Principles of Reactivity Energy and Chemical Reactions

Some chemical reactions begin as soon as the reactants come into contact and continue until at least one reactant (the limiting reactant) is completely consumed. If a candy Bear, which is mostly sugar, is dropped into hot, molten potassium chlorate, the sugar reacts very vigorously with the potassium chlorate C6H12O6(s) + 4KClO3(ℓ) → 6CO2(g) + 6H2O(g) + 4KCl(s) Reducing agent Oxidizing agent

or, if aluminum and bromine are combined, they react rapidly with evolution of a great deal of energy to give aluminum bromide. Other reactions happen much more slowly, but reactants are still converted almost completely to products. An example is the rusting of iron at room temperature. Two questions about chemical reactions are important to chemists: will a reaction occur? and, if it does, how fast will it go? In this chapter we wish to investigate some aspects of the first question: how can we predict? when a reaction will occur. We begin by categorizing reactions as product- or reactant- favored systems. If a system is product-favored, most of the reactants are eventually converted to products without outside intervention, although ‘‘eventually’’ may last a very long time. Product-favored reactions are also those that usually, but not always, evolve energy. You already know something about product-favored reactions for example : ● Reactions that lead to a precipitate. ● Reactions in which an acid and a base combine to form water. ● Reactions that produce a gas, such as the decomposition of a metal carbonate in acid ● Oxidation-reduction reactions, such as the combination of metals with oxygen or halogens to give metal oxides and halides. Some other reactions have virtually no tendency to occur by themselves. We label these reactions as reactant-favored systems.

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For example, the splitting of NaCl into its elements

Reactant-favored reaction : 2NaCl(s) → 2Na(s) + Cl2(g)

is exactly the opposite of the product-favored reaction of sodium and chlorine

Product- favored reaction : 2Na(s) + Cl2(g) → 2NaCl(s) The splitting of NaCl is therefore reactant-favored. Evidence of this is that deposits of salt, NaCl, have existed on earth for millions of years without forming the elements Na and Cl2. What do we mean by outside intervention? Usually it is some flow of energy. For example, at very high temperatures, small but significant quantities of NO can be formed from air.

N2(g) + O2(g) → 2NO(g) Such high-temperature conditions are found in power plants and automobile engines, and a large number of such sources can produce enough NO and other nitrogen oxides to cause significant air pollution problems. Salt can be decomposed to its elements by providing heat to melt it and electric energy to separate the ions and form the elements. 2NaCl(ℓ) → 2Na(ℓ) + Cl2(g)

In each case energy can cause a reactant-favored system to produce products. Energy is a central idea when describing a reaction as product- or reactant- favored. For this reason it is useful to know something about energy and its interactions with matter. The most common of these interactions is the transfer of energy as heat or thermal energy when chemical reactions occur. Transfer of heat is a major topic of thermodynamics, the science of heat and work the subject of this chapter.

ENERGY : ITS FORMS AND UNITS Just what is energy, and where does the energy we use come from? Energy is defined as the capacity to do work, and work is something you experience all the time. If you climb a mountain, you do some work against the force of gravity as you carry yourself and your equipment up the mountain’s side. You can do this work because you have the energy or capacity to do so, the energy having been provided by the food you have eaten. Food energy is chemical energy-energy stored in chemical compounds and released when the compounds undergo the chemical reactions of metabolism . Energy can be assigned to one of two classes: kinetic or potential. An object has kinetic energy because it is moving. Examples are : ● Thermal energy of atoms, molecules, or ions in motion at the sub- microscopic level. All matter has thermal energy because, according to the kinetic-molecular theory, the submicroscopic particles of matter are in constant motion. ● Mechanical energy of a macroscopic object like a moving baseball or automobile. ● Electric energy of electrons moving through a conductor. ● Sound, which corresponds to compression and expansion of the spaces between molecules. Potential energy is energy that results from an object’s position. Examples are : ● Chemical potential energy resulting from attractions among electrons and atomic nuclei in molecules ● Gravitational energy, such as that of a ball held well above the floor or that of water at the top of a waterfall ● Electrostatic energy, such as that of positive and negative ions a small distance apart. Potential energy is stored energy—it can be converted into kinetic energy. For example as water droplets fall over a waterfall, the potential energy of the water is converted into kinetic energy, and the drops move faster and faster. Similarly, kinetic energy can be converted to potential energy. The kinetic energy of falling water can turn a turbine to produce electricity.

Temperature , Heat, and the Conservation of Energy When you stand on a diving board, poised to dive into a swimming pool, you have considerable potential energy because of your position above the water. Once you jump off the board, some of that potential energy is converted progressively into kinetic energy, which depends on your mass (m) and velocity (v).

Kinetic energy of a body in motion = ½ (mass) . (velocity)2 = ½ m.v2

During the dive, your mass is constant, whereas the force of gravity accelerates your body to move faster and faster as you fall, so your velocity and kinetic energy increase. This happens at the expense of potential energy. At the moment you hit the water, your velocity is abruptly reduced, and much of your kinetic energy is converted to mechanical energy of the water, which splashes as your body moves it aside by doing work on it. Eventually you float on the surface, and the water becomes still again. If you could see them, however, you would find that the water molecules were moving a little faster in the vicinity of your dive; that is, the temperature of the water would be a little higher. This series of energy conversions, from potential to kinetic, and from kinetic to heat, illustrates the law of energy conservation, which states that energy can neither be created nor destroyed the total energy of the universe is constant. This law summarizes the results of a great many experiments in which heat, work, and other forms of energy transfer have been measured and the total energy found to be the same before and after an event. The law of energy conservation is the reason we have been careful not to say that when a lump of coal is burned, its energy has been used up. What has been used up is an energy resource: the coal’s capacity to transfer heat energy to its surroundings when it is burned. If the coal is burned in a power plant, its chemical energy can be changed to an equal quantity of energy in other forms. These are mainly electricity, which can be very useful, and thermal energy in the gases going up the smoke - stack. It is not the coal’s energy, but rather its ability to store energy and release it in a form that people can use that has been used up.

