Chapt06

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Unformatted text preview: Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6 CHAPTER Thermochemistry INTRODUCTION ON MAY 6, 1937, A GERMAN AIRSHIP FILLED WITH HYDROGEN GAS EX- PLODED IN THE STORMY SKY OVER ATTEMPT. LANDING HINDENBURG, OF THE LAKEHURST, NEW JERSEY, 97 PEOPLE ABOARD 6.1 THE NATURE OF ENERGY AND TYPES OF ENERGY DURING A THE 6.2 ENERGY CHANGES IN CHEMICAL REACTIONS AIRSHIP 35 DIED AS A RESULT OF THE DISASTER, WHICH PUT AN END TO A PROMISING NEW MODE OF TRANSPORTATION. 6.3 ENTHALPY THE HINDENBURG 6.4 CALORIMETRY WAS TO HAVE BEEN FILLED WITH HELIUM, A LIGHTWEIGHT, INERT GAS, RATHER THAN HIGHLY REACTIVE HYDROGEN, BUT HELIUM SALES TO GERMANY 6.5 STANDARD ENTHALPY OF FORMATION AND REACTION HAD BEEN SUSPENDED AS FEAR OF A NEW WORLD WAR ES- 6.6 HEAT OF SOLUTION AND DILUTION CALATED AND SO HYDROGEN WAS SUBSTITUTED. POSSIBLY 6.7 INTRODUCTION TO THERMODYNAMICS TRIGGERED BY LIGHTNING STRIKING A LEAKY VALVE ON THE SHIP’S SKIN, THE EXPLOSION AND FIRE THAT ENGULFED THE HINDENBURG WERE ACCOMPANIED BY THE RELEASE OF TREMENDOUS EN- ERGY, A BY-PRODUCT OF THE REACTION BETWEEN THE HYDROGEN IN THE SHIP AND OXYGEN IN THE AIR. THE DISASTER PROVIDES A DRA- MATIC EXAMPLE OF THE ENERGY CHANGES THAT OCCUR DURING CHEMICAL REACTIONS. EVERY CHEMICAL REACTION OBEYS TWO FUNDAMENTAL LAWS: THE LAW OF CONSERVATION OF MASS AND THE LAW OF CONSERVATION OF ENERGY. WE DISCUSSED THE MASS RELATIONSHIPS BETWEEN REACTANTS AND PRODUCTS IN CHAPTER 3; IN THIS CHAPTER WE WILL LOOK AT THE ENERGY CHANGES THAT ACCOMPANY CHEMICAL REACTIONS. 203 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 204 THERMOCHEMISTRY 6.1 Kinetic energy was introduced in Chapter 5. As the water falls over the dam, its potential energy is converted to kinetic energy. Use of this energy to generate electricity is called hydroelectric power. Back Forward THE NATURE OF ENERGY AND TYPES OF ENERGY “Energy” is a much-used term that represents a rather abstract concept. For instance, when we feel tired, we might say we haven’t any energy; and we read about the need to find alternatives to nonrenewable energy sources. Unlike matter, energy is known and recognized by its effects. It cannot be seen, touched, smelled, or weighed. Energy is usually defined as the capacity to do work. In Chapter 5 we defined work as “force distance,” but we will soon see that there are other kinds of work. All forms of energy are capable of doing work (that is, of exerting a force over a distance), but not all of them are equally relevant to chemistry. The energy contained in tidal waves, for example, can be harnessed to perform useful work, but the relationship between tidal waves and chemistry is minimal. Chemists define work as directed energy change resulting from a process. Kinetic energy—the energy produced by a moving object — is one form of energy that is of particular interest to chemists. Others include radiant energy, thermal energy, chemical energy, and potential energy. Radiant energy, or solar energy, comes from the sun and is Earth’s primary energy source. Solar energy heats the atmosphere and Earth’s surface, stimulates the growth of vegetation through the process known as photosynthesis, and influences global climate patterns. Thermal energy is the energy associated with the random motion of atoms and molecules. In general, thermal energy can be calculated from temperature measurements. The more vigorous the motion of the atoms and molecules in a sample of matter, the hotter the sample is and the greater its thermal energy. However, we need to distinguish carefully between thermal energy and temperature. A cup of coffee at 70°C has a higher temperature than a bathtub filled with warm water at 40°C, but much more thermal energy is stored in the bathtub water because it has a much larger volume and greater mass than the coffee and therefore more water molecules and more molecular motion. Chemical energy is stored within the structural units of chemical substances; its quantity is determined by the type and arrangement of constituent atoms. When substances participate in chemical reactions, chemical energy is released, stored, or converted to other forms of energy. Potential energy is energy available by virtue of an object’s position. For instance, because of its altitude, a rock at the top of a cliff has more potential energy and will make a bigger splash if it falls into the water below than a similar rock located part way down the cliff. Chemical energy can be considered a form of potential energy because it is associated with the relative positions and arrangements of atoms within a given substance. All forms of energy can be converted (as least in principle) from one form to another. We feel warm when we stand in sunlight because radiant energy is converted to thermal energy on our skin. When we exercise, chemical energy stored in our bodies is used to produce kinetic energy. When a ball starts to roll downhill, its potential energy is converted to kinetic energy. You can undoubtedly think of many other examples. Although energy can assume many different forms that are interconvertible, scientists have concluded that energy can be neither destroyed nor created. When one form of energy disappears, some other form of energy (of equal magnitude) must appear, and vice versa. This principle is summarized by the law of conservation of energy: the total quantity of energy in the universe is assumed constant. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.2 6.2 ENERGY CHANGES IN CHEMICAL REACTIONS 205 ENERGY CHANGES IN CHEMICAL REACTIONS Often the energy changes that take place during chemical reactions are of as much practical interest as the mass relationships we discussed in Chapter 3. For example, combustion reactions involving fuels such as natural gas and oil are carried out in daily life more for the thermal energy they release than for their products, which are water and carbon dioxide. Almost all chemical reactions absorb or produce (release) energy, generally in the form of heat. It is important to understand the distinction between thermal energy and heat. Heat is the transfer of thermal energy between two bodies that are at different temperatures. Thus we often speak of the “heat flow” from a hot object to a cold one. Although the term “heat” by itself implies the transfer of energy, we customarily talk of “heat absorbed” or “heat released” when describing the energy changes that occur during a process. Thermochemistry is the study of heat change in chemical reactions. In order to analyze energy changes associated with chemical reactions we must first define the system, or the specific part of the universe that is of interest to us. For chemists, systems usually include substances involved in chemical and physical changes. For example, in an acid-base neutralization experiment, the system may be a beaker containing 50 mL of HCl to which 50 mL of NaOH are added. The surroundings are the rest of the universe outside the system. There are three types of systems. An open system can exchange mass and energy, usually in the form of heat, with its surroundings. For example, an open system may consist of a quantity of water in an open container, as shown in Fig. 6.1(a). If we close the flask, as in Figure 6.1(b), so that no water vapor can escape from or condense into the container, we create a closed system, which allows the transfer of energy (heat) but not mass. By placing the water in a totally insulated container, we can construct an isolated system, which does not allow the transfer of either mass or energy, as shown in Figure 6.1(c). The combustion of hydrogen gas in oxygen is one of many chemical reactions that release considerable quantities of energy (Figure 6.2): 2H2(g) FIGURE 6.1 Three systems represented by water in a flask: (a) an open system, which allows the exchange of both energy and mass with surroundings; (b) a closed system, which allows the exchange of energy but not mass; and (c) an isolated system, which allows neither energy nor mass to be exchanged (here the flask is enclosed by a vacuum jacket). Forward Main Menu energy Water vapor Heat Heat (a) Back O2(g) 88n 2H2O(l) TOC (b) Study Guide TOC (c) Textbook Website MHHE Website 206 THERMOCHEMISTRY FIGURE 6.2 The Hindenburg disaster. The Hindenburg, a German airship filled with hydrogen gas, was destroyed in a spectacular fire at Lakehurst, New Jersey, in 1937. t No “ Exo- comes from the Greek word for “outside”, while endo- comes from the word for “within.” in le b ila a Av ” ion ers tV ex e-T In this case we label the reacting mixture (hydrogen, oxygen, and water molecules) the system and the rest of the universe the surroundings. Since energy cannot be created or destroyed, any energy lost by the system must be gained by the surroundings. Thus the heat generated by the combustion process is transferred from the system to its surroundings. This reaction is an example of an exothermic process which is any process that gives off heat — that is, transfers thermal energy to the surroundings. Figure 6.3(a) shows the energy change for the combustion of hydrogen gas. Now consider another reaction, the decomposition of mercury(II) oxide (HgO) at high temperatures: energy On heating, HgO decomposes to give Hg and O2. 6.3 We use Back in this text always to mean final initial. Forward 2HgO(s) 88n 2Hg(l) O2(g) This reaction is an endothermic process, in which heat has to be supplied to the system (that is, to HgO) by the surroundings [Figure 6.3(b)]. From Figure 6.3 you can see that in exothermic reactions, the total energy of the products is less than the total energy of the reactants. The difference is the heat supplied by the system to the surroundings. Just the opposite happens in endothermic reactions. Here, the difference between the energy of the products and the energy of the reactants is equal to the heat supplied to the system by the surroundings. ENTHALPY Most physical and chemical changes, including those that take place in living systems, occur in the constant-pressure conditions of our atmosphere. In the laboratory, for example, reactions are generally carried out in beakers, flasks, or test tubes that remain open to their surroundings and hence to a pressure of approximately one atmosphere (1 atm). To quantify the heat flow into or out of a system in a constant-pressure process, chemists use a property called enthalpy, represented by the symbol H and defined as E PV. Enthalpy is an extensive property; its magnitude depends on the amount of the substance present. It is impossible to determine the enthalpy of a substance, so it is the change in enthalpy, H, that we actually measure. (The Greek letter delta, , Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.3 2H2 (g) + O2 (g) 207 2Hg (l ) + O2 (g) Heat absorbed by the system from the surroundings Energy Heat given off by the system to the surroundings Energy FIGURE 6.3 (a) An exothermic process. (b) An endothermic process. Parts (a) and (b) are not drawn to the same scale; that is, the heat released in the formation of H2O from H2 and O2 is not equal to the heat absorbed in the decomposition of HgO. ENTHALPY 2H2O (l) 2HgO (s) (a) (b) symbolizes change.) The enthalpy of reaction, H, is the difference between the enthalpies of the products and the enthalpies of the reactants: H This analogy assumes that you will not overdraw your bank account. The enthalpy of a substance cannot be negative. H(products) H(reactants) (6.1) In other words, H represents the heat given off or absorbed during a reaction. The enthalpy of reaction can be positive or negative, depending on the process. For an endothermic process (heat absorbed by the system from the surroundings), H is positive (that is, H 0). For an exothermic process (heat released by the system to the surroundings), H is negative (that is, H 0). An analogy for enthalpy change is a change in the balance in your bank account. Suppose your initial balance is $100. After a transaction (deposit or withdrawal), the change in your bank balance, X, is given by X Xfinal Xinitial where X represents the bank balance. If you deposit $40 into your account, then X $140 $100 $40. This corresponds to an endothermic reaction. (The balance increases and so does the enthalpy of the system.) On the other hand, a withdrawal of $60 means X $40 $100 $60. The negative sign of X means your balance has decreased. Similarly a negative value of H reflects a decrease in enthalpy of the system as a result of an exothermic process. The difference