, its kinetic energy of motion is E = ½mv

2

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Introduction, page 3

12. Heat, Energy, and Chemical Bonds

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Introduction

Galileo's cannonball in the opening drawing converts potential energy to kinetic energy as it falls faster and faster. Heat is a form of energy, too, but in a degraded form. On the molecular level, heat is the uncoordinated vibration or motion of individual molecules. It is easy to convert work or other kinds of energy into heat. If you rub one hand hard against another on a cold day, the warmth generated by friction is just a conversion of the work that you do into heat. When Galileo's cannonball strikes the pavement, the potential energy that had first changed into kinetic energy of motion as it fell, is changed again into heat as it collides with the ground. The reverse transformation of heat into motion, as in an automobile engine, is accomplished only with difficulty; not all of the heat can be converted. We will come back to these ideas when we talk about order and disorder in the next chapter. For the moment, the important idea is that potential energy, kinetic energy, heat, and work all are different kinds of energy, and with certain limitations they can be interconverted from one form to another.

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Heat and Chemical Reactions, page 4

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

One can think of a chemical reaction as proceeding by the pulling apart of all of the atoms in the reactant molecules, and then the reassembling of these isolated atoms in new ways to make product molecules. The difference between bond energies of products and reactants, ∆E, is essentially the energy of the reaction:

∆E = E(products) - E(reactants) If the energy of the products is less than that of the initial molecules, the molecular energy decreases. The change in energy of the molecules and atoms during the reaction is negative, and heat is given off:

∆E = E(products) - E(reactants) = <0 Conversely, if the product molecules have more energy than the starting materials, heat is absorbed, the total molecular energy increases, and ∆E is positive.

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Heat and Chemical Reactions, page 5

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

When gases are produced in a chemical reaction, work must be done in pushing against the pressure of the atmosphere to make room for them. Conversely, if gases disappear or condense to form liquids or solids during a reaction, work is done on the reacting substances by the external world. If we make a small correction for any work that gases in a reaction might do in pushing against the atmosphere when they are generated, then we obtain a modified energy, the enthalpy, H. Enthalpy is defined formally by H = E + PV, in which PV is the product of pressure and volume. When a reaction takes place at a constant surrounding pressure, which is true for reactions conducted in the open at the surface of the Earth (e.g., at a laboratory bench), then the heat of reaction is a measure of the change in enthalpy, ∆H = H(products) - H(reactants). The difference between energy and enthalpy is only a few percent, and enthalpy should be thought of as a "corrected" energy, with allowance made for atmospheric pressure.

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Heat and Chemical Reactions, page 6

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Heat and Chemical Reactions

Whenever we talk about heats of reaction from now on, we shall be using enthalpies and not strictly molecular energies. You may think of H as standing for "heat" in reactions carried out at constant external pressure. As an example, when one mole of hydrogen gas and one half mole of oxygen gas react to produce one mole, or 18 grams, of liquid water, 68 kilocalories of heat are given off (see diagram). When this energy escapes as heat, the total molecular enthalpy decreases by 68 kcal: H (g) + ½O (g) → H O(l) + 68 kcal 2

2

2

or in the more conventional way of writing the heat of reaction, -1

H (g) + ½O (g) → H O(l) ∆H = -68 kcal mole H O 2

2

2

2

∆H is negative because the enthalpy of the chemicals taking part in the reaction decreases. Because heat is given off, this reaction is described as exothermic. A reaction that absorbs heat is endothermic. If we regard hydrogen gas, with a molecular weight of 2 grams per mole, as a fuel, then the energy yield of this reaction is 68/2 = 34 kcal per gram of fuel.

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Heat and Chemical Reactions, page 6

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Heat and Chemical Reactions

Whenever we talk about heats of reaction from now on, we shall be using enthalpies and not strictly molecular energies. You may think of H as standing for "heat" in reactions carried out at constant external pressure. As an example, when one mole of hydrogen gas and one half mole of oxygen gas react to produce one mole, or 18 grams, of liquid water, 68 kilocalories of heat are given off (see diagram). When this energy escapes as heat, the total molecular enthalpy decreases by 68 kcal: H (g) + ½O (g) → H O(l) + 68 kcal 2

2

2

or in the more conventional way of writing the heat of reaction, -1

H (g) + ½O (g) → H O(l) ∆H = -68 kcal mole H O 2

2

2

2

∆H is negative because the enthalpy of the chemicals taking part in the reaction decreases. Because heat is given off, this reaction is described as exothermic. A reaction that absorbs heat is endothermic. If we regard hydrogen gas, with a molecular weight of 2 grams per mole, as a fuel, then the energy yield of this reaction is 68/2 = 34 kcal per gram of fuel.