Thermal energy, or heat, is associated with temperature, but heat is not the same as temperature. Transferring energy by heating an object increases the object’s temperature, and the temperature increase can be measured with a thermometer. For example, when the thermometer is placed into hot water, heat is transferred from the water to the thermometer. The increased energy of the mercury atoms means that they are moving about more rapidly, which slightly increases the volume of the spaces between the atoms. Consequently, the mercury expands (as most substances do when heated), and the upper end of the column of mercury rises higher in the thermometer tube. Energy transfer, as heat, happens when two objects at different temperatures are brought into contact. Energy always transfers spontaneously from the hotter to the cooler object. For example, a piece of metal being heated in a Bunsen burner flame and a beaker containing cold water are two objects with different temperatures. When the hot metal is plunged into the cold water, heat is transferred from the metal to the water until the two objects reach the same temperature. Two important aspects of thermal energy should be understood: ● The more energy an atom or molecule has, the faster it moves. ● The total thermal energy in an object is the sum of the individual energies of all the atoms or molecules in that object. The average speed of motion of atoms and molecules is related to the temperature. The total thermal energy depends on temperature, the types of atoms or molecules, and the number of them in a sample. For a given substance thermal energy depends on temperature and the amount of substance. Thus a cup of steaming coffee may contain less thermal energy than a bath - tub full of warm water, even though the coffee is at a higher temperature. Under most circumstances, objects in a given region, such as your room, are at about the same temperature.

If an object such as a cup of coffee or tea is much hotter than this, it transfers energy by heating the rest of the objects in your room until the hot coffee cools off (and your room warms up a bit). If an object, say a cold glass of ice water, is much cooler than its surroundings in your room, heat is transferred to the water from everything else until it warms up (and your room cools off a little). Because the total amount of material in your room is much greater than in a cup of coffee or tea or a glass of ice water, the room temperature changes very little. Heat transfer occurs until everything is at the same temperature . Energy Units Some of the units in which energy is measured were originally designed to measure heat. A calorie was originally defined as the quantity of energy (transferred by heating) that is required to raise the temperature of 1.00 g of pure liquid water by 1.00 degree Celsius, from 14.5 ◌C to 15.5 ◌C. The calorie represents a very small quantity of heat, and because we often work with larger quantities of matter, the kilocalorie (1000 calories) is used instead. Most of us tend to think of heat as measured in calories, probably because we hear about dieting or read breakfast cereal boxes. Currently most chemists use the joule ( J), the SI unit of energy. One calorie is now defined as equivalent to 4.184 J. The joule is preferred because it is derived directly from the units used in the calculation of mechanical energy (i.e., potential and kinetic energy). If a 2.0-kg object (about 4 pounds) is moving with a velocity of 1.0 meter per second (roughly 2 mph), the kinetic energy is Kinetic energy = ½ m.v2 =½(2kg).(1m/s)2 = 1kg.m2/s2 = 1J

Like the calorie, the joule is often inconveniently small as a unit for many purposes in chemistry and other sciences, and so the kilo joule (kJ), which is equivalent to 1000 joules, is often used. For example, burning peanuts may produce 40,000 J or more per gram of peanut oil. Chemists find it more convenient to express large numbers such as these in kilojoules, that is, as 40 kJ/g in this case.

SPECIFIC HEAT AND THERMAL ENERGY TRANSFER In the 1770s Joseph Black studied the nature of heat and was the first to distinguish clearly between the hotness ( or temperature) of an object and its heat capacity. The specific heat capacity, C, of a substance is defined as the ratio of the heat supplied to some mass of the substance (say, 1.00 g) to the consequent rise in the substance’s temperature. Mathematically, we express this as Specific heat capacity (C)=quantity of heat supplied/(mass of object).(temperature change) C (J/g.K) = q (J) / m (g) . ∆T (K) where the quantity of heat transferred by heating is symbolized by q. In chemistry we shall use heat in units of joules, masses in grams, and temperatures in kelvins. The specific heat of a substance, which is often just called its specific heat, is therefore expressed in units of joules per grams times kelvins ( J/g . K). Specific heat is determined by experiment. For example, it has been found that 60.5 joules is required to change the temperature of 25.0 g of ethylene glycol (a compound used as antifreeze in automobile engines) by 1.00 K. This means the specific heat of the compound is Cethylene glycol = 60.5 J /(25 g) . (1 K) = 2.42 J/g.K

The specific heat of many substances have been determined experimentally. The wide differences in the specific heat of substances can be used to distinguish them from each other just as density can. Water has one of the highest known values of specific heat. For water, the specific heat is 4.184 J/g . K, whereas it is only about 0.45 J/ g . K for iron or 0.84 J/g . K for common glass. That is, it takes about nine times as much heat to raise the temperature of a gram of water 1 K as it does for iron, or about five times as much heat is required to increase the temperature of a gram of water by the same amount as a gram of glass. The high specific heat of water also informs us that a considerable quantity of thermal energy must be transferred out of the substance before it cools down appreciably. For example,1.00 g of water must give up 4.18 J of thermal energy to cool one kelvin, whereas 1.0 g of glass gives up only 0.84 J, or about one-fifth as much thermal energy as water for the same temperature change.