between this analogy and Equation (6.1) is that while you always know your exact bank balance, there is no way to know the enthalpies of individual products and reactants. In practice, we can only measure the difference in their values. Now let us apply the idea of enthalpy changes to two common processes, the first involving a physical change, the second a chemical change. THERMOCHEMICAL EQUATIONS At 0°C and a pressure of 1 atm, ice melts to form liquid water. Measurements show that for every mole of ice converted to liquid water under these conditions, 6.01 kilojoules (kJ) of energy are absorbed by the system (ice). Since H is a positive value, this is an endothermic process, as expected for an energy-absorbing change of melting ice (Figure 6.4). The equation for this physical change is Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 208 THERMOCHEMISTRY FIGURE 6.4 Melting 1 mole of ice at 0°C (an endothermic process) results in an enthalpy increase in the system of 6.01 k J. H2O (l ) Enthalpy Heat absorbed by the system from the surroundings H = 6.01 kJ H2O (s) H2O(s) 88n H2O(l) H 6.01 kJ As another example, consider the combustion of methane (CH4), the principal component of natural gas: CH4(g) 2O2(g) 88n CO2(g) 2H2O(l) H 890.4 kJ From experience we know that burning natural gas releases heat to its surroundings, so it is an exothermic process, and H must have a negative value. This enthalpy change is represented in Figure 6.5. The equations for the melting of ice and the combustion of methane are examples of thermochemical equations, which show the enthalpy changes as well as the mass relationships. The following guidelines are helpful in writing and interpreting thermochemical equations: The stoichiometric coefficients always refer to the number of moles of a substance. Thus, the equation for the melting of ice may be “read” as follows: When 1 mole of liquid water is formed from 1 mole of ice at 0°C, the enthalpy change is 6.01 kJ. For the combustion of methane, we interpret the equation this way: When 1 mole of gaseous methane reacts with 2 moles of gaseous oxygen to form 1 mole of gaseous carbon dioxide and 2 moles of liquid water, the enthalpy change is 890.4 kJ. • When we reverse an equation, we change the roles of reactants and products. Consequently, the magnitude of H for the equation remains the same, but its sign changes. For example, if a reaction consumes thermal energy from its surroundings • FIGURE 6.5 Burning 1 mole of methane in oxygen gas (an exothermic process) results in an enthalpy decrease in the system of 890.4 k J. CH4 (g) + 2O2 (g) Enthalpy Heat given off by the system to the surroundings H = –890.4 kJ CO2 (g) + 2H2O (l) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.4 CALORIMETRY 209 (that is, if it is endothermic), then the reverse reaction must release thermal energy back to its surroundings (that is, it must be exothermic) and the enthalpy change expression must also change its sign. Thus, reversing the melting of ice and the combustion of methane, the thermochemical equations are H2O(l ) 88n H2O(s) CO2(g) • H 2H2O(l) 88n CH4(g) 2O2(g) H 890.4 kJ and what was an endothermic process becomes exothermic, and vice versa. If we multiply both sides of a thermochemical equation by a factor n, then H must also change by the same factor. Thus, for the melting of ice, if n 2, then 2H2O(s) 88n 2H2O(l) • 6.01 kJ H 2(6.01 kJ) 12.0 kJ When writing thermochemical equations, we must always specify the physical states of all reactants and products, because they help determine the actual enthalpy changes. For example, in the equation for the combustion of methane, if we show water vapor rather than liquid water as a product, CH4(g) 2O2(g) 88n CO2(g) 2H2O(g) H 802.4 kJ the enthalpy change is 802.4 kJ rather than 890.4 kJ because 88.0 kJ are needed to convert 2 moles of liquid water to water vapor; that is, 2H2O(l) 88n 2H2O(g) H 88.0 kJ EXAMPLE 6.1 Given the thermochemical equation SO2(g) 1 2 O2(g) 88n SO3(g) calculate the heat evolved when 74.6 g of SO2 (molar mass verted to SO3. H 99.1 kJ 64.07 g/mol) is con- The thermochemical equation shows that for every mole of SO2 burned, 99.1 kJ of heat are given off (note the negative sign). Therefore, the heat produced when 74.6 g of SO2 reacts is Answer 74.6 g SO2 Similar problems: 6.42, 6.43. 1 mol SO2 64.07 g SO2 99.1 kJ 1 mol SO2 115 kJ PRACTICE EXERCISE Calculate the heat evolved when 266 g of white phosphorus (P4) burn in air according to the equation P4(s) 6.4 5O2(g) 88n P4O10(s) H 3013 kJ CALORIMETRY In the laboratory heat changes in physical and chemical processes are measured with a calorimeter, a closed container designed specifically for this purpose. Our discussion of calorimetry, the measurement of heat changes, will depend on an understanding of specific heat and heat capacity, so let us consider them first. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 210 THERMOCHEMISTRY SPECIFIC HEAT AND HEAT CAPACITY The specific heat (s) of a substance is the amount of heat required to raise the temperature of one gram of the substance by one degree Celsius. The heat capacity (C) of a substance is the amount of heat required to raise the temperature of a given quantity of the substance by one degree Celsius. Specific heat is an intensive property, whereas heat capacity is an extensive property. The relationship between the heat capacity and specific heat of a substance is C The dot between g and °C reminds us that both g and °C are in the denominator. ms (6.2) where m is the mass of the substance in grams. For example, the specific heat of water is 4.184 J/g °C, and the heat capacity of 60.0 g of water is (60.0 g)(4.184 J/g °C) TABLE 6.1 The Specific Heats of Some Common Substances SUBSTANCE Al Au C (graphite) C (diamond) Cu Fe Hg H2O C2H5OH (ethanol) SPECIFIC HEAT (J/g °C) 0.900 0.129 0.720 0.502 0.385 0.444 0.139 4.184 2.461 251 J/°C Note that specific heat has the units J/g °C and heat capacity has the units J/°C. Table 6.1 shows the specific heat of some common substances. If we know the specific heat and the amount of a substance, then the change in the sample’s temperature ( t) will tell us the amount of heat (q) that has been absorbed or released in a particular process. The equation for calculating the heat change is given by q ms t (6.3) q Ct (6.4) where m is the mass of the sample and t is the temperature change: t tfinal tinitial The sign convention for q is the same as that for enthalpy change; q is positive for endothermic processes and negative for exothermic processes. Example 6.2 illustrates the use of Equation (6.3). EXAMPLE 6.2 A 466-g sample of water is heated from 8.50°C to 74.60°C. Calculate the amount of heat absorbed by the water. Answer Using Equation (6.3), we write q ms t (466 g)(4.184 J/g °C)(74.60°C 1.29 8.50°C) 5 10 J 129 kJ Similar problems: 6.18, 6.19. PRACTICE EXERCISE An iron bar of mass 869 g cools from 94°C to 5°C. Calculate the heat released (in kilojoules) by the metal. CONSTANT-VOLUME CALORIMETRY Heat of combustion is usually measured by placing a known mass of a compound in a steel container called a constant-volume bomb calorimeter, which is filled with oxygen at about 30 atm of pressure. The closed bomb is immersed in a known amount of Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.4 FIGURE 6.6 A constant-volume bomb calorimeter. The calorimeter is filled with oxygen gas before it is placed in the bucket. The sample is ignited electrically, and the heat produced by the reaction can be accurately determined by measuring the temperature increase in the known amount of surrounding water. CALORIMETRY 211 Thermometer Stirrer Ignition wire Calorimeter bucket Insulated jacket Water O2 inlet Bomb Sample cup “Constant volume” refers to the volume of the container, which does not change during the reaction. Note that the container remains intact after the measurement. The term “bomb calorimeter” connotes the explosive nature of the reaction (on a small scale) in the presence of excess oxygen gas. water, as shown in Figure 6.6. The sample is ignited electrically, and the heat produced by the combustion reaction can be calculated accurately by recording the rise in temperature of the water. The heat given off by the sample is absorbed by the water and the calorimeter. The special design of the bomb calorimeter allows us to assume that no heat (or mass) is lost to the surroundings during the time it takes to make measurements. Therefore we can call the bomb calorimeter and the water in which it is submerged an isolated system. Because no heat enters or leaves the system throughout the process, we can write qsystem qwater qbomb qrxn (6.5) 0 where qwater, qbomb, and qrxn are the heat changes for the water, the bomb, and the reaction, respectively. Thus qrxn (qwater qbomb) (6.6) The quantity qwater is obtained by q qwater ms t (mwater)(4.184 J/g °C) t The product of the mass of the bomb and its specific heat is the heat capacity of the bomb, which remains constant for all experiments carried out in the bomb calorimeter: Cbomb Hence Back Forward Main Menu TOC qbomb Study Guide TOC mbomb sbomb Cbomb t Textbook Website MHHE Website 212 THERMOCHEMISTRY Note that because reactions in a bomb calorimeter occur under constant-volume rather than constant-pressure conditions, the heat changes do not correspond to the enthalpy change H (see Section 6.3). It is possible to correct the measured heat changes so that they correspond to H values, but the corrections usually are quite small, so we will not concern ourselves with the details of the correction procedure. Finally, it is interesting to note that the energy contents of food and fuel (usually expressed in calories where 1 cal 4.184 J) are measured with constant-volume calorimeters (see Chemistry in Action essay on p. 215.) Example 6.3 illustrates the determination of the heat of combustion of an organic compound. EXAMPLE 6.3 A quantity of 1.435 g of naphthalene (C10H8), a pungent-smelling substance used in moth repellents, was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose from 20.17°C to 25.84°C. If the mass of water surrounding the calorimeter was exactly 2000 g and the heat capacity of the bomb calorimeter was 1.80 kJ/°C, calculate the heat of combustion of naphthalene on a molar basis; that is, find the molar heat of combustion. Answer First we calculate the heat changes for the water and the bomb calorimeter. q qwater ms t (2000 g)(4.184 J/g °C)(25.84°C 4.74 qbomb (1.80 1.02 10 J 103 J/°C)(25.84°C 20.17°C) 4 10 J (Note that we changed 1.80 kJ/°C to 1.80 we write (4.74 104 J 5.76 qrxn 20.17°C) 4 103 J/°C.) Next, from Equation (6.6) 4 1.02 104 J) 10 J The molar mass of naphthalene is 128.2 g, so the heat of combustion of 1 mole of naphthalene is molar heat of combustion 5.76 104 J 1.435 g C10H8 128.2 g C10H8 1 mol C10H8 5.15 5.15 Similar problem: 6.21. 106 J/mol 103 kJ/mol PRACTICE EXERCISE A quantity of 1.922 g of methanol (CH3OH) was burned in a constant-volume bomb calorimeter. Consequently, the temperature of the water rose by 4.20°C. If the quantity of water surrounding the calorimeter was exactly 2000 g and the heat capacity of the calorimeter was 2.02 kJ/°C, calculate the molar heat of combustion of methanol. Finally, we note that the heat capacity of a bomb calorimeter is usually determined by burning in it a compound with an accurately known heat of combustion value. From the mass of the compound and the temperature increase, we can calculate the heat capacity of the calorimeter (see Problem 6.94). Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.4 213 CALORIMETRY CONSTANT-PRESSURE CALORIMETRY Thermometer Stirrer Styrofoam cups Reaction mixture A simpler device than the constant-volume calorimeter is the constant-pressure calorimeter used to determine the heat changes for noncombustion reactions. An operable constant-pressure calorimeter can be constructed from two Styrofoam coffee cups, as shown in Figure 6.7. This device measures the heat effects of a variety of reactions, such as acid-base neutralization, as well as the heat of solution and heat of dilution. Because the pressure is constant, the heat change for the process (qrxn) is equal to the enthalpy change (H). The measurements are similar to those of a constantvolume calorimeter — we need to know the heat capacity of the calorimeter, as well as the temperature change of the solution. Table 6.2 lists some reactions that have been studied with the constant-pressure calorimeter. As the following example shows, we can study the heat of acid-base reactions using a constant-pressure calorimeter. EXAMPLE 6.4 FIGURE 6.7 A constantpressure calorimeter made of two Styrofoam coffee cups. The outer cup helps to insulate the reacting mixture from the surroundings. Two solutions of known volume containing the reactants at the same temperature are carefully mixed in the calorimeter. The heat produced or absorbed by the reaction can be determined by measuring the temperature change. A quantity of 1.00 102 mL of 0.500 M HCl is mixed with 1.00 102 mL of 0.500 M NaOH in a constant-pressure calorimeter that has a heat capacity of 335 J/°C. The initial temperature of the HCl and NaOH solution is the same, 22.50°C, and the final temperature of the mixed solution is 24.90°C. Calculate the heat change for the neutralization reaction NaOH(aq) HCl(aq) 88n NaCl(aq) H2O(l) Assume that the densities and specific heats of the solutions are the same as for water (1.00 g/mL and 4.184 J/g °C, respectively). Answer Assuming that no heat is lost to the surroundings, we write qsystem qsoln qcalorimeter qrxn 0 or qrxn (qsoln qcalorimeter) Because the density of the solution is 1.00 g/mL, the mass of a 100-mL solution is 100 g. Thus qrxn (1.00 2.01 102 g 1.00 102 g)(4.184 J/g °C)(24.90°C 22.50°C) 3 10 J qcalorimeter (335 J/°C)(24.90°C 22.50°C) 804 J Next we write TABLE 6.2 Heats of Some Typical Reactions Measured at Constant Pressure TYPE OF REACTION EXAMPLE Heat Heat Heat Heat Heat HCl(aq) NaOH(aq) 88n NaCl(aq) H2O(l) H2O(l ) 88n H (aq) OH (aq) H2O(s) 88n H2O(l) H2O(l ) 88n H2O(g) MgCl2(s) 2Na(l) 88n 2NaCl(s) Mg(s) of of of of of neutralization ionization fusion vaporization reaction H (kJ) 56.2 56.2 6.01 44.0* 180.2 *Measured at 25°C. At 100°C, the value is 40.79 kJ. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 214 THERMOCHEMISTRY (2.01 103 J 2.81 qrxn 103 J 804 J) 2.81 kJ From the molarities given, we know there is 0.0500 mole of HCl in 1.00 102 g of the HCl solution and 0.0500 mole of NaOH in 1.00 102 g of the NaOH solution. Therefore the heat of neutralization when 1.00 mole of HCl reacts with 1.00 mole of NaOH is heat of neutralization 2.81 kJ 0.0500 mol 56.2 kJ/mol Because the reaction takes place at constant pressure, the heat given off is equal to the enthalpy change. Comment Similar problem: 6.22. PRACTICE EXERCISE A quantity of 4.00 102 mL of 0.600 M HNO3 is mixed with 4.00 102 mL of 0.300 M Ba(OH)2 in a constant-pressure calorimeter having a heat capacity of 387 J/°C. The initial temperature of both solutions is the same at 18.88°C. What is the final temperature of the solution? (Use the result in Example 6.4 for your calculation.) 6.5 Recall from Chapter 2 that different forms of the same element are called allotropes. Back Forward STANDARD ENTHALPY OF FORMATION AND REACTION So far we have learned that we can determine the enthalpy change that accompanies a reaction by measuring the heat absorbed or released (at constant pressure). From Equation (6.1) we see that H can also be calculated if we know the actual enthalpies of all reactants and products. However, as mentioned earlier, there is no way to measure the absolute value of the enthalpy of a substance. Only values relative to an arbitrary reference can be determined. This problem is similar to the one geographers face in expressing the elevations of specific mountains or valleys. Rather than trying to devise some type of “absolute” elevation scale (perhaps based on distance from the center of Earth?), by common agreement all geographic heights and depths are expressed relative to sea level, an arbitrary reference with a defined elevation of “zero” meters or feet. Similarly, chemists have agreed on an arbitrary reference point for enthalpy. The “sea level” reference point for all enthalpy expressions is called the standard o enthalpy of formation ( Hf ), which is defined as the heat change that results when one mole of a compound is formed from its elements at a pressure of 1 atm. Elements are said to be in the standard state at 1 atm, hence the term “standard enthalpy.” The superscript “o” represents standard-state conditions (1 atm), and the subscript “f ” stands for formation. Although the standard state does not specify a temperature, we will always use Ho values measured at 25°C. f Table 6.3 lists the standard enthalpies of formation for a number of elements and compounds. (For a more complete list of Ho values, see Appendix 3.) By convenf tion, the standard enthalpy of formation of any element in its most stable form is zero. Take oxygen as an example. Molecular oxygen (O2) is more stable than the other allotropic form of oxygen, ozone (O3), at 1 atm and 25°C. Thus we can write Ho(O2) 0, but Ho(O3) 0. Similarly, graphite is a more stable allotropic form of f f Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.5 STANDARD ENTHALPY OF FORMATION AND REACTION 215 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Back Fuel Values of Foods and Other Substances The food we eat is broken down, or metabolized, in stages by a group of complex biological molecules called enzymes. Most of the energy released at each stage is captured for function and growth. One interesting aspect of metabolism is that the overall change in energy is the same as it is in combustion. For example, the total enthalpy change for the conversion of glucose (C6H12O6) to carbon dioxide and water is the same whether we burn the substance in air or digest it in our bodies: C6H12O6(s) 6O2(g) 88n 6CO2(g) 6H2O(l ) H° 2801 kJ The important difference between metabolism and combustion is that the latter is usually a one-step, high-temperature process. Consequently, much of the energy released by combustion is lost to the surroundings. Various foods have different compositions and hence different energy contents. The energy content of food is generally measured in calories. The calorie (cal ) is a non-SI unit of energy that is equivalent to 4.184 J: 1 cal Fuel Values of Foods and Some Common Fuels SUBSTANCE Hcombustion (kJ/g) Apple Beef Beer Bread Butter Cheese Eggs Milk Potatoes 2.0 8.0 1.5 11.0 34.0 18.0 6.0 3.0 3.0 Charcoal Coal Gasoline Kerosene Natural gas Wood 35.0 30.0 34.0 37.0 50.0 20.0 4.184 J In the context of nutrition, however, the calorie we speak of (sometimes called a “big calorie”) is actually equal to a kilocalorie; that is, 1 Cal 1000 cal 4184 J Note the use of a capital “C” to represent the “big calorie.” The bomb calorimeter described in Section 6.4 is ideally suited for measuring the energy content, or “fuel value,” of foods. Fuel values are just the enthalpies of combustion (see table). In order to be analyzed in a bomb calorimeter, food must be dried first because most foods contain a considerable amount of water. Since the composition of particular foods is often not known, fuel values are expressed in terms of k J/g rather than k J/mol. Forward Main Menu TOC The labels on food packages reveal the calorie content of the food inside. Study Guide TOC Textbook Website MHHE Website 216 THERMOCHEMISTRY TABLE 6.3 Standard Enthalpies of Formation of Some Inorganic Substances at 25°C SUBSTANCE Ho (kJ/mol) f Ag(s) AgCl(s) Al(s) Al2O3(s) Br2(l) HBr(g) C(graphite) C(diamond) CO(g) CO2(g) Ca(s) CaO(s) CaCO3(s) Cl2(g) HCl(g) Cu(s) CuO(s) F2(g) HF(g) H(g) H2(g) H2O(g) H2O(l) 0. 127.04 0. 1669.8 0. 36.2 0. 1.90 110.5 393.5 0. 635.6 1206.9 0. 92.3 0. 155.2 0. 268.61 218.2 0. 241.8 285.8 Hfo(kJ/mol) SUBSTANCE H2O2(l) Hg(l) I2(s) HI(g) Mg(s) MgO(s) MgCO3(s) N2(g) NH3(g) NO(g) NO2(g) N2O4(g) N2O(g) O(g) O2(g) O3(g) S(rhombic) S(monoclinic) SO2(g) SO3(g) H2S(g) ZnO(s) 187.6 0. 0. 25.94 0. 601.8 1112.9 0. 46.3 90.4 33.85 9.66 81.56 249.4 0. 142.2 0. 0.30 296.1 395.2 20.15 347.98 carbon than diamond at 1 atm and 25°C, so Ho(C, graphite) 0 and Ho(C, diaf f mond) 0. The importance of the standard enthalpies of formation is that once we know their values, we can calculate the standard enthalpy of reaction, Ho , defined as the enrxn thalpy of a reaction carried out at 1 atm. For example, consider the hypothetical reaction aA bB 88n cC dD where a, b, c, and d are stoichiometric coefficients. For this reaction Ho is given by rxn Ho rxn [c Ho(C) f d Ho(D)] f [a Ho(A) f b Ho(B)] f (6.7) where a, b, c, and d are in moles. We can generalize Equation (6.7) as follows: Ho rxn Graphite (top) and diamond (bottom). Back Forward n Ho(products) f m Ho(reactants) f (6.8) where m and n denote the stoichiometric coefficients for the reactants and products, and (sigma) means “the sum of.” In order to use Equation (6.7) to calculate Ho we must know the Ho values f rxn of the compounds that take part in the reaction. To determine these values we can apply the direct method or the indirect method. Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.5 217 STANDARD ENTHALPY OF FORMATION AND REACTION This method of measuring Ho works for compounds that can be f readily synthesized from their elements. Suppose we want to know the enthalpy of formation of carbon dioxide. We must measure the enthalpy of the reaction when carbon (graphite) and molecular oxygen in their standard states are converted to carbon dioxide in its standard state: The Direct Method. C(graphite) Ho rxn O2(g) 88n CO2(g) 393.5 kJ As we know from experience, this combustion easily goes to completion. From Equation (6.7) we can write Ho rxn (1 mol) Ho(CO2, g) f [(1 mol) Ho(C, graphite) f (1 mol) Ho(O2, g)] f 393.5 kJ Because both graphite and O2 are stable allotropic forms, it follows that graphite) and Ho(O2, g) are zero. Therefore f Ho rxn (1 mol) Ho(CO2, g) f Ho(CO2, f or g) Ho(C, f 393.5 kJ 393.5 kJ/mol Note that arbitrarily assigning zero Ho for each element in its most stable form f at the standard state does not affect our calculations in any way. Remember, in thermochemistry we are interested only in enthalpy changes, because they can be determined experimentally whereas the absolute enthalpy values cannot. The choice of a zero “reference level” for enthalpy makes calculations easier to handle. Again referring to the terrestrial altitude analogy, we find that Mt. Everest is 8708 ft higher than Mt. McKinley. This difference in altitude is unaffected by the decision to set sea level at 0 ft or at 1000 ft. Other compounds that can be studied by the direct method are SF6, P4O10, and CS2. The equations representing their syntheses are S(rhombic) 4P(white) C(graphite) 3F2(g) 88n SF6(g) 5O2(g) 88n P4O10(s) 2S(rhombic) 88n CS2(l) Note that S(rhombic) and P(white) are the most stable allotropes of sulfur and phosphorus, respectively, at 1 atm and 25°C, so their Ho values are zero. f White phosphorus burns in air to form P4O10. Many compounds cannot be directly synthesized from their elements. In some cases, the reaction proceeds too slowly, or side reactions produce substances other than the desired compound. In these cases Ho can be determined by an f indirect approach, which is based on the law of heat summation (or simply Hess’s law). Hess’s law† can be stated as follows: When reactants are converted to products, the change in enthalpy is the same whether the reaction takes place in one step or in a series of steps. In other words, if we can break down the reaction of interest into a series of reactions for which Ho can be measured, we can calculate Ho for the overrxn rxn all reaction. A useful analogy of Hess’s law is as follows. Suppose you go from the first floor to the sixth floor of a building by elevator. The gain in your gravitational potential enThe Indirect Method. † Germain Henri Hess (1802 – 1850). Swiss chemist. Hess was born in Switzerland but spent most of his life in Russia. For formulating Hess’s law, he is called the father of thermochemistry. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 218 THERMOCHEMISTRY ergy (which corresponds to the enthalpy change for the overall process) is the same whether you go directly there or stop at each floor on your way up (breaking the reaction into a series of steps). Let’s say we are interested in the standard enthalpy of formation of methane (CH4). We might represent the synthesis of CH4 from its elements as C(graphite) 2H2(g) 88n CH4(g) However, this reaction does not take place as written, so we cannot measure the enthalpy change directly. We must employ an indirect route, based on Hess’s law. To begin with, the following reactions involving C, H2, and CH4 with O2 have all been studied and the Ho values are accurately known rxn (a) C(graphite) 2H2(g) (b) CH4(g) (c) O2(g) 88n CO2(g) Ho rxn 393.5 kJ O2(g) 88n 2H2O(l) Ho rxn Ho rxn 571.6 kJ 2O2(g) 88n CO2(g) 2H2O(l) 890.4 kJ Since we want to obtain one equation containing only C and H2 as reactants and CH4 as product, we must reverse (c) to get Remember to reverse the sign of H when you reverse an equation. (d) CO2(g) 2H2O(l) 88n CH4(g) 2O2(g) Ho rxn 890.4 kJ The next step is to add Equations (a), (b), and (d): (a) C(graphite) O2(g) 88n CO2(g) Ho rxn 393.5 kJ (b) 2H2(g) O2(g) 88n 2H2O(l) Ho rxn 571.6 kJ (d) CO2(g) 2H2O(l) 88n CH4(g) (e) C(graphite) 2H2(g) 88n CH4(g) 2O2(g) Ho rxn Ho rxn 890.4 kJ 74.7 kJ As you can see, all the nonessential species (O2, CO2, and H2O) cancel in this operation. Because the above equation represents the synthesis of 1 mole of CH4 from its elements, we have Ho(CH4) 74.7 kJ/mol. f The general rule in applying Hess’s law is that we should arrange a series of chemical equations (corresponding to a series of steps) in such a way that, when added together, all species will cancel except for the reactants and products that appear in the overall reaction. This means that we want the elements on the left and the compound of interest on the right of the arrow. To achieve this, we often need to multiply some or all of the equations representing the individual steps by the appropriate coefficients. The following example shows this approach. EXAMPLE 6.5 Calculate the standard enthalpy of formation of acetylene (C2H2) from its elements: 2C(graphite) H2(g) 88n C2H2(g) The equations for each step and the corresponding enthalpy changes are: (a) H2(g) (b) 2C2H2(g) (c) Back Forward Main Menu O2(g) 88n CO2(g) Ho rxn 393.5 kJ O2(g) 88n H2O(l) Ho rxn 285.8 kJ 5O2(g) 88n 4CO2(g) Ho rxn 2598.8 kJ C(graphite) TOC 1 2 Study Guide TOC 2H2O(l) Textbook Website MHHE Website 6.5 219 STANDARD ENTHALPY OF FORMATION AND REACTION Answer Since we want to obtain one equation containing C and H2 as reactants and C2H2 as product, we need to eliminate O2, CO2, and H2O from equations (a), (b), and (c). We note that (c) contains 5 moles of O2, 4 moles of CO2, and 2 moles of H2O. First we reverse (c) to get C2H2 on the product side: (d) 4CO2(g) 2H2O(l) 88n 2C2H2(g) Ho rxn 5O2(g) Next, we multiply (a) by 4 and (b) by 2 and carry out the addition of 4(a) 2(b) (d): 2H2(g) 4CO2(g) 4C(graphite) 2C(graphite) 4O2(g) 88n 4CO2(g) Ho rxn 1574.0 kJ O2(g) 88n 2H2O(l ) Ho rxn 571.6 kJ 2H2O(l) 88n 2C2H2(g) 4C(graphite) or 2598.8 kJ Ho rxn Ho rxn Ho rxn 2598.8 kJ 5O2(g) 2H2(g) 88n 2C2H2(g) H2(g) 88n C2H2(g) 453.2 kJ 226.6 kJ Since the above equation represents the synthesis of C2H2 from its elements, we have Ho(C2H2) Ho /mol 226.6 kJ/mol. (When the equation is divided by f rxn o 2, the value of Hrxn is halved.) An oxyacetylene torch has a high flame temperature (3000°C) and is used to weld metals. PRACTICE EXERCISE Calculate the standard enthalpy of formation of carbon disulfide (CS2) from its elements, given that C(graphite) O2(g) 88n CO2(g) Ho rxn 393.5 kJ S(rhombic) O2(g) 88n SO2(g) Ho rxn 296.1 kJ 3O2(g) 88n CO2(g) Ho rxn 1072 kJ CS2(l) Similar problems: 6.46, 6.47. 2SO2(g) We can calculate the enthalpy of reactions from values of following example. Ho as shown in the f EXAMPLE 6.6 Pentaborane-9, B5H9, is a colorless, highly reactive liquid that will burst into flame or even explode when exposed to oxygen. The reaction is 2B5H9(l) 12O2(g) 88n 5B2O3(s) 9H2O(l) Pentaborane-9 was considered as a potential rocket fuel in the 1950s because it produces a large amount of heat per gram. However, the solid B2O3 formed by the combustion of B5H9 is an abrasive that would quickly destroy the nozzle of the rocket, and so the idea was abandoned. Calculate the kilojoules of heat released per gram of the compound reacted with oxygen. The standard enthalpy of formation of B5H9 is 73.2 kJ/mol. Using the Ho values in Appendix 3 and Equation (6.7), we write f Answer Ho rxn [(5 mol) Ho(B2O3) f (9 mol) Ho(H2O)] f [(2 mol) Ho(B5H9) f [(5 mol)( 1263.6 kJ/mol) (12 mol) Ho(O2)] f (9 mol)( 285.8 kJ/mol)] [(2 mol)(73.2 kJ/mol) (12 mol)(0)] 9036.6 kJ Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 220 THERMOCHEMISTRY This is the amount of heat released for every 2 moles of B5H9 reacted. The heat released per gram of B5H9 reacted is heat released per gram of B5H9 1 mol B5H9 63.12 g B5H9 9036.6 kJ 2 mol B5H9 71.58 kJ/g B5H9 Similar problems: 6.42. PRACTICE EXERCISE Benzene (C6H6) burns in air to produce carbon dioxide and liquid water. Calculate the heat released (in kilojoules) per gram of the compound reacted with oxygen. The standard enthalpy of formation of benzene is 49.04 kJ/mol. B5H9 is called a high-energy compound because of the large amount of energy it releases upon combustion. In general, compounds that have positive Ho values tend f to release more heat as a result of combustion and to be less stable than those with negative Ho values. f 6.6 HEAT OF SOLUTION AND DILUTION Although we have focused so far on the thermal energy effects resulting from chemical reactions, many physical processes, such as the melting of ice or the condensation of a vapor, also involve the absorption or release of heat. Enthalpy changes occur as well when a solute dissolves in a solvent or when a solution is diluted. Let us look at these two related physical processes, involving heat of solution and heat of dilution. HEAT OF SOLUTION In the vast majority of cases, dissolving a solute in a solvent produces measurable heat change. At constant pressure, the heat change is equal to the enthalpy change. The heat of solution, or enthalpy of solution, Hsoln , is the heat generated or absorbed when a certain amount of solute dissolves in a certain amount of solvent. The quantity Hsoln represents the difference between the enthalpy of the final solution and the enthalpies of its original components (that is, solute and solvent) before they are mixed. Thus Hsoln Hsoln Hcomponents (6.9) Neither Hsoln nor Hcomponents can be measured, but their difference, Hsoln, can be readily determined in a constant-pressure calorimeter. Like other enthalpy changes, Hsoln is positive for endothermic (heat-absorbing) processes and negative for exothermic (heat-generating) processes. Consider the heat of solution of a process in which an ionic compound is the solute and water is the solvent. For example, what happens when solid NaCl dissolves in water? In solid NaCl, the Na and Cl ions are held together by strong positive-negative (electrostatic) forces, but when a small crystal of NaCl dissolves in water, the threedimensional network of ions breaks into its individual units. (The structure of solid NaCl is shown in Figure 2.12.) The separated Na and Cl ions are stabilized in solution by their interaction with water molecules (see Figure 4.2). These ions are said to be hydrated. In this case water plays a role similar to that of a good electrical insulator. Water molecules shield the ions (Na and Cl ) from each other and effectively Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.6 HEAT OF SOLUTION AND DILUTION 221 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry How a Bombardier Beetle Defends Itself Survival techniques of insects and small animals in a fiercely competitive environment take many forms. For example, chameleons have developed the ability to change color to match their surroundings and the butterfly Limenitis has evolved into a form that mimics the poisonous and unpleasant-tasting monarch butterfly (Danaus). A less passive defense mechanism is employed by bombardier beetles (Brachinus), which repel predators with a “chemical spray.” The bombardier beetle has a pair of glands at the tip of its abdomen. Each gland consists of two compartments. The inner compartment contains an aqueous solution of hydroquinone and hydrogen peroxide, and the outer compartment holds a mixture of enzymes. (Enzymes are biological molecules that can speed up a reaction.) When threatened, the beetle squeezes some fluid from the inner compartment into the outer compartment, where, in the presence of the enzymes, an exothermic reaction takes place: (a) C6H4(OH)2(aq) hydroquinone H2O2(aq) 88n Therefore, we write C6H4O2(aq) quinone (c) H2O2(aq) 88n H2O(l ) (d) H2(g) 1 2 1 2 H2(g) H° 177 k J O2(g) O2(g) 88n H2O(l ) H° H° 94.6 k J 286 k J Recalling Hess’s law, we find that the heat of reaction for (a) is simply the sum of those for (b), (c), and (d). Ho b Ho c (177 Ho a 2H2O(l ) To estimate the heat of reaction, let us consider the following steps: (b) C6H4(OH)2(aq) 88n C6H4O2(aq) A bombardier beetle discharging a chemical spray. 94.6 Ho d 286) k J 204 k J The large amount of heat generated is sufficient to bring the mixture to its boiling point. By rotating the tip of its abdomen, the beetle can quickly discharge the vapor in the form of a fine mist toward an unsuspecting predator. In addition to the thermal effect, the quinones also act as a repellent to other insects and animals. One bombardier beetle carries enough reagents to produce 20 to 30 discharges in quick succession, each with an audible detonation. reduce the electrostatic attraction that held them together in the solid state. The heat of solution is defined by the following process: HO 2 NaCl(s) 88n Na (aq) Cl (aq) Hsoln ? Dissolving an ionic compound such as NaCl in water involves complex interactions among the solute and solvent species. However, for the sake of analysis we can imagine that the solution process takes place in two separate steps, illustrated in Figure 6.8. Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 222 THERMOCHEMISTRY FIGURE 6.8 The solution process for NaCl. The process can be considered to occur in two separate steps: (1) separation of ions from the crystal state to the gaseous state, and (2) hydration of the gaseous ions. The heat of solution is equal to the energy changes for these two steps, Hsoln U Hhydr. – – + + – – + – + + – + Na+ and Cl– ions in the gaseous state n tio ol ra /m yd kJ f h 84 7 to ea = – ∆ H Hh Step 2 yd tti c 78 e en e 8 kJ rgy /m ol r – + La Step 1 = U + – + – + – – + – + + – + – – +– + Heat of solution ∆ Hsoln = 4 kJ/mol – + + – Na+ and Cl– ions in the solid state Hydrated Na+ and Cl– ions First, the Na and Cl ions in the solid crystal are separated from each other and converted to the gaseous state: energy Lattice energy is a positive quantity. NaCl(s) 88n Na (g) Cl (g) The energy required to completely separate one mole of a solid ionic compound into gaseous ions is called lattice energy (U). The lattice energy of NaCl is 788 kJ/mol. In other words, we would need to supply 788 kJ of energy to break 1 mole of solid NaCl into 1 mole of Na ions and 1 mole of Cl ions. Next, the “gaseous” Na and Cl ions enter the water and become hydrated: Na (g) HO 2 Cl (g) 88n Na (aq) Cl (aq) energy The enthalpy change associated with the hydration process is called the heat of hydration, Hhydr. (Heat of hydration is a negative quantity for cations and anions.) Applying Hess’s law, it is possible to consider Hsoln as the sum of two related quantities, lattice energy (U) and heat of hydration ( Hhydr): NaCl(s) 88n Na (g) Na (g) HO 2 Cl (g) 88n Na (aq) H2O NaCl(s) 88n Na (aq) Cl (g) U Cl (aq) Hhydr Cl (aq) Hsoln 788 kJ 784 kJ 4 kJ Therefore, when 1 mole of NaCl dissolves in water, 4 kJ of heat will be absorbed from Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.7 TABLE 6.4 Heats of Solution of Some Ionic Compounds COMPOUND Hsoln (kJ/mol) LiCl CaCl2 NaCl KCl NH4Cl NH4NO3 37.1exothermic 82.8 4.0 17.2 endothermic 15.2 26.2 Generations of chemistry students have been reminded of the safe procedure for diluting acids by the venerable saying, “Do as you oughter, add acid to water.” 