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Heat and Chemical Reactions, page 7

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Heat and Chemical Reactions

This is a higher value than for any other chemical fuel burned in O2. If hydrogen could be stored and handled more easily, we probably would use it instead of more conventional hydrocarbon fuels. If the water in the combustion remained as a gas rather than condensing to a liquid, less heat would be given off (right): -1

H (g) + ½O (g) → H O(g) ∆H = -58 kcal mole H O 2

2

2

2

The difference is the enthalpy or heat of vaporization: H 0(l) → H 0(g) ∆H = +10kcal mole 2

-1

2

o

During vaporization at 25 C, water molecules absorb 10 kcal of heat energy per mole, so the enthalpy increases. This is necessary for liquids in general because gas molecules move faster and have more energy (and enthalpy) than molecules in a liquid. The same amount of heat energy is given off again when water vapor condenses: H O(g) → H O(l) ∆H = -10 kcal mole 2

-1

2

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Heat and Chemical Reactions, page 8

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

Water has the highest heat of vaporization per gram of any known liquid. A comparison of water with a few other common liquids is given below. The reason for the specially high value for water is the network of hydrogen bonds that holds the water molecules together, even in the liquid state. In the "iceberg" or "flickering cluster" picture of liquid water structure, small clusters of water molecules are held together by hydrogen bonds in a more or less icelike manner (right). Bonds are made and broken continually, and one particular molecule may be bonded to others at one instant and free the next. Throughout the liquid a certain percentage of the possible hydrogen bonds will be formed at a given moment. These bonds must be broken when the liquid is vaporized, and this requires energy. The result is a higher heat of vaporization than for liquids without hydrogen bonds.

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Heat and Chemical Reactions, page 8

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

Water has the highest heat of vaporization per gram of any known liquid. A comparison of water with a few other common liquids is given below. The reason for the specially high value for water is the network of hydrogen bonds that holds the water molecules together, even in the liquid state. In the "iceberg" or "flickering cluster" picture of liquid water structure, small clusters of water molecules are held together by hydrogen bonds in a more or less icelike manner (right). Bonds are made and broken continually, and one particular molecule may be bonded to others at one instant and free the next. Throughout the liquid a certain percentage of the possible hydrogen bonds will be formed at a given moment. These bonds must be broken when the liquid is vaporized, and this requires energy. The result is a higher heat of vaporization than for liquids without hydrogen bonds.

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Heat and Chemical Reactions, page 9

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Heat and Chemical Reactions

In the series O-S-Se-Te down Group VIA of the periodic table, hydrogen bonding becomes unimportant, because the atoms of S, Se, and Te are too large and their negative charge is too diffuse to attract a proton of a neighboring molecule strongly. Hence H S, H Se, and H Te have more 2

2

2

"normal" boiling points and heats of vaporization (see right, and next page). This anomalously high heat of vaporization of water has major consequences for life on this planet. Evaporation of ocean water in the tropics keeps the equatorial regions from being as hot as they would be otherwise, and the heat removed warms the more polar regions when the water vapor condenses. Liquid water therefore acts as a heat reservoir, moderating the extremes of temperature both at the equator and near the poles. The evaporation of ocean water at the equator appears to be approximately 2.3 meters of depth per year, corresponding to a removal of 1.3 trillion (1,300,000,000,000) kilocalories of heat per square kilometer of surface area!

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Heat and Chemical Reaction, page 10

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

To bring such a giant number into comprehensible range, a person not doing heavy labor normally needs the equivalent of 2000 kcal of energy from food per day. The energy removed annually by evaporation from every square kilometer of tropical ocean would correspond to enough energy to keep 2 million people going for a year. Notice that we can add two of the previous reactions to yield the third:

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Heat and Chemical Reactions, page 11

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

The heats of reaction also add in the same way. In general, if we can add or subtract two reactions to obtain a third, then the heat of this third reaction is obtained by adding or subtracting the heats of the first two. For those who particularly enjoy filing things away by name, this is Hess' law of heat summation. It actually is a natural consequence of the first law of thermodynamics. (In the example above, the reaction of hydrogen and oxygen gases to produce water vapor, followed by condensation of the vapor in a second step, must lead to exactly the same enthalpy change as the reaction of hydrogen and oxygen gases to produce liquid water directly. Hence the heats of the first two reactions must add to yield the heat of the direct process.) We must be careful that the two reactions added really do yield the third, and that the heats used are for the reactions as they are written. For example, we could multiply the hydrogen combustion reaction by two to eliminate the fractional coefficient before the oxygen, but then the heat of reaction also would have to be doubled: 2H (g) + 0 (g) → 2H 0(g) ∆H=-116 kcal 2

2

2

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Heat and Chemical Reactions, page 12

12. Heat, Energy, and Chemical Bonds

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Heat and Chemical Reactions

The heat given off is 116 kcal for the molar reaction as written; that is, per two moles of hydrogen gas and one mole of oxygen used, and two moles of water vapor formed. If, without thinking, we were to add to this reaction, the reaction H O(g) → H O(l) ∆H = -10 kcal mole 2

-1

2

then the result would be not what we want at all: 2H (g) + O (g) → H O(g) + H O(l) ∆H = -126 kcal 2

2

2

2

Although the equation would be true, it would not be particularly useful. One must keep the stoichiometry (the relative numbers of reactant and product molecules) correct for the heats of reaction to be meaningful.