The very high specific heat of water is important to all of us, because many joules must be absorbed by a large body of water to raise its temperature just a degree or so. Conversely, a great deal of energy must be lost before the temperature of the water drops by more than a degree. Thus, a lake can store an enormous quantity of energy, and so bodies of water have a profound influence on our weather. The equation defining specific heat informs us that the larger the specific heat, the more thermal energy a substance can store. Furthermore, substances with a high specific heat often cool down more slowly than those with a smaller specific heat. If you wrap some bread in aluminum foil and heat it in an oven, you know you can remove the foil after taking the bread from the oven without burning your fingers, even though the bread is very hot. This is also the reason that a chain of fast food restaurants warns you that the filling of an apple pie can be much warmer than the paper wrapper or the pie crust . Knowing the specific heat of a substance allows you to calculate the heat transferred to or from that substance if the mass of substance and its temperature change are also known. Conversely, you can calculate the temperature change that should occur when a given quantity of heat is transferred to or from a sample of known mass. The equation that defines specific heat allows you to do this, but it may be more convenient to rewrite it in the following form Heat transferred = (specific heat).(mass).(temperature change of substance) q = ( C in J/g.K ).( m in g ).( ∆T in K ) For example, the thermal energy required to warm a 10.0 g piece of copper from 25 ◌C (298 K) to 325 ◌ C (598 K) is q = 10 g x 0.385 J/g.K x (598 – 298)K = 1160 J To this point, we have emphasized only the quantity of heat transferred. Equation, however, allows a demonstration not only of the quantity of heat but also the direction in which it is transferred. When ΔT is determined as in the previous calculation, that is, as : ∆T = Final temperature – Initial temperature it has an algebraic sign: positive (+) for T increase (Tf > Ti) and negative (-) for a decrease (Tf < Ti). If the temperature of the substance increases, ΔT has a positive (+) sign and so does q. This means heat was transferred to the substance(as in the example of heating a piece of copper).

The opposite case, a decrease in the temperature of the substance, means ΔT has a negative (-) sign and does q ; heat was transferred from the substance. Let us find the number of joules of heat energy transferred from a cup of coffee to your body and the surrounding air if the temperature of a cup of coffee held in your hand drops from 60.0 ◌C (333.2 K) to 37.0 ◌C (310.2 K) (normal body temperature). Assume the coffee has a mass of 250. g; also assume the specific heat of coffee is the same as water. The quantity of heat transferred can be obtained from Equation. Heat transferred = q = 4.184 J/g.K x (250 g) x (310 – 333)K = 24100 J = 24.1 kJ Notice that the final answer has a negative value. In this case heat energy is transferred from the coffee to the surroundings, and the temperature of the coffee decreases.

Units of Specific Heat In this textbook we have given specific heat values in units of J/g . K. This means that units of joules are divided by units of grams and kelvins. An alternative method of writing this is J g-1 K-1. Recall that 1/10 is the same as 10-1, so 1/g is the same as g-1, for example. Specific heat values given in handbooks of chemistry are often molar specific heat. For example, liquid water has a specific heat of 4.184 J/g . K, or 75.4 J/mol . K. 4.184 J/g.K x 18 g/mol = 75.4 J/mol.K E X E R C I S E 6.3 Using Specific Heat If 24.1 kJ is used to warm a piece of aluminum with a mass of 250. g, what is the final temperature of the aluminum if its initial temperature is 5.0 ◌C. The specific heat of aluminum is 0.902 J/g . K. Using the Concept of Energy Conservation In Example 6.3 you learned that the thermal energy transferred from a hot piece of metal was equal to the thermal energy transferred to the water. We expressed this as qmetal =-qwater, where the negative sign reflects the fact that qmetal has a negative value (qmetal = -9820 J). Another way to look at this is to write the equation as qmetal + qwater = 0. This tells us that net value of the energy transferred must be zero; energy must be conserved.

ENERGY AND CHANGES OF STATE So far we have described transfers of energy that occur between objects as a result of temperature differences. But energy transfers also occur when matter is transformed from one form to another in the course of a physical or chemical change. When a solid melts, its atoms, molecules, or ions move about vigorously enough to break free of the constraints imposed by their neighbors in the solid. When a liquid boils, particles move much farther apart from one another. In both cases attractive forces among the particles must be overcome, which requires an input of energy. Joseph Black was the first to recognize that heat is associated with changes of state, which always take place at constant temperature. The quantity of heat required to melt ice at 0 ◌C is 333 J/g and is called the heat of fusion of the ice. (500 kJ) required to raise the temperature of a 1.00-kg block of iron from 0 ◌C to 1100 ◌C, at which it glows red hot, leads only to the melting of 1.50 kg of ice, at which point the ice and the melted water both have a temperature of 0 ◌C. 500 x 103 J x 1 g / 333 J = 1.5 kg ice melted The quantity of heat required to convert liquid water to vapor, called the heat of vaporization, is 2260 J/g at 100 ◌C. What quantity of water can be vaporized if 500. kJ of heat is transferred to water at 100 ◌C ? Only 221 grams 500 x 103 J x 1 g / 2260 J = 221 g water vaporized Heat transfer to an object can lead to a temperature or a phase change (or both). Here the same quantity of heat (500 kJ) has been transferred to a one-kilogram block of iron at 0 ◌C as has been transferred to a two-kilogram block of ice at 0 ◌C. The temperature of the iron block has increased by 1100 K and makes the block red hot. On the other hand, this quantity of heat has led only to the melting of 1.5 kg of ice, and the ice and melted water are still at 0 ◌C.

Figure 6.9 A graph illustrating the quantity of heat absorbed and the consequence temperature change as 500. g of water is warmed from -50 ◌C to 200 ◌C. The graph also Illustrates the heat evolved when steam at 200 ◌C is cooled to -50 ◌C.