6.7 INTRODUCTION TO THERMODYNAMICS 223 the surroundings. We would observe this effect by noting that the beaker containing the solution becomes slightly colder. Table 6.4 lists the Hsoln of several ionic compounds. Depending on the nature of the cation and anion involved, Hsoln for an ionic compound may be either negative (exothermic) or positive (endothermic). HEAT OF DILUTION When a previously prepared solution is diluted, that is, when more solvent is added to lower the overall concentration of the solute, additional heat is usually given off or absorbed. The heat of dilution is the heat change associated with the dilution process. If a certain solution process is endothermic and the solution is subsequently diluted, more heat will be absorbed by the same solution from the surroundings. The converse holds true for an exothermic solution process—more heat will be liberated if additional solvent is added to dilute the solution. Therefore, always be cautious when working on a dilution procedure in the laboratory. Because of its highly exothermic heat of dilution, concentrated sulfuric acid (H2SO4) poses a particularly hazardous problem if its concentration must be reduced by mixing it with additional water. Concentrated H2SO4 is composed of 98 percent acid and 2 percent water by mass. Diluting it with water releases considerable heat to the surroundings. This process is so exothermic that you must never attempt to dilute the concentrated acid by adding water to it. The heat generated could cause the acid solution to boil and splatter. The recommended procedure is to add the concentrated acid slowly to the water (while constantly stirring). INTRODUCTION TO THERMODYNAMICS Thermochemistry is part of a broader subject called thermodynamics, which is the scientific study of the interconversion of heat and other kinds of energy. The laws of thermodynamics provide useful guidelines for understanding the energetics and directions of processes. In this section we will concentrate on the first law of thermodynamics, which is particularly relevant to the study of thermochemistry. We will continue our discussion of thermodynamics in Chapter 18. In thermodynamics, we study changes in the state of a system, which is defined by the values of all relevant macroscopic properties, for example, composition, energy, temperature, pressure, and volume. Energy, pressure, volume, and temperature are said to be state functions—properties that are determined by the state of the system, regardless of how that condition was achieved. In other words, when the state of a system changes, the magnitude of change in any state function depends only on the initial and final states of the system and not on how the change is accomplished. The state of a given amount of a gas is specified by its volume, pressure, and temperature. Consider a gas at 2 atm, 300 K, and 1 L (the initial state). Suppose a process is carried out at constant temperature such that the gas pressure decreases to 1 atm. According to Boyle’s law, its volume must increase to 2 L. The final state then corresponds to 1 atm, 300 K, and 2 L. The change in volume ( V ) is V Vf Vi 2L 1L 1L where Vi and Vf denote the initial and final volume, respectively. No matter how we ar- Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 224 THERMOCHEMISTRY FIGURE 6.9 The gain in gravitational potential energy that occurs when a person climbs from the base to the top of a mountain is independent of the path taken. Recall that an object possesses potential energy by virtue of its position or chemical composition. rive at the final state (for example, the pressure of the gas can be increased first and then decreased to 1 atm), the change in volume is always 1 L. Thus the volume of a gas is a state function. In a similar manner we could show that pressure and temperature are also state functions. Energy is another state function. Using potential energy as an example, we find that the net increase in gravitational potential energy when we go from the same starting point to the top of a mountain is always the same, regardless of how we get there (Figure 6.9). THE FIRST LAW OF THERMODYNAMICS The first law of thermodynamics, which is based on the law of conservation of energy, states that energy can be converted from one form to another, but cannot be created or destroyed. How do we know this is so? It would be impossible to prove the validity of the first law of thermodynamics if we had to determine the total energy content of the universe. Even determining the total energy content of 1 g of iron, say, would be extremely difficult. Fortunately, we can test the validity of the first law by measuring only the change in the internal energy of a system between its initial state and its final state. The change in internal energy E is given by E Ef Ei where Ei and Ef are the internal energies of the system in the initial and final states, respectively. The internal energy of a system has two components: kinetic energy and potential energy. The kinetic energy component consists of various types of molecular motion and the movement of electrons within molecules. Potential energy is determined by the attractive interactions between electrons and nuclei and by repulsive interactions between electrons and between nuclei in individual molecules, as well as by interaction between molecules. It is impossible to measure all these contributions accurately, so we cannot calculate the total energy of a system with any certainty. Changes in energy, on the other hand, can be determined experimentally. Consider the reaction between 1 mole of sulfur and 1 mole of oxygen gas to produce 1 mole of sulfur dioxide: S(s) Sulfur burns in air to form SO2. Back Forward O2(g) 88n SO2(g) In this case the system is composed of the reactant molecules S and O2 and the product molecules SO2. We do not know the internal energy content of either the reactant Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.7 INTRODUCTION TO THERMODYNAMICS 225 molecules or the product molecules, but we can accurately measure the change in energy content, E, given by E E(product) E(reactants) energy content of 1 mol SO2(g) [energy content of 1 mol S(s) 1 mol O2(g)] We find that this reaction gives off heat. Therefore, the energy of the product is less than that of the reactants, and E is negative. Interpreting the release of heat in the above reaction to mean that some of the chemical energy contained in the reactants has been converted to thermal energy, we conclude that the transfer of energy from the reactants to the surroundings does not change the total energy of the universe. That is, the sum of the energy changes must be zero: Esys or Esurr Esys 0 Esurr where the subscripts “sys” and “surr” denote system and surroundings, respectively. Thus, if one system undergoes an energy change Esys, the rest of the universe, or the surroundings, must undergo a change in energy that is equal in magnitude but opposite in sign; energy gained in one place must have been lost somewhere else. Furthermore, because energy can be changed from one form to another, the energy lost by one system can be gained by another system in a different form. For example, the energy lost by burning oil in a power plant may ultimately turn up in our homes as electrical energy, heat, light, and so on. In chemistry, we are normally interested in the changes associated with the system (which may be a flask containing reactants and products), not with its surroundings. Therefore, a more useful form of the first law is E For convenience, we sometimes omit the word “internal” when discussing the energy of a system. q w (6.10) (We drop the subscript “sys” for simplicity.) Equation (6.10) says that the change in the internal energy E of a system is the sum of the heat exchange q between the system and the surroundings and the work done w on (or by) the system. Using the sign convention for thermochemical processes (see Section 6.2), we find that q is positive for an endothermic process and negative for an exothermic process. The sign convention for work is that w is positive for work done on the system by the surroundings and negative for work done by the system on the surroundings. Equation (6.10) may seem abstract, but it is actually quite logical. If a system loses heat to the surroundings or does work on the surroundings, we would expect its internal energy to decrease since both are energy-depleting processes. Conversely, if heat is added to the system or if work is done on the system, then the internal energy of the system would increase, as predicted by Equation (6.10). Table 6.5 summarizes the sign conventions for q and w. TABLE 6.5 Sign Conventions for Work and Heat PROCESS SIGN Work done by the system on the surroundings Work done on the system by the surroundings Heat absorbed by the system from the surroundings (endothermic process) Heat absorbed by the surroundings from the system (exothermic process) Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 226 THERMOCHEMISTRY FIGURE 6.10 The expansion of a gas against a constant external pressure (such as atmospheric pressure). The gas is in a cylinder fitted with a weightless movable piston. The work done is given by P V. P P ∆V WORK AND HEAT We have seen that work can be defined as force F multiplied by distance d: w Fd In thermodynamics, work has a broader meaning that includes mechanical work (for example, a crane lifting a steel beam), electrical work (a battery supplying electrons to light the bulb of a flashlight), and so on. In this section we will concentrate on mechanical work; in Chapter 19 we will discuss the nature of electrical work. A useful example of mechanical work is the expansion of a gas (Figure 6.10). Suppose that a gas is in a cylinder fitted with a weightless, frictionless movable piston, at a certain temperature, pressure, and volume. As it expands, the gas pushes the piston upward against a constant opposing external atmospheric pressure P. The work done by the gas on the surroundings is w PV (6.11) where V, the change in volume, is given by Vf Vi. The minus sign in Equation (6.11) takes care of the convention for w. For gas expansion, V 0, so P V is a negative quantity. For gas compression (work done on the system), V 0, and P V is a positive quantity. Equation (6.11) derived from the fact that pressure volume can be expressed as (force/area) volume; that is, P V F d2 pressure d3 Fd w volume where F is the opposing force and d has the dimension of length, d 2 has the dimensions of area, and d 3 has the dimensions of volume. Thus the product of pressure and volume is equal to force times distance, or work. You can see that for a given increase in volume (that is, for a certain value of V ), the work done depends on the magnitude of the external, opposing pressure P. If P is zero (that is, if the gas is expanding against a vacuum), the work done must also be zero. If P is some positive, nonzero value, then the work done is given by P V. According to Equation (6.11), the units for work done by or on a gas are liters atmospheres. To express the work done in the more familiar unit of joules, we use the conversion factor Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 6.7 INTRODUCTION TO THERMODYNAMICS 1 L atm 227 101.3 J Example 6.7 shows how to calculate work done by an expanding gas. EXAMPLE 6.7 A certain gas expands in volume from 2.0 L to 6.0 L at constant temperature. Calculate the work done by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 1.2 atm. Answer (a) Since the external pressure is zero, the work w done is w PV 0(6.0 2.0) L 0 (b) The external, opposing pressure is 1.2 atm, so w PV (1.2 atm)(6.0 2.0) L 4.8 atm L 4.8 L atm To convert the answer to joules, we write w 4.8 L atm 4.9 Similar problem: 6.86. 101.3 J 1 L atm 102 J PRACTICE EXERCISE A gas expands from 264 mL to 971 mL at constant temperature. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 4.00 atm. Because the temperature is kept constant, you can use Boyle’s law to show that the final pressure is the same in (a) and (b). Example 6.7 shows that work is not a state function. Although the initial and final states are the same in (a) and (b), the amount of work done is different because the external, opposing pressures are different. We cannot write w wf wi for a change. Work done depends not only on the initial state and final state, but also on how the process is carried out. The other component of internal energy is q, heat. Like work, heat is not a state function. Suppose that the change can be brought about in a system in two different ways. In one case the work done is zero, so that we have E q1 w1 q1 In the second case work is done and heat is transferred, so that E We use lowercase letters (such as w and q) to represent thermodynamic quantities that are not state functions. Back Forward Main Menu q2 w2 Since E is the same for both cases (because the initial and final states are the same in both cases), it follows that q1 q2. This simple example shows that the heat associated with a given process, like work, depends on how the process is carried out; that is, we cannot write q qf qi. It is important to note that although neither heat nor TOC Study Guide TOC Textbook Website MHHE Website 228 THERMOCHEMISTRY work is a state function, their sum (q w) is equal to E, and as we saw earlier, E is a state function. In summary, heat and work are not state functions because they are not properties of a system. They manifest themselves only during a process (during a change). Thus their values depend on the path of the process and vary accordingly. The following example shows an application of the first law. EXAMPLE 6.8 The work done when a gas is compressed in a cylinder like that shown in Figure 6.10 is 462 J. During this process, there is a heat transfer of 128 J from the gas to the surroundings. Calculate the energy change for this process. Compression is work done on the gas, so the sign for w is positive: w 462 J. From the direction of heat transfer (system to surroundings), we know that q is negative: Answer E q w 128 J 462 J 334 J Similar problem: 6.59. As a result of the compression and heat transfer, the energy of the gas increases by 334 J. PRACTICE EXERCISE A gas expands and does P-V work on the surroundings equal to 279 J. At the same time, it absorbs 216 J of heat from the surroundings. What is the change in energy of the system? ENTHALPY AND THE FIRST LAW OF THERMODYNAMICS How do we apply the first law of thermodynamics to processes carried out under constant-pressure and constant-volume conditions? If a reaction is run at constant volume, then V 0 and no work will result from this change. From Equation (6.10) it follows that the change in energy is equal to the heat change: E q w qv We add the subscript “v” to remind us that this is a constant-volume process. Bomb calorimetry, discussed in Section 6.4, is a constant-volume process. Constant-volume conditions are often inconvenient and sometimes impossible to achieve. Most reactions occur under conditions of constant pressure (usually atmospheric pressure). If such a reaction results in a net increase in the number of moles of a gas, then the system does work on the surroundings (expansion). This follows from the fact that the gas formed must push the atmosphere back. Conversely, if more gas molecules are consumed than are produced, work is done on the system by the surroundings (compression). On p. 206 it was stated that the heat change under constant-pressure conditions, qp, is equal to the enthalpy change. Therefore, for a constant-pressure process E q qp Back Forward Main Menu TOC Study Guide TOC w PV Textbook Website MHHE Website 6.7 FIGURE 6.11 (a) A beaker of water inside a cylinder fitted with a movable piston. The pressure inside is equal to the atmospheric pressure. (b) After the sodium metal has reacted with water, hydrogen gas pushes the piston upward (doing work on the surroundings) until the pressure inside is again equal to that outside. INTRODUCTION TO THERMODYNAMICS 229 P P Air + water vapor + H2 gas Air + water vapor (a) (b) or qp E PV E PV We define the enthalpy of a system as H (6.12) where E is the internal energy of the system and P and V are the pressure and volume of the system, respectively. Since E and the product PV have energy units, enthalpy also has the energy units. Furthermore, because E, P, and V are all state functions, the changes in (E PV ) depend only on the initial and final states. It follows that the change in H also depends only on the initial and final states; that is, H is a state function. The fact that H is a state function allows us to explain Hess’s law, discussed in Section 6.5. Because H depends only on the initial and final state (that is, only on the nature of reactants and products), the enthalpy change for a given reaction is the same whether it takes one step or many steps. For any process, the change in enthalpy is given by H E (PV ) (6.13) PV (6.14) If pressure is held constant, then H E Comparing Equation (6.14) with the expression for qp above, we see that H qp. Let us consider a reaction carried out under constant-pressure conditions. When a small piece of sodium metal is added to a beaker of water, the reaction that takes place is The heat or enthalpy change for this reaction can be measured in a constant-pressure calorimeter. 2Na(s) 2H2O(l) 88n 2NaOH(aq) H2(g) H 367.5 kJ One of the products is gaseous hydrogen, which must push back air to enter the atmosphere. Consequently, some of the energy produced by the reaction is used to do the work of pushing back a volume of air ( V ) against atmospheric pressure (P) (Figure 6.11). To calculate the change in internal energy, we rearrange Equation (6.14) as follows: E H PV If we assume the temperature to be 25°C and ignore the small change in the volume of the solution, we can show that the volume of 1 mole of H2 gas at 1.0 atm is 24.5 L, so that P V 24.5 L atm or 2.5 kJ. Finally, Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website 230 THERMOCHEMISTRY E 367.5 kJ 2.5 kJ 370.0 kJ This calculation shows that E and H are approximately the same. The reason H is smaller than E is that some of the internal energy released is used to do gas expansion work, so less heat is evolved. For reactions that do not involve gases, V is very small and so E is practically the same as H. Another way to calculate the internal energy change of a gaseous reaction is to assume ideal gas behavior and constant temperature. In this case, E H (PV ) H (nRT ) H RT n (6.15) where n is defined as n number of moles of product gases number of moles of reactant gases The following example applies this equation. EXAMPLE 6.9 Calculate the change in internal energy when 2 moles of CO are converted to 2 moles of CO2 at 1 atm and 25°C: 2CO(g) Answer: O2(g) 88n 2CO2(g) H° 566.0 kJ The total change in the number of moles of gases is given by n 2 mol 3 mol 1 mol From Equation (6.15) we write Carbon monoxide burns in air to form carbon dioxide. E° H° RT n 566.0 kJ (8.314 J/K mol)(298 K)( 1 mol) 563.5 kJ Note that we used 8.314 J/K mol for R (see Appendix 2) because the units for E and H are in joules. Comment Similar problems: 6.67, 6.68. PRACTICE EXERCISE What is E° for the formation of 1 mole of CO at 1 atm and 25°C? C(graphite) SUMMARY OF KEY EQUATIONS • C ms (6.2) • q ms t (6.3) • q C t (6.4) • Ho n Ho(products) f rxn o m Hf (reactants) • E q w (6.10) •w Back Forward Main Menu P V (6.11) TOC 1 2 O2(g) 88n CO(g) H° 110.5 kJ Definition of heat capacity. Calculating heat change in terms of specific heat. Calculating heat change in terms of heat capacity. Calculating standard enthalpy of reaction. (6.8) Mathematical statement of the first law of thermodynamics. Calculating work done in gas expansion or compression. Study Guide TOC Textbook Website MHHE Website SUMMARY OF KEY EQUATIONS 231 Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry in Action Chemistry Making Snow and Inflating a Bicycle Tire Many phenomena in everyday life can be explained by the first law of thermodynamics. Here we will discuss two examples of interest to lovers of the outdoors. • MAKING SNOW If you are an avid downhill skier, you have probably skied on artificial snow. How is this stuff made in quantities large enough to meet the needs of skiers on snowless days? The secret of snowmaking is in the equation E q w. A snowmaking machine contains a mixture of compressed air and water vapor at about 20 atm. Because of the large difference in pressure between the tank and the outside atmosphere, when the mixture is sprayed into the atmosphere it expands so rapidly that, as a good approximation, no heat exchange occurs between the system (air and water) and its surroundings; that is, q 0. (In thermodynamics, such a process is called an adiabatic process.) Thus we write E q w Because the system does work on the surroundings, w is a negative quantity, and there is a decrease in the system’s energy. Kinetic energy is part of the total energy of the A snowmaking machine in operation. system. In Section 5.7 we saw that the average ki• INFLATING A BICYCLE TIRE netic energy of a gas is directly proportional to the If you have ever pumped air into a bicycle tire, you absolute temperature [Equation (5.11)]. It follows, probably noticed a warming effect at the valve stem. therefore, that the change in energy E is given by This phenomenon, too, can be explained by the first law of thermodynamics. The action of the pump comE CT presses the air inside the pump and the tire. The where C is the proportionality constant. Because E process is rapid enough to be treated as approxiis negative, T must also be negative, and it is this mately adiabatic, so that q 0 and E w. Because cooling effect (or the decrease in the kinetic energy work is done on the gas in this case (it is being comof the water molecules) that is responsible for the forpressed), w is positive, and there is an increase in enmation of snow. Although we need only water to form ergy. Hence, the temperature of the system increases snow, the presence of air, which also cools on exalso, according to the equation pansion, helps to lower the temperature of the water E CT vapor. • H E PV (6.12) •H E P V (6.14) • Back w Forward Main Menu E H TOC RT n (6.15) Definition of enthalpy. Calculating enthalpy (or energy) change for a constant-pressure process. Calculating enthalpy (or energy) change for a constant-pressure process. Study Guide TOC Textbook Website MHHE Website 232 THERMOCHEMISTRY SUMMARY OF FACTS AND CONCEPTS 1. Energy is the capacity to do work. There are many forms of energy and they are interconvertible. The law of conservation of energy states that the total amount of energy in the universe is constant. 2. A process that gives off heat to the surroundings is exothermic; a process that absorbs heat from the surroundings is endothermic. 3. The change in enthalpy ( H, usually given in kilojoules) is a measure of the heat of reaction (or any other process) at constant pressure. 4. Constant-volume and constant-pressure calorimeters are used to measure heat changes that occur in physical and chemical processes. 5. Hess’s law states that the overall enthalpy change in a reaction is equal to the sum of enthalpy changes for individual steps in the overall reaction. 6. The standard enthalpy of a reaction can be calculated from the standard enthalpies of formation of reactants and products. 7. The heat of solution of an ionic compound in water is the sum of the lattice energy of the compound and the heat of hydration. The relative magnitudes of these two quantities determine whether the solution process is endothermic or exothermic. The heat of dilution is the heat absorbed or evolved when a solution is diluted. 8. The state of a system is defined by properties such as composition, volume, temperature, and pressure. These properties are called state functions. 9. The change in a state function for a system depends only on the initial and final states of the system, and not on the path by which the change is accomplished. Energy is a state function; work and heat are not. 10. Energy can be converted from one form to another, but it cannot be created or destroyed (first law of thermodynamics). In chemistry we are concerned mainly with thermal energy, electrical energy, and mechanical energy, which is usually associated with pressurevolume work. 11. Enthalpy is a state function. A change in enthalpy H is equal to E P V for a constant-pressure process. KEY WORDS Calorimetry, p. 209 Chemical energy, p. 204 Closed system, p. 205 Endothermic process, p. 206 Energy, p. 204 Enthalpy, p. 206 Enthalpy of reaction, p. 209 Enthalpy of solution ( Hsoln), p. 220 Exothermic process, p. 206 First law of thermodynamics, p. 224 Heat, p. 205 Heat capacity, p. 210 Heat of dilution, p. 223 Heat of hydration ( Hhydr), p. 222 Heat of solution ( Hsoln), p. 220 Hess’s law, p. 217 Isolated system, p. 205 Lattice energy, p. 222 Law of conservation of energy, p. 204 QUESTIONS AND PROBLEMS THE NATURE OF ENERGY AND TYPES OF ENERGY Review Questions 6.1 Define and give an example of each of the following terms: thermal energy, chemical energy, potential energy, kinetic energy, law of conservation of energy. Back Forward Main Menu TOC Open system, p. 205 Potential energy, p. 204 Radiant energy, p. 204 Specific heat(s), p. 210 Standard enthalpy of formation ( Ho), p. 214 f Standard enthalpy of reaction ( Ho ), p. 216 rxn Standard state, p. 214 State function, p. 223 State of a system, p. 223 Surroundings, p. 205 System, p. 205 Thermal energy, p. 204 Thermochemical equation, p. 208 Thermochemistry, p. 205 Thermodynamics, p. 223 Work, p. 204 6.2 What units of energy are commonly used in chemistry? 6.3 A truck traveling at 60 kilometers per hour is brought to a complete stop at a traffic light. Does this change in velocity violate the law of conservation of energy? Explain. Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 6.4 Describe the energy conversions that occur in the following processes: (a) You throw a softball up into the air and catch it. (b) You switch on a flashlight. (c) You ride the ski lift to the top of a hill and then ski down. (d) You strike a match and let it burn down. ENERGY CHANGES IN CHEMICAL REACTIONS 6.5 What is heat? How does heat differ from thermal energy? Under what condition is heat transferred from one system to another? 