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Fuels, Combustion, and Energy, page 13

12. Heat, Energy, and Chemical Bonds

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Fuels, Combustion, and Energy

We mentioned previously that hydrogen gas was the most efficient of all fuels in terms of heat emitted per gram of fuel burned. Gasoline is less efficient by nearly a factor of three, as the table at the right shows. Hydrogen gas releases 34 kcal per gram upon combustion in air; gasoline yields less than 12 kcal per gram. Fats, the main energy-storage system in animals, produce 9.5 kcal per gram upon combustion, and are almost as efficient in energy storage as gasoline. Starch, the main energy-storage molecule in plants, is a long-chain polymer of glucose. Glucose can store only 3.7 kcal per gram, so on a weight basis, starch is an inefficient material for storing energy. Then why is the entire photosynthetic and energy-storage machinery of green plants based on starch?

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Fuels, Combustion, and Energy, page 14

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Fuels, Combustion, and Energy

The answer is that weight considerations are of minor concern to a stationary plant. There is no advantage in a lightweight fuel to an organism that is not going anywhere. The chemistry of assembling starch chains from glucose and recovering glucose again is particularly simple, whereas the metabolism of fats and fatty acids is more complicated.

Plants store energy in starch because it is especially easy to get the energy in and out. Animals, which must move about and carry their fuel supplies with them, find the low energy-per-gram feature of starch to be a disadvantage. They put up with the more complicated chemistry of fats to achieve an energy-storage efficiency only slightly less than that of gasoline.

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Fuels, Combustion, and Energy, page 15

12. Heat, Energy, and Chemical Bonds

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Fuels, Combustion, and Energy

The one big advantage of fats over gasoline is that fats are solids. It would be almost as disadvantageous for us to carry around liquid gasoline as an energy reserve as it would be to carry huge gas bags of hydrogen. (An organism that evolved bags of hydrogen gas for energy storage also would have come a long way in solving its locomotion problem. But thunderstorms and lightning would be hard on such a creature.) Even animals cannot get along without the rapid-access feature of starch as an energy-storage agent. Animals have glycogen in their bloodstream as a quickenergy source, which serves as a buffer between the immediate energy needs of the organism and the slow production of energy from fats. Glycogen, or "animal starch," is a branched-chain polymer of glucose molecules very similar to plant starch. The table of heats of combustion on page 14 also points out that we could do much better than to burn our fuels in oxygen. Combustion of methane in 0 2 produces only 13 kcal of heat per mole of methane, whereas combustion in F 2

would yield 25 kcal, nearly twice as much. Chlorine gas is less favorable as an oxidant, yielding only 6.5 kcal per mole of methane consumed.

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Fuels, Combustion, and Energy, page 15

12. Heat, Energy, and Chemical Bonds

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Fuels, Combustion, and Energy

The one big advantage of fats over gasoline is that fats are solids. It would be almost as disadvantageous for us to carry around liquid gasoline as an energy reserve as it would be to carry huge gas bags of hydrogen. (An organism that evolved bags of hydrogen gas for energy storage also would have come a long way in solving its locomotion problem. But thunderstorms and lightning would be hard on such a creature.) Even animals cannot get along without the rapid-access feature of starch as an energy-storage agent. Animals have glycogen in their bloodstream as a quickenergy source, which serves as a buffer between the immediate energy needs of the organism and the slow production of energy from fats. Glycogen, or "animal starch," is a branched-chain polymer of glucose molecules very similar to plant starch. The table of heats of combustion on page 14 also points out that we could do much better than to burn our fuels in oxygen. Combustion of methane in 0 2 produces only 13 kcal of heat per mole of methane, whereas combustion in F 2

would yield 25 kcal, nearly twice as much. Chlorine gas is less favorable as an oxidant, yielding only 6.5 kcal per mole of methane consumed.

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Fuels, Combustion, and Energy, Page 16

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Fuels, Combustion, and Energy

These differences are purely a function of the electronegativities of F, O, and Cl, that is, the strength with which each atom pulls electrons toward itself. After combustion, the electrons that were shared equally between carbon and hydrogen in methane are drawn toward the Cl atoms when Cl is the 2

combustion medium, drawn more strongly toward O atoms, and most strongly of all toward F when F is the combustion medium. 2

The tighter the electrons are held after combustion, the more stable the product molecules are, and the more energy is given off as heat.

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Fuels, Combustion, and Energy, page 17

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Fuels, Combustion, and Energy

We can use the spring analogy below to illustrate why combustions in strongly electronegative media are exothermic, or heat-emitting. Let the ball in the middle symbolize the electrons, drawn either toward oxygen at the left, or carbon and hydrogen at the right. The stronger spring to the left represents the greater electronegativity of oxygen, and the weak spring at the right, the lesser electronegativities of carbon and hydrogen.