Figure 6.9 illustrates the quantity of heat absorbed and the consequent temperature change as 500. g of water is warmed from -50 ◌C to 200 ◌C. First, the temperature of the ice increases as heat is added. On reaching 0 ◌C, however, the temperature remains constant as sufficient heat is absorbed to melt the ice to liquid water. When all the ice has melted, the liquid absorbs heat and is warmed to 100 ◌C, the boiling point of water. The temperature is again constant as heat is absorbed to completely convert the liquid to vapor, that is, to steam. Any further heat absorbed heats the steam to 200 ◌C. ENTHALPY Carbon dioxide undergoes a change of state from solid to gas, a process called sublimation, at -78 ◌C: CO2(s, -78oC) → CO2(g, -78oC) heat

To describe the change from solid to gas in thermodynamic terms, we first need to extend the earlier discussion of heat as energy transferred between objects. In thermodynamics one of the ‘‘objects’’ is usually of primary concern. It may be a substance involved in a change of state or substances undergoing a reaction. It is called the system. The other ‘‘object’’ is called the surroundings and includes everything outside the system that can exchange energy with the system. A system may be contained within an actual physical boundary, such as a flask or a cell in your body. Alternatively, the boundary may be purely imaginary. Suppose our system consists of solid and gaseous CO2 at -78 ◌C, that the surroundings are also at -78 ◌C, and that one mole CO2 (solid) is transformed into one mole CO2 (gas). Just as the potential energy of a ball is greater when it is farther above the earth, which attracts it, the potential energy of a CO2 molecule is greater when it is farther from another CO2 molecule, again because an attractive force must be overcome when the two are separated. Thus, the energy of a mole of CO2(g) is greater than the energy of a mole of CO2(s). Where does this energy come from? That depends on the circumstances. Let us begin with the case in which CO2(s) and CO2(g) are inside a rigid container for one molecule to escape from solid to gas can be supplied by neighboring molecules which consequently vibrate a little less about their position in the crystal lattice. Phase Change Direction of Heat Transfer - Sign of qv(system) -Type of Change

ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ CO2(solid)→CO2(gas) Surrounding →System Positive Endothermic CO2(gas)→CO2(solid) System→ Surrounding Negative Exothermic As more and more molecules escape, the motion of remaining solid-state molecules diminishes more and more, which would lead to a lowering of the temperature. Atoms and molecules in the surroundings (the rigid container) are in contact with CO2 molecules in the solid, however, and energy exchange can occur. As soon as the temperature of the solid CO2 drops slightly below the temperature

of the surroundings, more energy transfers into the solid CO2 than transfers out, and the temperature of the solid CO2 is restored to -78 ◌C. For the temperature not to drop below -78 ◌C, the heat transfer from the surroundings must exactly equal the energy required to overcome attractive forces as solid CO2sublimes. What we have just described is transfer of energy as heat from the surroundings to the system while the process CO2 (s) → CO2 (g) takes place. In such a case the temperature of the system at first decreases, but then thermal energy is transferred from the surroundings and constant temperature is maintained. When heat must be transferred into a system to maintain constant temperature as a process occurs, the process is said to be endothermic. In contrast, the reverse process, converting CO2 gas to solid CO2 (dry ice), is an exothermic process; heat is transferred out of the sample of CO2 to the surroundings when some of the CO2(g) condenses to a solid. In summary, when the endothermic process CO2(s)→ CO2(g) occurs inside a rigid container, the only exchange of energy is the heat transferred into the system. Thus, by conservation of energy, the heat absorbed at constant volume, qv, must equal the change in the energy of the system, ΔE. If the process does not occur in a rigid container the situation is a bit more complicated, because two occur, not just one.

Crushed dry ice has been placed in the plastic bag.

(c) The bag has been closed. Some dry ice has sublimed , converting solid CO2 at -78 ◌C to gaseous CO2 at -78 ◌C. The gas has filled the bag and has done work by raising a book that has been placed on the bag. (d) A similar experiment but without the book. The expanding CO2 in this case has pushed aside the air that formerly occupied the space taken up by the inflated bag. Pushing aside the atmosphere requires work, just as raising a book does. As shown in Figure 6.11c, CO2 (s) vaporizing into a flexible container can raise a book above a table top. Raising a book is an example of doing work. Not only is energy transferred into the solid CO2 as heat, but energy is also transferred out of the CO2 gas as work. Work is done whenever something is caused to move against an opposing force. In this case the gas has to work against the weight of the book to raise it. Because energy is now transferred in two ways between system and surroundings, it is no longer true that ΔE = q. Energy transferred in or out of the system by work, symbolized by w, must also be taken into account. The change in energy of the system is now: ∆E = q + W That is, the energy of the system is changed by the quantity of energy transferred as heat and by the quantity of energy transferred by work. This is a statement of the first law of thermodynamics, also called the law of energy conservation. Another way of saying the same thing is that the total amount of energy in the universe is constant. Energy may be transferred as work or heat, but no energy can be lost, nor can heat or work be obtained from nothing. All the energy transferred between a system and its surroundings must be accounted for as heat and work. [This is true as long as no other energy transfer, such as the radiant energy of a glowing light stick is involved.] The first law is important in all aspects of our lives; it plays a central role in our models of weather, in designing a power plant, or in understanding why diet and exercise together lead to weight loss. Even if the book had not been on top of the plastic bag in Figure 6.11c, work would have been done by the expanding gas.

This is because whenever a gas expands into the atmosphere it has to push back the atmosphere itself. Instead of raising a book, the expanding gas moves a part of the atmosphere. For the process shown in Figure 6.11d, the energy transferred from the surroundings has two effects: (1) it overcomes the forces holding the molecules together in the solid state at -78 ◌C, and (2) it does work on the atmosphere as the gas expands. The energy that allowed the system to perform work on its surroundings had to come from somewhere, and that ‘‘somewhere’’ was the energy, q, transferred as heat to the system from its surroundings. The system converted part of that heat into work. Therefore, the first law of thermodynamics tells us that the energy change, ΔE, for the CO2 system must be less than q. The reason for this is that heat is transferred into the system from its surroundings (q has a positive sign), and work is done by the system on the surroundings (w has a negative sign). That is, ΔE = q + w = (q, positive number) + (w, negative number) and so: ΔE < q In plants and animals, as well as in the laboratory, reactions usually occur at equals a quantity called the enthalpy change, symbolized by ΔH (‘‘delta H’’). Thus, ΔH = qp = heat transferred to or from a system at constant pressure. Because ΔE = q + w, then at constant pressure ΔE = ΔH +w, showing that ΔH accounts for all the energy transferred except the amount that does the work of pushing back the atmosphere (which is usually small compared with ΔH). Even if pressure is not constant, ΔH = (enthalpy of the system at the end) - (enthalpy of the system at the start) = Hfinal ــــ Hinitial

Figure 6.13 Enthalpy diagram for the Interco version of solid CO2 and CO2 gas at constant pressure.