6.6 Explain the following terms: thermochemistry, system, surroundings, open system, closed system, isolated system, exothermic process, endothermic process. 6.7 Stoichiometry is based on the law of conservation of mass. On what law is thermochemistry based? 6.8 Describe two exothermic processes and two endothermic processes. ENTHALPY Review Questions 6.9 Write an expression for the enthalpy of a reaction in terms of the enthalpies of products and reactants. Under what condition is the heat of a reaction equal to the enthalpy change of the same reaction? 6.10 In writing thermochemical equations, why is it important to indicate the physical state (that is, gaseous, liquid, solid, or aqueous) of each substance? 6.11 Explain the following thermochemical equation: 5O2(g) 88n 4NO(g) 6H2O(g) H 905 J 6.12 Consider the following reaction: 2CH3OH(l) 3O2(g) 88n 4H2O(l) 2CO2(g) H 1452.8 kJ What is the value of H if (a) the equation is multiplied throughout by 2, (b) the direction of the reaction is reversed so that the products become the reactants and vice versa, (c) water vapor instead of liquid water is formed as the product? CALORIMETRY Review Questions 6.13 What is the difference between specific heat and heat capacity? What are the units for these two quantities? Which is the intensive property and which is the extensive property? 6.14 Consider two metals A and B, each having a mass of 100 g and an initial temperature of 20°C. The specific heat of A is larger than that of B. Under the same Back Forward Main Menu TOC heating conditions, which metal would take longer to reach a temperature of 21°C? 6.15 Define calorimetry and describe two commonly used calorimeters. 6.16 In a calorimetric measurement, why is it important that we know the heat capacity of the calorimeter? Problems Review Questions 4NH3(g) 233 6.17 A piece of silver of mass 362 g has a heat capacity of 85.7 J/°C. What is the specific heat of silver? 6.18 A 6.22-kg piece of copper metal is heated from 20.5°C to 324.3°C. Calculate the heat absorbed (in kJ) by the metal. 6.19 Calculate the amount of heat liberated (in kJ) from 366 g of mercury when it cools from 77.0°C to 12.0°C. 6.20 A sheet of gold weighing 10.0 g and at a temperature of 18.0°C is placed flat on a sheet of iron weighing 20.0 g and at a temperature of 55.6°C. What is the final temperature of the combined metals? Assume that no heat is lost to the surroundings. (Hint: The heat gained by the gold must be equal to the heat lost by the iron.) 6.21 A 0.1375-g sample of solid magnesium is burned in a constant-volume bomb calorimeter that has a heat capacity of 1769 J/°C. The calorimeter contains exactly 300 g of water, and the temperature increases by 1.126°C. Calculate the heat given off by the burning Mg, in kJ/g and in kJ/mol. 6.22 A quantity of 2.00 102 mL of 0.862 M HCl is mixed with 2.00 102 mL of 0.431 M Ba(OH)2 in a constant-pressure calorimeter that has a heat capacity of 453 J/°C. The initial temperature of the HCl and Ba(OH)2 solutions is the same at 20.48°C. For the process H (aq) OH (aq) 88n H2O(l) the heat of neutralization is 56.2 kJ. What is the final temperature of the mixed solution? STANDARD ENTHALPY OF FORMATION AND REACTION Review Questions 6.23 What is meant by the standard-state condition? 6.24 How are the standard enthalpies of an element and of a compound determined? 6.25 What is meant by the standard enthalpy of a reaction? 6.26 Write the equation for calculating the enthalpy of a reaction. Define all the terms. 6.27 State Hess’s law. Explain, with one example, the usefulness of Hess’s law in thermochemistry. 6.28 Describe how chemists use Hess’s law to determine the Ho of a compound by measuring its heat (enf thalpy) of combustion. Study Guide TOC Textbook Website MHHE Website 234 THERMOCHEMISTRY Problems 6.29 Which of the following standard enthalpy of formation values is not zero at 25°C? Na(s), Ne(g), CH4(g), S8(s), Hg(l), H(g). 6.30 The Ho values of the two allotropes of oxygen, O2 f and O3, are 0 and 142.2 kJ/mol, respectively, at 25°C. Which is the more stable form at this temperature? 6.31 Which is the more negative quantity at 25°C: Ho f for H2O(l ) or Ho for H2O(g)? f 6.32 Predict the value of Ho (greater than, less than, or f equal to zero) for these elements at 25°C: (a) Br2(g); Br2(l), (b) I2(g); I2(s). 6.33 In general, compounds with negative Ho values are f more stable than those with positive Ho values. f H2O2(l) has a negative Ho (see Table 6.3). Why, f then, does H2O2(l) have a tendency to decompose to H2O(l) and O2(g)? 6.34 Suggest ways (with appropriate equations) that would allow you to measure the Ho values of Ag2O(s) and f CaCl2(s) from their elements. No calculations are necessary. 6.35 Calculate the heat of decomposition for this process at constant pressure and 25°C: CaCO3(s) 88n CaO(s) CO2(g) (Look up the standard enthalpy of formation of the reactant and products in Table 6.3.) 6.36 The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to H ions; that is, Ho[H (aq)] 0. f (a) For the following reaction HCl(g) 88n H (aq) Cl (aq) H° 74.9 kJ calculate Ho for the Cl ions. (b) Given that Ho f f for OH ions is 229.6 kJ/mol, calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as HCl) is titrated by 1 mole of a strong base (such as KOH) at 25°C. 6.37 Calculate the heats of combustion for the following reactions from the standard enthalpies of formation listed in Appendix 3: (a) 2H2(g) O2(g) 88n 2H2O(l) (b) 2C2H2(g) 5O2(g) 88n 4CO2(g) 2H2O(l) 6.38 Calculate the heats of combustion for the following reactions from the standard enthalpies of formation listed in Appendix 3: (a) C2H4(g) 3O2(g) 88n 2CO2(g) 2H2O(l) (b) 2H2S(g) 3O2(g) 88n 2H2O(l) 2SO2(g) 6.39 Methanol, ethanol, and n-propanol are three common alcohols. When 1.00 g of each of these alcohols is burned in air, heat is liberated as follows: 22.6 kJ; (b) ethanol (a) methanol (CH3OH), (C2H5OH), 29.7 kJ; (c) n-propanol (C3H7OH), Back Forward Main Menu TOC 33.4 kJ. Calculate the heats of combustion of these alcohols in kJ/mol. 6.40 The standard enthalpy change for the following reaction is 436.4 kJ: H2(g) 88n H(g) H(g) Calculate the standard enthalpy of formation of atomic hydrogen (H). 6.41 From the standard enthalpies of formation, calculate Ho for the reaction rxn C6H12(l) 9O2(g) 88n 6CO2(g) 6H2O(l) Ho f For C6H12(l), 151.9 kJ/mol. 6.42 The first step in the industrial recovery of zinc from the zinc sulfide ore is roasting, that is, the conversion of ZnS to ZnO by heating: 2ZnS(s) 3O2(g) 88n 2ZnO(s) 2SO2(g) Ho rxn 879 kJ Calculate the heat evolved (in kJ) per gram of ZnS roasted. 6.43 Determine the amount of heat (in kJ) given off when 1.26 104 g of ammonia are produced according to the equation N2(g) 3H2(g) 88n 2NH3(g) Ho rxn 92.6 kJ Assume that the reaction takes place under standardstate conditions at 25°C. 6.44 At 850°C, CaCO3 undergoes substantial decomposition to yield CaO and CO2. Assuming that the Ho f values of the reactant and products are the same at 850°C as they are at 25°C, calculate the enthalpy change (in kJ) if 66.8 g of CO2 are produced in one reaction. 6.45 From these data, S(rhombic) Ho rxn 296.06 kJ O2(g) 88n SO2(g) Ho rxn 296.36 kJ O2(g) 88n SO2(g) S(monoclinic) calculate the enthalpy change for the transformation S(rhombic) 88n S(monoclinic) (Monoclinic and rhombic are different allotropic forms of elemental sulfur.) 6.46 From the following data, C(graphite) H2(g) 1 2 2C2H6(g) O2(g) 88n CO2(g) O2(g) 88n H2O(l) 7O2(g) 88n 4CO2(g) Ho rxn 393.5 kJ Ho rxn 285.8 kJ 6H2O(l) Ho 3119.6 kJ rxn calculate the enthalpy change for the reaction Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS 2C(graphite) 3H2(g) 88n C2H6(g) 6.47 From the following heats of combustion, 3 2 CH3OH(l) 1 2 H2(g) 2H2O(l) Ho 726.4 kJ rxn O2(g) 88n CO2(g) C(graphite) O2(g) 88n CO2(g) Ho rxn 393.5 kJ Ho rxn 285.8 kJ O2(g) 88n H2O(l) calculate the enthalpy of formation of methanol (CH3OH) from its elements: C(graphite) 2H2(g) 1 2 O2(g) 88n CH3OH(l) 6.48 Calculate the standard enthalpy change for the reaction 2Al(s) Fe2O3(s) 88n 2Fe(s) Al2O3(s) given that 2Al(s) 3 2 O2(g) 88n Al2O3(s) 2Fe(s) 3 2 O2(g) 88n Fe2O3(s) Ho rxn Ho rxn 1601 kJ 821 kJ HEAT OF SOLUTION AND DILUTION Review Questions 6.49 Define the following terms: enthalpy of solution, hydration, heat of hydration, lattice energy, heat of dilution. 6.50 Why is the lattice of a solid always a positive quantity? Why is the hydration of ions always a negative quantity? 6.51 Consider two ionic compounds A and B. A has a larger lattice energy than B. Which of the two compounds is more stable? 6.52 Mg2 is a smaller cation than Na and also carries more positive charge. Which of the two species has a larger hydration energy (in kJ/mol)? Explain. 6.53 Consider the dissolution of an ionic compound such as potassium fluoride in water. Break the process into the following steps: separation of the cations and anions in the vapor phase and the hydration of the ions in the aqueous medium. Discuss the energy changes associated with each step. How does the heat of solution of KF depend on the relative magnitudes of these two quantities? On what law is the relationship based? 6.54 Why is it dangerous to add water to a concentrated acid such as sulfuric acid in a dilution process? 235 6.56 Explain what is meant by a state function. Give two examples of quantities that are state functions and two that are not. 6.57 The internal energy of an ideal gas depends only on its temperature. Do a first-law analysis of the following process. A sample of an ideal gas is allowed to expand at constant temperature against atmospheric pressure. (a) Does the gas do work on its surroundings? (b) Is there heat exchange between the system and the surroundings? If so, in which direction? (c) What is E for the gas for this process? 6.58 At constant pressure, in which of the following reactions is work done by the system on the surroundings? By the surroundings on the system? In which of them is no work done? (a) Hg(l) 88n Hg(g) (b) 3O2(g) 88n 2O3(g) (c) CuSO4 5H2O(s) 88n CuSO4(s) 5H2O(g) (d) H2(g) F2(g) 88n 2HF(g) Problems 6.59 A gas expands and does P-V work on the surroundings equal to 325 J. At the same time, it absorbs 127 J of heat from the surroundings. Calculate the change in energy of the gas. 6.60 The work done to compress a gas is 74 J. As a result, 26 J of heat is given off to the surroundings. Calculate the change in energy of the gas. 6.61 Calculate the work done when 50.0 g of tin are dissolved in excess acid at 1.00 atm and 25°C: Sn(s) 2H (aq) 88n Sn2 (aq) H2(g) Assume ideal gas behavior. 6.62 Calculate the work done in joules when 1.0 mole of water vaporizes at 1.0 atm and 100°C. Assume that the volume of liquid water is negligible compared with that of steam at 100°C and ideal gas behavior. Additional Problems 6.63 The convention of arbitrarily assigning a zero enthalpy value for the most stable form of each element in the standard state of 25°C is a convenient way of dealing with enthalpies of reactions. Explain why this convention cannot be applied to nuclear reactions. 6.64 Consider the following two reactions: 6.55 On what law is the first law of thermodynamics based? Explain the sign conventions in the equation E q w. Forward Main Menu H1 Ho rxn H2 Determine the enthalpy change for the process Review Questions Back Ho rxn A 88n C FIRST LAW OF THERMODYNAMICS A 88n 2B TOC 2B 88n C 6.65 The standard enthalpy change H° for the thermal decomposition of silver nitrate according to the fol- Study Guide TOC Textbook Website MHHE Website 236 THERMOCHEMISTRY lowing equation is 6.72 You are given the following data: 78.67 kJ: 1 2 AgNO3(s) 88n AgNO2(s) H2(g) 88n 2H(g) The standard enthalpy of formation of AgNO3(s) is 123.02 kJ/mol. Calculate the standard enthalpy of formation of AgNO2(s). 6.66 Hydrazine, N2H4, decomposes according to the following reaction: 3N2H4(l) 88n 4NH3(g) N2(g) (a) Given that the standard enthalpy of formation of hydrazine is 50.42 kJ/mol, calculate H° for its decomposition. (b) Both hydrazine and ammonia burn in oxygen to produce H2O(l) and N2(g). Write balanced equations for each of these processes and calculate H° for each of them. On a mass basis (per kg), would hydrazine or ammonia be the better fuel? 6.67 Consider the reaction N2(g) 3H2(g) 88n 2NH3(g) Ho rxn 92.6 kJ If 2.0 moles of N2 react with 6.0 moles of H2 to form NH3, calculate the work done (in joules) against a pressure of 1.0 atm at 25°C. What is E for this reaction? Assume the reaction goes to completion. 