If water and CO were to react to form methane and O , electrons 2

2

would be moved away from oxygen, against oxygen's natural electronegativity. This corresponds to using energy to push the ball to the right, thereby stretching the strong spring. This energy is stored in the spring (or in the methane molecules) as potential, or latent, energy.

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Fuels, Combustion, and Energy, page 17

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Fuels, Combustion, and Energy

We can use the spring analogy below to illustrate why combustions in strongly electronegative media are exothermic, or heat-emitting. Let the ball in the middle symbolize the electrons, drawn either toward oxygen at the left, or carbon and hydrogen at the right. The stronger spring to the left represents the greater electronegativity of oxygen, and the weak spring at the right, the lesser electronegativities of carbon and hydrogen.

If water and CO were to react to form methane and O , electrons 2

2

would be moved away from oxygen, against oxygen's natural electronegativity. This corresponds to using energy to push the ball to the right, thereby stretching the strong spring. This energy is stored in the spring (or in the methane molecules) as potential, or latent, energy.

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Fuels, Combustion, and Energy, page 17

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Fuels, Combustion, and Energy

We can use the spring analogy below to illustrate why combustions in strongly electronegative media are exothermic, or heat-emitting. Let the ball in the middle symbolize the electrons, drawn either toward oxygen at the left, or carbon and hydrogen at the right. The stronger spring to the left represents the greater electronegativity of oxygen, and the weak spring at the right, the lesser electronegativities of carbon and hydrogen.

If water and CO were to react to form methane and O , electrons 2

2

would be moved away from oxygen, against oxygen's natural electronegativity. This corresponds to using energy to push the ball to the right, thereby stretching the strong spring. This energy is stored in the spring (or in the methane molecules) as potential, or latent, energy.

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Fuels, Combustion, and Energy, page 18

12. Heat, Energy, and Chemical Bonds

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Fuels, Combustion, and Energy

If methane is combined later with a strongly electronegative element, oxygen, and the electrons are permitted to relax toward more electronegative atoms again, the stored energy is released. In our spring model, combustion is analogous to releasing the constraints on the ball and letting the strong spring pull it back. In this model the spring is stronger for F than for O, so the energy released on combustion in fluorine is greater. One reason why all life on this planet uses a second-best oxidant, of course, is the natural rarity of fluorine, which in turn goes all the way back to elemental synthesis and the fact that fluorine has an odd rather than an even atomic number.

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Heats of Formation, page 19

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Heats of Formation

To summarize what has been said so far about heats of reaction, heats are always quoted per mole of an individual reactant or product. For the ethanol combustion in the table on the previous page, C H OH(l) + 3O (g) → 2CO (g) + 3H O(l) 2

5

2

2

2

the enthalpy change, or heat of reaction, can be quoted in three equivalent ways:

∆H=-327 kcal per mole of ethanol consumed, ∆H=-164 kcal per mole of CO produced, and 2 ∆H=-109 kcal per mole of O used, or liquid water produced, since the numbers of moles are the same.

2

Moreover, each of the above heats would be different if any one of the reactants or products was in a different physical state: ethanol vapor, water vapor, ice, solid CO , and so on. The reverse reaction, synthesis of ethanol 2

from carbon dioxide and water, would have the same heat but with opposite sign: +327 kcal per mole of ethanol produced.

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Heats of Formation, page 20

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Heats of Formation

Heats of reactions are additive in exactly the same way that the reactions they belong to are additive. From the information in the heat of combustion table, we can calculate how much heat would be given off by the combustion of a mole of water in fluorine gas, even though this reaction is not found in the table:

This calculation tells us that 60 calories more heat are obtained by burning hydrogen in F (Reaction 1) than in O (reverse of Reaction 2). Even the 2 2 "garbage" from O combustion (H O) still is a fuel for F combustion (Reaction 2

2

2

3), because F is more electronegative than O. This additivity of heats of reaction is a tremendous labour-saving device. One might think that, to have available complete heat of reaction data for all possible chemical reactions, it would be necessary to measure and tabulate all of these heats. This is not the case. It is necessary to tabulate heats only for the minimum number of reactions from which all other reactions can be obtained by suitable combinations.

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Heats of Formation, page 21

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Heats of Formation

There are several choices that could have been made; one example is the tabulation of the combustion reactions of all chemical compounds with O . The 2

choice that was made and agreed upon by chemists is that of tabulating the heat of formation of every substance from its elements in standard reference states. The standard state of an element is the state in which it ordinarily is found - solid, liquid, or gas - at room temperature (298K) and one atmosphere pressure. Therefore the heat of formation of any element in its standard state is zero by definition. If we have the heats of formation for all chemical compounds, all other reactions between compounds can be built up by the proper combination of formation reactions, and the heats of reaction can be built up in the same way. Example. Calculate the heat of combustion of ethyl alcohol (ethanol) from heats of formation of reactants and products. Solution. The balanced equation for combustion of ethanol is C H OH(l) + 3O (g) → 2CO (g) + 3H O(l) 2