If the enthalpy of the final system is greater than that of the initial system (as when solid CO2 changes to CO2 vapor at -78 ◌C), the enthalpy has increased, and the

process has a positive ΔH; qp must also be positive and the process is endothermic. Conversely, if the enthalpy of the final system is less than that of the initial system, heat has transferred out of the system, and ΔH is negative; the process is exothermic. Another endothermic process is the evaporation of water to water vapor at 25 ◌C. The evaporation of one mole of water requires 44 kJ. H2O(ℓ) → H2O(g) ; ∆Hv = + 44 kJ/mol (change is endothermic) 44 kJ of heat energy transferred from surrounding to the liquid water . But what about water vapor condensing to form liquid again?

If 44.0 kJ of heat energy is required to break the attractions between H2O molecules in one mole of the liquid so they can move into the gas phase, the same quantity of energy (44.0 kJ per mole) is regained when the molecules in the vapor condense to form the liquid. Condensation of water is exothermic; 44.0 kJ per mole is transferred from the water to the surroundings when a mole of H2O molecules condenses to liquid: H2O(g) → H2O(ℓ) ; ∆Hc = - 44 kJ (change is exothermic) 44 kJ of heat energy transferred to surrounding from the vapor water . Some key ideas that apply to all of thermodynamics may be summarized here. ● When heat transfer occurs (at constant pressure) from a system to its surroundings, as when a vapor condenses to a liquid or solid, the process is exothermic with respect to the system and ΔH(qp) has a negative value. Conversely, when heat is absorbed from the surroundings, as when a solid or liquid changes to a vapor, the process is endothermic with respect to the system and ΔH(qp) has a positive value. ● For changes that are the reverse of each other, the ΔH values are numerically the same, but their signs are opposite. Thus, for evaporation of water, ΔH=+44.0 kJ/mol, whereas for the condensation of water ΔH=-44.0 kJ/mol. ● The change in energy or enthalpy is directly proportional to the quantity of material undergoing a change. If two moles of water are evaporated, twice as much heat energy or 88.0 kJ is required. ● The value of ΔH is always associated with a balanced equation for which the coefficients are read as moles, so that the equation shows the macroscopic amount of material to which the value of ΔH applies. Thus, for the evaporation of two moles of water : 2H2O(ℓ) → 2H2O(g) ; ∆H = 88 kJ

ENTHALPY CHANGES FOR CHEMICAL REACTIONS The enthalpy change can be determined for any physical or chemical change. For a chemical reaction the products represent the ‘‘final system’’ and the reactants the ‘‘initial system.’’ Therefore, ∆H = Hproducts – Hreactants

Like changes of state, chemical reactions can be exothermic or endothermic. Many reactions [such as the Al + Br2 reaction and the reactions of elements with air] are exothermic. To learn more about enthalpy changes that accompany chemical reactions, consider the decomposition of one mole of water vapor to its elements. H2O(g) → H2(g) + ½O2(g) ; ∆H = +242 kJ/mol (change endothermic) Decomposition of 1 mol water vapor requires 242 kJ of energy to be transferred in from surrounding .

(Note that to write an equation for the decomposition of one mole of H2O, it is necessary to use a fractional coefficient for O2. This is acceptable in thermo- chemistry because coefficients are always taken to mean moles and not molecules.) The left side of Figure 6.14 shows that the enthalpy of the products of this reaction is greater than that of the reactants. Figure 6.14 Enthalpy diagram for the Interco version of water vapor, H2O(g), liquid water, H2O(ℓ), and the elements.

Water vapor would have to absorb 242 kJ of energy (at constant pressure) from the surroundings as it decomposes to its elements. That is, the enthalpy change for the endothermic decomposition of water is ΔH=+242 kJ per mole of water. The decomposition of water can be reversed; hydrogen and oxygen can combine to form water. The quantity of heat energy involved in this oxidation reduction reaction (Fig. 6.14)

H2(g) + ½O2(g) → H2O(g) ; ∆H=-242kJ/mol (change is exothermic) is the same as for the decomposition reaction except that the combination of the elements is exothermic. That is, ΔH =-242 kJ per mole of water vapor formed. As in the case of phase changes, some key ideas about enthalpy changes for reactions can be summarized: ● When heat is evolved or transferred (at p=const.) to the surroundings by an exothermic reaction, ΔH has a negative value. Conversely, when heat is absorbed from the surroundings by an endothermic reaction, ΔH has a positive value. ● For chemical reactions that are the reverse of each other, the ∆H values are numerically the same, but their signs are opposite. ● The change in energy or enthalpy is directly proportional to the quantity of material undergoing a change. If 2 moles of water are decomposed, twice as much heat energy, or 484 kJ, is required. ● In thermodynamics, the value of ΔH is always associated with a balanced equation for which the coefficients are read as moles, so that the equation shows the macroscopic amount of material to which the value of ΔH applies. Thus, for the decomposition of 2 moles of water vapor H2O(g) → 2H2(g) + O2(g) ; ∆H=+484 kJ Because energy is transferred when a substance undergoes a change of state, the quantity of energy associated with a chemical reaction must depend on the physical state (solid, liquid, or gas) of the reactants and products. For example, the decomposition of one mole of liquid water to H2 and O2

H2O(ℓ) → H2(g) + ½O2(g) ; ∆H=+286 kJ/mol requires more energy than the decomposition of one mole of water vapor, as shown in Figure 6.14.