5

2

2

2

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Heats of Formation, page 22

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Heats of Formation

The individual formation reactions for the substances appearing in the combustion process and their molar heats of formation, ∆H

o

are: 298

1. 2C(s) + 3H (g) + ½O (g) → C H OH(l) -66.4 kcal 2 2 2 6 2. O (g) → O (g) (already in standard state) 0 2 2 3. C(s) + O (g) → CO (g) -94.1 kcal 2 2 4. H (g) + ½O (g) → H O(l) -68.3 kcal 2

2

2

The standard state for carbon is solid graphite, and that for all other elements on the left side of the equations is the gas at one atmosphere. The superscript zero on ∆H indicates standard states for reactants and products, and the subscript 298 (often omitted and assumed) indicates that the heats quoted are for reactions begun at 298K, with the products brought back to the starting temperature at the conclusion of the reaction. The heats are expressed, as they usually are, as kilocalories per mole of compound formed. Values for some representative substances are shown in this table.

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Heats of Formation, page 23

12. Heat, Energy, and Chemical Bonds

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Heats of Formation

How can we calculate the heat of combustion of ethanol from these data? Notice that the combustion reaction can be obtained by the following combination of formation Reactions 1 through 4: combustion reaction = (products) - (reactants) combustion = 2(R3) + 3(R4) - (R1) - 3(R2) From the additivity principle, it follows that the heat of combustion of ethanol is found from the corresponding combination of heats of formation: heat of reaction = 2(-94.1) + 3(-68.3) - (-66.4) - 3(0) = -326.7 kcal Notice that we can manipulate these heats of formation as if they were absolute enthalpies of molecules, rather than enthalpies of formation of the molecules from elements. Everything pertaining to the elements from which they came cancels out on both sides of the equation.

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Heats of Formation, page 24

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Heats of Formation

For a general reaction of the type A + 4B → 3C + 2D, in which A, B, C, and D 0

are chemical substances, the heat of reaction can be calculated from ∆H = 0 0 0 0 0 3∆H + 2∆H - ∆H - 4∆H , in which ∆H is the heat of formation of C D A B C Compound C from its elements in their standard states. For example, if the reaction is the oxidation of glucose by nitrate, as is carried out by some nitrate-respiring bacteria 5C H O +24KNO → 6CO +24KHCO +12N +18H O 6

12

6

3

2

3

2

2

then the heat of reaction is obtained from the following combination of heats of formation of products and reactants: Nitrogen fixing nodules on a bean root formed by the Rhizobium etli bacterium

Tables of standard heats of formation can be found in the CRC Handbook of Chemistry and Physics, in Lange's Handbook of Chemistry. With such tables, the heats of all reactions involving the compounds can be calculated, including reactions that never have been carried out, or that for various reasons we cannot carry out easily in the laboratory.

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Energy in a Dissolving Salt, page 26

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12. Heat, Energy, and Chemical Bonds

-- Jump to --

Energy in a Dissolving Salt

A delicate balance determines whether the dissolving of a salt crystal in water will absorb or give off heat. Energy is required to pull apart the oppositely charged ions in the crystal, and this is

Predictions about heats of solution (∆H ) become tricky, because sol 5% errors in theoretical calculations of lattice and hydration energies will completely reverse a prediction. This table makes it

called the lattice energy, ∆H . lat In contrast, energy is liberated when the separated ions attract polar water molecules around themselves, and this is the heat of

look as if the ∆H

sol

values were obtained from ∆H

lat

and ∆H

.

hyd

hydration, ∆H . The observed heat effect on solution depends hyd on which factor predominates. Typical examples are as follows:

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Energy in a Dissolving Salt, page 26

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12. Heat, Energy, and Chemical Bonds

-- Jump to --

Energy in a Dissolving Salt

A delicate balance determines whether the dissolving of a salt crystal in water will absorb or give off heat. Energy is required to pull apart the oppositely charged ions in the crystal, and this is

Predictions about heats of solution (∆H ) become tricky, because sol 5% errors in theoretical calculations of lattice and hydration energies will completely reverse a prediction. This table makes it

called the lattice energy, ∆H . lat In contrast, energy is liberated when the separated ions attract polar water molecules around themselves, and this is the heat of

look as if the ∆H

sol

values were obtained from ∆H

lat

and ∆H

.

hyd

hydration, ∆H . The observed heat effect on solution depends hyd on which factor predominates. Typical examples are as follows:

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Energy in a Dissolving Salt, page 26

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12. Heat, Energy, and Chemical Bonds

-- Jump to --

Energy in a Dissolving Salt

A delicate balance determines whether the dissolving of a salt crystal in water will absorb or give off heat. Energy is required to pull apart the oppositely charged ions in the crystal, and this is

Predictions about heats of solution (∆H ) become tricky, because sol 5% errors in theoretical calculations of lattice and hydration energies will completely reverse a prediction. This table makes it

called the lattice energy, ∆H . lat In contrast, energy is liberated when the separated ions attract polar water molecules around themselves, and this is the heat of

look as if the ∆H

sol

values were obtained from ∆H

lat

and ∆H

.