The heat of vaporization, 44 kJ/mol, must be added to the enthalpy change for the decomposition of H2O(g) to give the value for the decomposition of H2O(ℓ). Enthalpy changes for reactions have many practical applications. For instance, when enthalpies of combustion are known, the quantity of heat transferred by the combustion of a given mass of a fuel such as propane, C3H8, can be calculated. Suppose you are designing a heating system, and you want to know how much heat is provided by burning 454 grams (1 pound) of propane gas in a furnace. The exothermic reaction that occurs is C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(ℓ) ; ∆H = -2220 kJ/mol

and the enthalpy change is ΔH=-2220 kJ per mole propane burned. How much heat is transferred to the surroundings by burning 454 g of C3H8 gas? The first step is to find the number of moles of propane present in the sample. N = W/M =454 g / 44 g.mol-1 = 10.3 mol Then you can multiply by the amount of heat transferred per mole of gas. 10.3 mol x 2220 kJ/mol = 22900kJ0.3 This is a substantial quantity of energy when you compare it with the fact that, when your body completely ‘‘burns’’ 454 g of milk, only about 1400 kJ is evolved, in part because milk is largely water.

HESS’S LAW Knowing how much heat is transferred as a chemical process occurs is very important. First of all, the direction of heat transfer is an important clue that helps predict in which direction a chemical reaction will go because, at room temperature, most exothermic reactions are product-favored.

Second, we can use ΔH to calculate the heat obtainable when a fuel is burned, as was done. Third, when reactions are carried out on a larger scale, as in a chemical plant.

We therefore want to know ΔH values for as many reactions as possible. For many reactions direct experimental measurements can be made by using a device called a calorimeter, but for many other reactions this is not a simple task. Besides, it would be very time-consuming to measure values for every conceivable reaction, and it would take a great deal of space to tabulate so many values. Fortunately there is a better way. It is based on the fact that mass and energy are conserved in chemical reactions. Energy conservation is the basis of Hess’s law, which states that, if a reaction is the sum of two or more other reactions, then ΔH for the overall process must be the sum of the ΔH values of the constituent reactions. For example, as illustrated in the preceding section (see Figure 6.14) for the decomposition of liquid water into the elements H2(g) and O2(g) (with all substances at 25 ◌C), the two successive changes are (1) the vaporization of liquid water and (2) the decomposition of water vapor to the elements (with all substances at 25 ◌C). The equation and the ΔH value for the overall process can be found by adding the equations and the ΔH value for the overall process can be found by adding the equations and the ΔH values for the two steps: (1) H2O(ℓ) → H2O(g) ; ∆H1= +44 kJ/mol (2) H2O(g) → H2(g) + ½O2(g) ; ∆H2 = +242 kJ/mol ـــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ (1)+(2) H2O(ℓ) → H2(g) + ½O2(g) ; ∆H3 = +286 kJ/mol

Here, H2O(g) is a product of the first reaction and a reactant in the second. Thus, as in adding two algebraic equations in which the same quantity or term appears on both sides of the equation , H2O(g) can be canceled out. The net result is an equation for the overall reaction and its associated enthalpy change. Hess’s law is useful because it enables us to find the enthalpy change for a reaction or state change that cannot be measured conveniently.

Suppose you want to know the enthalpy change for the formation of carbon monoxide, CO, from the elements, solid carbon (as graphite) and oxygen gas. C(s) + ½O2(g) → CO(g) ; ∆H1 = ?

Experimentally this is not easy to do; it is difficult to keep the carbon from burning completely to carbon dioxide. The way to solve this is to recognize that we can more easily determine by experiments the enthalpy change for the conversion of carbon to CO2 and for the conversion of CO to CO2. C(s) + O2(g) → CO2(g) ; ∆H3 = -394 kJ/mol CO(g) + ½O2(g) → CO2(g) ; ∆H2 = -283 kJ/mol

Figure 6.15 One mole of carbon dioxide can be formed from carbon and oxygen in a direct reaction (ΔHnet=-393.5 kJ) or in two steps. Hess’s law states that the sum of the enthalpy changes for the two steps must equal that for the direct reaction. That is, ΔH1 + ΔH2 = ΔHnet. Knowing any two of these values, therefore, allows us to calculate the third.

As the diagram in Figure 6.15 makes clear, the formation of CO is the first of two steps in going from carbon to carbon dioxide. We know the enthalpy change for the second step (ΔH2), and we know the enthalpy change for the net reaction (ΔHnet). Using Hess’s law, therefore, we can solve for the enthalpy change for the first step (ΔH1). This can be done readily, because ΔH for the net reaction is the sum of the ΔH values for the first and second steps. C(s) + ½O2(g) → CO(g) ; ∆H1 = ? CO(g) + ½O2(g → CO2(g) ; ∆H2 = -283 kJ/mol ــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــــ

C(s) + O2(g) → CO2(g) ; ∆H3 = -394 kJ/mol ∆H3 = ∆H1 + ∆H2

∆H1 = -394 – (-283) = -111 kJ/mol

E X A M P L E 6.6 Using Hess’s Law Suppose we want to know the enthalpy change for the formation of methane , CH4, from solid carbon (as graphite) and hydrogen gas. C(s) + 2H2(g) → CH4(g) ; ∆H = ? The enthalpy change for the direct combination of the elements would be extremely difficult to measure in the laboratory. We can measure ΔH, however, when the elements and methane burn in oxygen . (1) C(s) + O2(g) → CO2(g) ; ∆H = -394 kJ/mol (2) H2(g) + ½O2(g) → H2O(ℓ) ; ∆H = -286 kJ/mol (3) CH4(g) + 2O2(g) → CO2(g) + 2H2O(ℓ) ; ∆H = -890 kJ/mol Solution The three reactions (1, 2, and 3), as they are now written, cannot be added together to obtain the equation for the formation of CH4 from its elements. CH4 should be a product, but it is a reactant in equation (3). The solution to this is to reverse equation (3). At the same time, the sign of ΔH for the reaction is reversed. If a reaction is exothermic in one direction (the combustion of methane evolves energy), its reverse must be endothermic. (3)' CO2(g) + 2H2O(g) → CH4(g) + 2O2(g) ; ∆H = -∆H3 = +890kJ/mol Next, we see that two moles of H2O(ℓ) is required as a product, whereas equation (2) is written for only one mole of water. We therefore multiply the stoichiometric coefficients in (2) by 2 and multiply the value of ΔH by 2. 2(2) 2H2(g) + O2(g) → 2H2O(ℓ) ; 2∆H2 = 2mol(-286kJ/mol)= -572 kJ With these modifications, we can add the three equations to give the equation for the formation of methane from its elements. (1) C(s) + O2(g) → CO2(g) ; ∆H = -394 kJ/mol 2(2) 2H2(g) + O2(g) → 2H2O(ℓ) ; ∆H = -572 kJ/mol (3)' CO2(g) + 2H2O(ℓ) → CH4(g) + 2O2(g) ; ∆H = +890 kJ/mol

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C(s) + 2H2(g) → CH4(g) ; ∆Hnet = -75 kJ/mol

C(s) + 2H2(g) → CH4(g) ; ∆Hnet = -75 kJ/mol

The solution to this problem is shown diagrammatically in Figure 6.16.

Figure 6.16 Methane may be formed directly from the elements. Alternatively, the carbon may be burned to CO2 (g) and the hydrogen burned to water. Then CO2 and H2O are combined to form CH4 (and the O2 used in the combustion is returned). Hess’s law states that the enthalpy change for the direct reaction (ΔHnet) is the sum of the enthalpy changes along the alternative path (ΔHnet = ∆H1+2∆H2+(-ΔH3). See Example 6.6 for a more complete designation of the ∆H values.

STATE FUNCTIONS

Hess’s law works because the enthalpy change for a reaction is a state function, a quantity whose value is determined only by the state of the system. The enthalpy change for a chemical or physical change does not depend on the path you choose to go from the initial conditions to the final conditions. No matter how you go from reactants to products in a reaction the net heat evolved or required (at constant pressure) is always the same. This point is illustrated in Figure 6.15 for the formation of CO2 from its elements and in Figure 6.16 for the formation of methane. For the reaction C(s) + 2H2(g) → CH4(g) ; ∆Hnet = -75 kJ/mol

the enthalpy change for the direct reaction (ΔHdirect=-74.8 kJ) is the same as the sum of the enthalpy changes along the indirect pathway that goes through carbon dioxide and water.

Many commonly measured quantities, such as the pressure of a gas, the volume of a gas or liquid, the temperature of a substance, and the size of your bank account, are state functions. The volume of a balloon is also a state function. You can blow up a balloon to a large volume and then let some air out to arrive at the desired volume. Alternatively, you can blow up the balloon in stages, adding tiny amounts of air at each stage. The final volume does not depend on how you got there . An infinite number of ways exist for how to arrive at the final state, but the final value depends only on the size of the balloon, and not on the path taken from the initial to the final state. Because enthalpy is a state function, in principle there is an absolute enthalpy for the reactants (Hinitial = Hreactants) and for the products (Hfinal = Hproducts). The difference between these enthalpies is the change for the system. ΔHreaction = Hfinal - Hinitial = Hproducts - Hreactants Because the reaction starts and finishes at the same place no matter which pathway is chosen, ΔHreaction must always be independent of pathway. Unlike volume, temperature, pressure, or energy, however, the absolute enthalpy of a substance is not usually determined, only its change in a chemical or physical process. The thermal energy evolved or required in a chemical process, for example, is a reflection of the difference in enthalpy between the reactants and products. We determine only whether the enthalpy of the products is greater than or less than that of the reactants by some amount.

to derive a great many other ΔH values by adding together appropriate ones as described. Because ΔH values depend on temperature and pressure, it is necessary to specify both of these when a table of data is made. Usually a pressure of 1 bar and a temperature of 25 ◌C are specified, although other conditions of temperature are used. The standard state of an element or a compound is the most stable form of the substance in the physical state in which it exists at 1 bar and the specified temperature. Thus, at 25 ◌C and a pressure of 1 bar, the standard state for hydrogen is the gaseous state, H2(g), and for sodium chloride, it is the solid state, NaCl(s). For carbon, which can exist in any one of three solid states at 1 bar and 25 ◌C, the most stable form, graphite, is selected as the standard. When a reaction occurs with all the reactants and products in their standard states under standard conditions, the observed, or calculated, enthalpy change is known as the standard enthalpy change of reaction, ΔH ◌, where the superscript ◌◌ indicates standard conditions. All the reactions we have discussed to this point have followed this convention, so all the ΔH values should have ◌ attached. The standard enthalpy change of a reaction for the formation of one mole of a compound directly from its elements is called the standard molar enthalpy of formation, ΔH ◌ f , where the subscript f indicates that one mole of the compound in question has been formed in its standard state from its elements, also in their standard states. Some of the reactions already discussed define standard molar enthalpies of formation. H2(g) + ½O2(g) → H2O(ℓ) ; ∆Hf

o = -286 kJ/mol

C(s) + 2H2(g) → CH4(g) ; ∆Hfo = -75 kJ/mol

STANDARD ENTHALPIES OF FORMATION

Hess’s law makes it possible for us to tabulate ΔH values for a few reactions and

As another example, the following equation shows that 277.7 kJ is evolved if graphite, the standard state form for carbon, is combined with gaseous hydrogen and oxygen to form one mole of ethanol at 298 K.