hyd

hydration, ∆H . The observed heat effect on solution depends hyd on which factor predominates. Typical examples are as follows:

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Energy in a Dissolving Salt, page 26

12. Heat, Energy, and Chemical Bonds

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-- Jump to --

The Energy in a Dissolving Salt

In fact, the heats of solution are the experimentally measured quantities, and it is only with their help that we can get reasonable estimates of heats of hydration. The lattice energies are on somewhat firmer theoretical grounds. These "small" heats of solution are large enough to have obvious physical consequences. The cooling when ordinary table salt is dissolved in water is small, but can be felt if one makes a concentrated solution and uses an aluminum tumbler. Ammonium chloride absorbs so much heat when it dissolves that hoar frost may form on the outside of the beaker. In contrast, sodium hydroxide generates so much heat that the mixing beaker may become too hot to touch. To a certain extent, we can account for the trends in heats that we see in the table. Ammonium chloride has a weaker lattice energy than NaCl, because the 4+

+

NH ion is larger than Na and the binding attractions in the crystal are weaker. Unfortunately, the hydration energy also decreases with increasing ionic size, and it is difficult to predict whether lattice energy or hydration energy will show the greater change with larger ions. -

-

The heats of hydration of Cl and OH ions are similar, so in the comparison of NaOH with NaCl, the dominant effect comes from weaker crystal forces of NaOH in comparison with NaCl. The crystal structure of NaOH is in fact a badly distorted NaCl structure, with the distortion probably arising from the fact that -

the OH ions are nonspherical. It is possible that this distortion makes the NaOH lattice easier to pull apart.

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Bond Energies, page 27

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12. Heat, Energy, and Chemical Bonds

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Bond Energies

We began this chapter with a mention of making and breaking of bonds, but quickly began talking as if heats of reactions were nothing more than experimentally observable numbers, to be manipulated in whatever way was useful. Where do these heats come from? We can come a long way toward understanding chemical reactions by thinking of heats of reactions solely in terms of individual bond energies in molecules. The water molecule, H-O-H, has two O-H bonds. How much energy is required to tear these bonds apart and form isolated H and O atoms? This process is represented diagrammatically in the picture story below. We can make an experimental measurement of the standard heat of the reaction -1

H O(g) → H (g) + ½O (g) ∆H = +57.8 kcal mole H O 2

2

2

2

This is not precisely what we were after, however. After the water molecule is pulled into H and O atoms, the situation is complicated by the formation of H-H and O=O bonds.

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Bond Energies, page 27

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12. Heat, Energy, and Chemical Bonds

-- Jump to --

Bond Energies

We began this chapter with a mention of making and breaking of bonds, but quickly began talking as if heats of reactions were nothing more than experimentally observable numbers, to be manipulated in whatever way was useful. Where do these heats come from? We can come a long way toward understanding chemical reactions by thinking of heats of reactions solely in terms of individual bond energies in molecules. The water molecule, H-O-H, has two O-H bonds. How much energy is required to tear these bonds apart and form isolated H and O atoms? This process is represented diagrammatically in the picture story below. We can make an experimental measurement of the standard heat of the reaction -1

H O(g) → H (g) + ½O (g) ∆H = +57.8 kcal mole H O 2

2

2

2

This is not precisely what we were after, however. After the water molecule is pulled into H and O atoms, the situation is complicated by the formation of H-H and O=O bonds.

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Bond Energies, page 28

12. Heat, Energy, and Chemical Bonds

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Bond Energies

We must correct for the energy absorbed in taking these diatomic molecules apart: H (g) → 2H(g, atoms) 2 O (g) → 2O(g, atoms)

∆H = +52.1 kcal per mole of H atoms ∆H = +59.2 kcal per mole of O atoms 2 The desired process and its ∆H can be determined from the

This is what we were after: the energy required to tear a water molecule apart, not into elements in their standard states, but into isolated atoms. Half of this total energy can then be ascribed to each bond, giving an O-H bond energy of 110.6 kcal per mole of bonds.

previous reactions: From calculations such as this, we can obtain the table of bond energies shown above. These bond energies are not perfect for any one molecule, but are the best average values for a great number of molecules with the same kind of bond.

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Bond Energies, page 29

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12. Heat, Energy, and Chemical Bonds

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Bond Energies

The heats of atomization of elements are heats of dissociation of diatomic gas molecules such as H , N , O , or F , or of vaporization of solids such as 2

2

2

2

graphite and sulfur into atoms. Since the standard state of sulfur is a solid made up of packed S rings, atomization would entail first breaking van der 8 Waals forces and evaporating the S molecules, then taking them apart into 8

isolated sulfur atoms. Notice how the heats of atomization agree roughly with the number of bonds formed by each atom: four bonds in graphite, a triple N≡N bond in N , a double 2 bond in O=O, and a single bond in F-F. The single bond in H-H is almost as strong as the double bond in O=O because the H atoms are small and can get close to one another. The C=C double bond is not quite twice as strong as a single bond because the second shared electron pair does not have as favorable geometry for bonding as does the first pair. For the same reason, a triple bond is not three times as strong as a single bond. Carbon binds to hydrogen more strongly than to another carbon because of hydrogen's small size and the closer approach of atomic centers.

Sulphur S8 ring

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Heats of Reactions from Bond Energies, page 30

12. Heat, Energy, and Chemical Bonds

Page 1 of 1

-- Jump to --

Heats of Reactions from Bond Energies

How good are these approximate bond-energy values? How close can we come to reproducing measurable heats of reaction? As an example, let us calculate the heat of formation of ethanol vapor: 2C(graphite) + 3H2(g) + ½O2(g) → CH3CH2OH(g)

This is the heat of the hypothetical reaction

CH3CH2OH(g) → 2C(g) + 6H(g) + O(g) ∆H = +771.7 kcal -1 mole ethanol We must add to this the heat involved in atomizing solid graphite, and H2 and O2 gases: Ethanol molecule (C2H5OH) This sum is the heat of the reaction 2C(s) + 3H2(g) + ½O2(g) → 2C(g) + 6H(g) + O(g) ∆H = +715.1 kcal Page 30 of 39

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Heats of Reactions from Bond Energies, page 31

12. Heat, Energy, and Chemical Bonds

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Heats of Reactions from Bond Energies

We can get the formation reaction we are seeking by subtracting the first reaction from this one: 2C(s) + 3H (g) + ½O (g) → CH CH OH(g) 2 2 3 2 ∆H = (+715.1) - (+771.7) = -56.6 kcal per mole of ethanol Compare this with the value of -56.3 kcal mole-1 from the table of measured heats of formation given on Page 22. Accuracy to within one kilocalorie is considered quite good. We can see what is going on physically by means of the energy level diagram opposite. The isolated C, H, and O atoms are in a state 715.1 kcal higher in energy than graphite crystals and H and O gases, because 715.1 kcal are required to 2

2

produce the separated atoms. In turn, because the calculation from bond energies showed that 771.7 kcal are required to tear an ethanol molecule apart, the intact molecule must be 771.7 kcal lower in energy than the separated atoms. The observable heat of formation of ethanol molecules from elements (not atoms), unfortunately, is a small difference between two large numbers:

∆H = +715.1 - 771.7 = -56.6 kcal mole-1 This is one reason for the relative inaccuracy of bond-energy calculations. A small percent error in a number the size of 700 means a much bigger percent error in the difference of 57.

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Benzene and Resonance, page 32

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12. Heat, Energy, and Chemical Bonds

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Benzene and Resonance

One can calculate heats of formation from bond-energy values for hundreds of molecules, and never be in error more than a kilocalorie or two. However, in those cases where the discrepancy is large, one can learn something new about the nature of chemical bonding. Benzene is a good illustration of this. Let us try to calculate the standard heat of formation of benzene, C H . As we 6

6

saw in Chapter 9, benzene is an example of a molecule for which simple single bond and double bond ideas are inadequate, and structures using them, such as the Kekulé structures at the right, are wrong. Benzene has six electrons delocalized around the ring. From a bond-energy viewpoint, how bad is the localized Kekulé model? If we provisionally accept the Kekulé structures, then benzene has three C-C single bonds, three C=C double bonds, and six C-H single bonds. The energy involved in taking one mole of benzene molecules apart into atoms is

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Benzene and Resonance, page 33

12. Heat, Energy, and Chemical Bonds

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-- Jump to --

Benzene and Resonance

The energy needed to make the same number of isolated atoms from graphite and H gas is 2

The difference between these two numbers should be the heat of formation, as shown below: 0

∆H = 1343 - 1283 = +60 kcal per mole of benzene By this calculation, heat is absorbed, not emitted, when benzene is formed from its elements. We encounter trouble as soon as we -1

try to compare this calculated value of +60 kcal mole with reality. It is true that heat is absorbed when benzene is made -1

from carbon and hydrogen, but only 20 kcal mole , not 60.

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Benzene and Resonance, page 33

12. Heat, Energy, and Chemical Bonds

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Benzene and Resonance

The energy needed to make the same number of isolated atoms from graphite and H gas is 2

The difference between these two numbers should be the heat of formation, as shown below: 0

∆H = 1343 - 1283 = +60 kcal per mole of benzene By this calculation, heat is absorbed, not emitted, when benzene is formed from its elements. We encounter trouble as soon as we -1

try to compare this calculated value of +60 kcal mole with reality. It is true that heat is absorbed when benzene is made -1

from carbon and hydrogen, but only 20 kcal mole , not 60.

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Benzene and Resonance, page 34

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12. Heat, Energy, and Chemical Bonds

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Benzene and Resonance

As the second part of the energy-level diagram shows, -1

the "real" benzene molecule is 40 kcal mole more stable than the Kekulé bond model would predict. We saw the reason in terms of bonding in Chapter 9: In the benzene molecule, six electrons are delocalized around the ring (see diagram), and all six C-C bonds are the same length, intermediate between that of a single and double bond. Each bond has a partial double bond character, and the molecule as a whole gains 40 kcal -1

mole of added stability.