2C(graphite) + 3H2(g) + ½O2(g) → C2H5OH(ℓ) ; ∆Hof= -278 kJ/mol

Finally, it is important to understand that a standard molar enthalpy of formation (ΔH ◌ f) is just a special case of an enthalpy change for a reaction. In contrast with the three previous reactions, the enthalpy change for the following reaction is not an enthalpy of formation; calcium carbonate has been formed from other compounds, not directly from its elements. CaO(s) + CO2(g) → CaCO3(s) ; ∆Ho

rxn= -178 kJ/mol

Therefore, the enthalpy change is given the more general symbol of ΔH ◌rxn. The enthalpy change for the following reaction is also not a standard enthalpy of formation.

P4(s) + 6Cl2(g) → 4PCl3(ℓ) ; ∆Horxn= -1279 kJ

Here a compound has been formed from its elements, but more than one mole of the compound has been formed, and so ΔH ◌rxn = 4 ΔH ◌ f [PCl3(ℓ)]. Standard enthalpies of formation for the elements in their standard states are zero, because forming an element in its standard state from the same element in its standard state involves no chemical or physical change. Be sure to notice also that most ΔH ◌ f values are negative. That is, the process of forming most compounds from their elements is exothermic. Recall that we previously stated that at room temperature most exothermic reactions are product-favored. Thus, forming compounds from their elements (under standard conditions) is generally a product-favored process.

Reactions of oxygen with metals are usually exothermic and product-favored 2Al(s) + 3/2O2(g) → Al2O3(s) ; ∆Ho

f= -1675 kJ/mol and carbon generally combines with other elements in reactions with negative enthalpy changes. C(s) + O2(g) → CO2(g) ; ∆Ho

f= -394 kJ/mol 2C(s) + 3H2(g) + ½O2(g) → C2H5OH(ℓ) ; ∆Ho

f= -1676 kJ One important exception to our general observation is ethyne (commonly called acetylene), a compound with a strongly endothermic standard enthalpy of formation.

2C(s) + H2(g) → C2H2(g) ; ∆Hof= +227 kJ/mol

It is interesting that acetylene is a good fuel and that it is the starting point for manufacturing many other carbon-containing compounds.

DETERMINING ENTHALPIES OF REACTION Calorimetry

The heat evolved by a chemical reaction can be determined by a technique called calorimetry. When finding heats of combustion or the caloric value of foods, the measurement is often done in a combustion calorimeter (Figure 6.17). Figure 6.17 A combustion calorimeter. A combustible sample is burned in pure oxygen in a steel ‘‘bomb.’’ The heat generated by the reaction warms the bomb and the water surrounding it. By measuring the temperature increase,

the heat evolved by the reaction can be determined .

A weighed sample of a combustible solid or liquid is placed in a dish that is encased in a ‘‘bomb,’’ a cylinder about the size of a large fruit juice can with heavy steel walls and ends . The bomb is then placed in a water-filled container with well-insulated walls. After filling the bomb with pure oxygen, the mixture of oxygen and sample is ignited, usually by an electric spark. The heat generated when the sample burns warms the bomb and the water around it, with both coming to the same temperature. In this configuration, the oxygen and the compound represent the system, and the bomb and water around it are the surroundings. From the law of energy conservation, we can say that Heat transferred from the system = heat transferred into the surroundings Heat evolved by the reaction = heat absorbed by water and bomb qreaction = - (qwater + qbomb) where qreaction has a negative value because the combustion reaction is exothermic . The temperature change of the water, which is also equal to the change for the bomb, is measured. Using these experimental measurements, the total quantity of heat absorbed by the water and the bomb (qwater + qbomb) can be calculated from the specific heat of the bomb and the water. According to the previous equation, this total gives the heat evolved by combustion of the compound. Because the bomb is rigid, the heat transfer is measured at constant volume and is therefore equivalent to DE, the change in energy. As explained earlier (see Section 6.4), ΔE = qv , whereas the change in enthalpy is the heat evolved or required at constant pressure, that is, ΔH = qp. Because ΔE and ΔH are related in a relatively simple way, however, ΔH values can be calculated from ΔE values found in bomb calorimetry experiments.

What experiments would you do to find the standard enthalpy of formation of solid calcium hydroxide? Ca(s) + O2(g) + H2(g) → Ca(OH)2(s) ; ∆Ho

f = ? We know from experience that the enthalpy change for the reaction cannot be measured directly in any convenient manner. It is therefore necessary to break down the formation of calcium hydroxide into a series of reactions that can be added together to give the equation for the formation of solid Ca(OH)2 from the elements and whose enthalpy changes can be determined in a calorimeter

● Calcium is a metal and so reacts with oxygen to produce the oxide, CaO. Ca(s) + ½O2(g) → CaO(s) ; ∆Ho

rxn= ∆Hof = -635 kJ/mol

The enthalpy change for this reaction can be determined in a calorimeter (or found in a table of ΔH ◌f values). ●ΔH ◌f for the formation of water from hydrogen and oxygen is well known. H2(g) + ½O2(g) → H2O(ℓ) ; ∆Ho

rxn= ∆Hof = -286 kJ/mol

● CaO is a basic oxide; it reacts with water to give calcium hydroxide, Ca(OH)2(s). CaO(s) + H2O(ℓ) → Ca(OH)2(s) ; ∆Ho

rxn= -65 kJ/mol Once CaO has been formed from the direct reaction of the metal with oxygen, the enthalpy change for its reaction with water can be determined in a calorimeter. Adding the three reactions above together, we see that they give us the enthalpy of formation of solid Ca(OH)2. Ca(s) + ½O2(g) → CaO(s) ; ∆Ho

f = -635 kJ/mol

H2(g) + ½O2(g) → H2O(ℓ) ; ∆Hof = -286 kJ/mol

CaO(s) + H2O(ℓ) → Ca(OH)2(s) ; ∆Horxn= -65 kJ/mol

Using Hess’s Law

Using Hess’s Law

Using Hess’s Law

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