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Why is Fire Hot?, page 35

12. Heat, Energy, and Chemical Bonds

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-- Jump to --

Why is Fire Hot?

We can pull everything in this chapter together and bring it to a close by taking a second look at a question that was raised in Chapter 6 and again at the beginning of this chapter: Why is fire hot? The answer given previously was that oxidations give off heat because electrons in the product molecules are shifted toward the electronegative O or F atoms, and the molecules are more stable as a result. Now, with the aid of bond energies, we can stop being qualitative and put numbers to our argument. The bond-energy values in the table on Page 22 support the assertion that the more the electrons in a bond are shifted toward an electronegative atom, the more stable the bond is. Single-bond energies between H and some other elements are

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Why is Fire Hot?, page 35

12. Heat, Energy, and Chemical Bonds

Page 1 of 1

-- Jump to --

Why is Fire Hot?

We can pull everything in this chapter together and bring it to a close by taking a second look at a question that was raised in Chapter 6 and again at the beginning of this chapter: Why is fire hot? The answer given previously was that oxidations give off heat because electrons in the product molecules are shifted toward the electronegative O or F atoms, and the molecules are more stable as a result. Now, with the aid of bond energies, we can stop being qualitative and put numbers to our argument. The bond-energy values in the table on Page 22 support the assertion that the more the electrons in a bond are shifted toward an electronegative atom, the more stable the bond is. Single-bond energies between H and some other elements are

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Why is Fire Hot?, page 36

12. Heat, Energy, and Chemical Bonds

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Why is Fire Hot?

The same trend appears in bonds between carbon and other atoms:

The spring model seems reasonable so far. A bond-energy calculation of the heat of combustion of methane yields no surprises:

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Why is Fire Hot?, page 37

12. Heat, Energy, and Chemical Bonds

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Why is Fire Hot?

The observed heat of combustion can be calculated from the heats of formation listed on Page 22:

This is as good agreement with experiment as one could hope for when bondenergy values are used. So far we only have been manipulating numbers. Let us look at these numbers more closely and see what they tell us about shifts of electrons and energy, by comparing bond breaking and remaking in reactant and product molecules.

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Why is Fire Hot?, page 38

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12. Heat, Energy, and Chemical Bonds

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Why is Fire Hot?

Part of the stability of the C=O bonds in CO2 arises from resonance stabilization involving bond structures of the type

Had there been no resonance or delocalization in CO2, the -1

C=O bonds would have strengths of 174 kcal mole , as they do in other non-delocalized molecules, rather than the 192 -1 kcal mole that is observed in CO2. Without delocalization, the calculation in Part E would have been

We now can summarize the three components in the molar heat of combustion of methane:

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Why is Fire Hot?, page 39

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12. Heat, Energy, and Chemical Bonds

-- Jump to --

Why is Fire Hot?

Eighty-one percent of the heat of combustion of methane comes from the electronegativity of oxygen, and 19% from the special stability of product CO 2

molecules caused by the delocalization of electrons. Combustions of all types with oxygen are very exothermic because oxygen is quite electronegative and draws electrons to it (a feature shared with fluorine), and the oxide of carbon is given extra stability by delocalization (not shared by F). The special electronic properties of CO make up, in part, for the scarcity of 2

F in the universe and our having to settle for a second-best oxidant. Voltaire's Dr. Pangloss was right; this is almost the best of all possible worlds!

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Questions, page i

12. Heat, Energy, and Chemical Bonds

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-- Jump to --

Questions

1. How can potential energy be changed into kinetic energy? Give an example of the reverse process, changing kinetic energy into potential energy. 2. How can work be converted into potential energy? How does a water mill use potential energy, and what does it convert it into? Why does a water mill convert some of this energy into heat, and where does the heat appear? 3. Give an example of the conversion of heat into work, and of work into heat. 4. If the burning process gives off real heat, why do chemists say that the heat of reaction is negative? 5. Why are energy, E, and enthalpy, H, different? Which quantity represents the heat of a reaction carried out in a constant-pressure situation such as the open air of a laboratory? 6. How might you carry out an experiment such that the energy, E, rather than H, represented the heat of reaction? 7. What is the standard heat of formation of a substance? 8. Why is the standard heat of formation of a liquid always more negative than that of the corresponding gas? Illustrate with an energy-level diagram.

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Questions, page iii

12. Heat, Energy, and Chemical Bonds

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-- Jump to --

Questions

16. If both of the factors of Question 15 are large energy quantities, why is a prediction of solubilities of salts so difficult? 17. What is bond energy? Why is the standard heat of formation of a molecule not the same as the sum of bond energies of all of the bonds in the molecule? What is the missing component? 18. Why does a simple bond-energy calculation fail for benzene? 19. How does your answer to Question 18 also pertain to the heat of combustion of carbon-containing substances?

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