6.68 Consider the reaction H2(g) Cl2(g) 88n 2HCl(g) Ho rxn 184.6 kJ If 3 moles of H2 react with 3 moles of Cl2 to form HCl, calculate the work done (in joules) against a pressure of 1.0 atm at 25°C. What is E for this reaction? Assume the reaction goes to completion. 6.69 Photosynthesis produces glucose, C6H12O6, and oxygen from carbon dioxide and water: 6CO2 6H2O 88n C6H12O6 6O2 (a) How would you determine experimentally the Ho value for this reaction? (b) Solar radiation prorxn duces about 7.0 1014 kg glucose a year on Earth. What is the corresponding H° change? 6.70 A 2.10-mole sample of crystalline acetic acid, initially at 17.0°C, is allowed to melt at 17.0°C and is then heated to 118.1°C (its normal boiling point) at 1.00 atm. The sample is allowed to vaporize at 118.1°C and is then rapidly quenched to 17.0°C, so that it recrystallizes. Calculate H° for the total process as described. 6.71 Calculate the work done in joules by the reaction 2Na(s) 2H2O(l) 88n 2NaOH(aq) H2(g) when 0.34 g of Na reacts with water to form hydrogen gas at 0°C and 1.0 atm. Back Forward Main Menu H° 436.4 kJ Br2(g) 88n 2Br(g) H° 192.5 kJ Br2(g) 88n 2HBr(g) H° 104 kJ O2(g) TOC H2(g) Calculate H° for the reaction H(g) Br(g) 88n HBr(g) 6.73 Methanol (CH3OH) is an organic solvent and is also used as a fuel in some automobile engines. From the following data, calculate the standard enthalpy of formation of methanol: 2CH3OH(l) 3O2(g) 88n 2CO2(g) 4H2O(l) Ho 1452.8 kJ rxn 6.74 A 44.0-g sample of an unknown metal at 99.0°C was placed in a constant-pressure calorimeter containing 80.0 g of water at 24.0°C. The final temperature of the system was found to be 28.4°C. Calculate the specific heat of the metal. (The heat capacity of the calorimeter is 12.4 J/°C.) 6.75 A 1.00-mole sample of ammonia at 14.0 atm and 25°C in a cylinder fitted with a movable piston expands against a constant external pressure of 1.00 atm. At equilibrium, the pressure and volume of the gas are 1.00 atm and 23.5 L, respectively. (a) Calculate the final temperature of the sample. (b) Calculate q, w, and E for the process. The specific heat of ammonia is 0.0258 J/g °C. 6.76 Producer gas (carbon monoxide) is prepared by passing air over red-hot coke: C(s) 1 2 O2(g) 88n CO(g) Water gas (mixture of carbon monoxide and hydrogen) is prepared by passing steam over red-hot coke: C(s) H2O(g) 88n CO(g) H2(g) For many years, both producer gas and water gas were used as fuels in industry and for domestic cooking. The large-scale preparation of these gases was carried out alternately, that is, first producer gas, then water gas, and so on. Using thermochemical reasoning, explain why this procedure was chosen. 6.77 Compare the heat produced by the complete combustion of 1 mole of methane (CH4) with a mole of water gas (0.50 mole H2 and 0.50 mole CO) under the same conditions. On the basis of your answer, would you prefer methane over water gas as a fuel? Can you suggest two other reasons why methane is preferable to water gas as a fuel? 6.78 The so-called hydrogen economy is based on hydrogen produced from water using solar energy. The gas Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS is then burned as a fuel: 2H2(g) 6.79 6.80 6.81 6.82 6.83 O2(g) 88n 2H2O(l) A primary advantage of hydrogen as a fuel is that it is nonpolluting. A major disadvantage is that it is a gas and therefore is harder to store than liquids or solids. Calculate the volume of hydrogen gas at 25°C and 1.00 atm required to produce an amount of energy equivalent to that produced by the combustion of a gallon of octane (C8H18). The density of octane is 2.66 kg/gal, and its standard enthalpy of formation is 249.9 kJ/mol. Ethanol (C2H5OH) and gasoline (assumed to be all octane, C8H18) are both used as automobile fuel. If gasoline is selling for $1.20/gal, what would the price of ethanol have to be in order to provide the same amount of heat per dollar? The density and Ho of f octane are 0.7025 g/mL and 249.9 kJ/mol and of ethanol are 0.7894 g/mL and 277.0 kJ/mol, respectively. 1 gal 3.785 L. What volume of ethane (C2H6), measured at 23.0°C and 752 mmHg, would be required to heat 855 g of water from 25.0°C to 98.0°C? If energy is conserved, how can there be an energy crisis? The heat of vaporization of a liquid ( Hvap) is the energy required to vaporize 1.00 g of the liquid at its boiling point. In one experiment, 60.0 g of liquid nitrogen (boiling point 196°C) are poured into a Styrofoam cup containing 2.00 102 g of water at 55.3°C. Calculate the molar heat of vaporization of liquid nitrogen if the final temperature of the water is 41.0°C. Explain the cooling effect experienced when ethanol is rubbed on your skin, given that C2H5OH(l) 88n C2H5OH(g) H° Back Forward Main Menu TOC until its pressure is equal to the external pressure, what would its final volume be before it stopped expanding, and what would be the work done? 6.87 Calculate the standard enthalpy of formation for diamond, given that C(graphite) O2(g) 88n CO2(g) H° 393.5 kJ C(diamond) O2(g) 88n CO2(g) H° 395.4 kJ 6.88 (a) For most efficient use, refrigerator freezer compartments should be fully packed with food. What is the thermochemical basis for this recommendation? (b) Starting at the same temperature, tea and coffee remain hot longer in a thermal flask than chicken noodle soup. Explain. 6.89 Calculate the standard enthalpy change for the fermentation process. (See Problem 3.68.) 6.90 Portable hot packs are available for skiers and people engaged in other outdoor activities in a cold climate. The air-permeable paper packet contains a mixture of powdered iron, sodium chloride, and other components, all moistened by a little water. The exothermic reaction that produces the heat is a very common one — the rusting of iron: 4Fe(s) 6.91 42.2 kJ 6.84 For which of the following reactions does Ho rxn Ho? f (a) H2(g) S(rhombic) 88n H2S(g) (b) C(diamond) O2(g) 88n CO2(g) (c) H2(g) CuO(s) 88n H2O(l) Cu(s) (d) O(g) O2(g) 88n O3(g) 6.85 Calculate the work done (in joules) when 1.0 mole of water is frozen at 0°C and 1.0 atm. The volumes of one mole of water and ice at 0°C are 0.0180 L and 0.0196 L, respectively. 6.86 A quantity of 0.020 mole of a gas initially at 0.050 L and 20°C undergoes a constant-temperature expansion until its volume is 0.50 L. Calculate the work done (in joules) by the gas if it expands (a) against a vacuum and (b) against a constant pressure of 0.20 atm. (c) If the gas in (b) is allowed to expand unchecked 237 6.92 6.93 6.94 3O2(g) 88n 2Fe2O3(s) When the outside plastic envelope is removed, O2 molecules penetrate the paper, causing the reaction to begin. A typical packet contains 250 g of iron to warm your hands or feet for up to 4 hours. How much heat (in kJ) is produced by this reaction? (Hint: See Appendix 3 for Ho values.) f A person ate 0.50 pound of cheese (an energy intake of 4000 kJ). Suppose that none of the energy was stored in his body. What mass (in grams) of water would he need to perspire in order to maintain his original temperature? (It takes 44.0 kJ to vaporize 1 mole of water.) The total volume of the Pacific Ocean is estimated to be 7.2 108 km3. A medium-sized atomic bomb produces 1.0 1015 J of energy upon explosion. Calculate the number of atomic bombs needed to release enough energy to raise the temperature of the water in the Pacific Ocean by 1°C. A 19.2-g quantity of dry ice (solid carbon dioxide) is allowed to sublime (evaporate) in an apparatus like the one shown in Figure 6.10. Calculate the expansion work done against a constant external pressure of 0.995 atm and at a constant temperature of 22°C. Assume that the initial volume of dry ice is negligible and that CO2 behaves like an ideal gas. The enthalpy of combustion of benzoic acid (C6H5COOH) is commonly used as the standard for calibrating constant-volume bomb calorimeters; its Study Guide TOC Textbook Website MHHE Website 238 THERMOCHEMISTRY value has been accurately determined to be 3226.7 kJ/mol. When 1.9862 g of benzoic acid are burned, the temperature rises from 21.84°C to 25.67°C. What is the heat capacity of the calorimeter? (Assume that the quantity of water surrounding the calorimeter is exactly 2000 g.) 6.95 Lime is a term that includes calcium oxide (CaO, also called quicklime) and calcium hydroxide [Ca(OH)2, also called slaked lime]. It is used in the steel industry to remove acidic impurities, in air-pollution control to remove acidic oxides such as SO2, and in water treatment. Quicklime is made industrially by heating limestone (CaCO3) above 2000°C: CaCO3(s) 88n CaO(s) CO2(g) H° 177.8 kJ Slaked lime is produced by treating quicklime with water: CaO(s) H2O(l) 88n Ca(OH)2(s) H° 65.2 kJ The exothermic reaction of quicklime with water and the rather small specific heats of both quicklime (0.946 J/g °C) and slaked lime (1.20 J g °C) make it hazardous to store and transport lime in vessels made of wood. Wooden sailing ships carrying lime would occasionally catch fire when water leaked into the hold. (a) If a 500-g sample of water reacts with an equimolar amount of CaO (both at an initial temperature of 25°C), what is the final temperature of the product, Ca(OH)2? Assume that the product absorbs all of the heat released in the reaction. (b) Given that the standard enthalpies of formation of CaO and H2O are 635.6 kJ/mol and 285.8 kJ/mol, respectively, calculate the standard enthalpy of formation of Ca(OH)2. 6.96 Calcium oxide (CaO) is used to remove sulfur dioxide generated by coal-burning power stations: 2CaO(s) 2SO2(g) O2(g) 88n 2CaSO4(s) Calculate the enthalpy change for this process if 6.6 105 g of SO2 are removed by this process every day. 6.97 Glauber ’s salt, sodium sulfate decahydrate (Na2SO4 10H2O), undergoes a phase transition (that is, melting or freezing) at a convenient temperature of about 32°C: Na2SO4 10H2O(s) 88n Na2SO4 10H2O(l) H° Forward Main Menu TOC Zn(s) 2Ag (aq) 88n Zn2 (aq) 2Ag(s) the temperature rises from 19.25°C to 22.17°C. If the heat capacity of the calorimeter is 98.6 J/°C, calculate the enthalpy change for the above reaction on a molar basis. Assume that the density and specific heat of the solution are the same as those for water, and ignore the specific heats of the metals. 6.100 (a) A person drinks four glasses of cold water (3.0°C) every day. The volume of each glass is 2.5 102 mL. How much heat (in kJ) does the body have to supply to raise the temperature of the water to 37°C, the body temperature? (b) How much heat would your body lose if you were to ingest 8.0 102 g of snow at 0°C to quench thirst? (The amount of heat necessary to melt snow is 6.01 kJ/mol.) 6.101 A driver ’s manual states that the stopping distance quadruples as the speed doubles; that is, if it takes 30 ft to stop a car moving at 25 mph then it would take 120 ft to stop a car moving at 50 mph. Justify this statement by using mechanics and the first law of thermodynamics. [Assume that when a car is stopped, its kinetic energy ( 1 mu2) is totally converted 2 to heat.] 6.102 At 25°C the standard enthalpy of formation of HF(aq) is 320.1 kJ/mol; of OH (aq), it is 229.6 kJ/mol; of F (aq), it is 329.1 kJ/mol; and of H2O(l), it is 285.8 kJ/mol. (a) Calculate the standard enthalpy of neutralization of HF(aq): HF(aq) 74.4 kJ As a result, this compound is used to regulate the temperature in homes. It is placed in plastic bags in the ceiling of a room. During the day, the endothermic melting process absorbs heat from the surroundings, cooling the room. At night, it gives off heat as it Back freezes. Calculate the mass of Glauber ’s salt in kilograms needed to lower the temperature of air in a room by 8.2°C at 1.0 atm. The dimensions of the room are 2.80 m 10.6 m 17.2 m, the specific heat of air is 1.2 J/g °C, and the molar mass of air may be taken as 29.0 g/mol. 6.98 A balloon 16 m in diameter is inflated with helium at 18°C. (a) Calculate the mass of He in the balloon, assuming ideal behavior. (b) Calculate the work done (in joules) during the inflation process if the atmospheric pressure is 98.7 kPa. 6.99 An excess of zinc metal is added to 50.0 mL of a 0.100 M AgNO3 solution in a constant-pressure calorimeter like the one pictured in Figure 6.7. As a result of the reaction OH (aq) 88n F (aq) H2O(l) (b) Using the value of 56.2 kJ as the standard enthalpy change for the reaction H (aq) OH (aq) 88n H2O(l) calculate the standard enthalpy change for the reaction Study Guide TOC Textbook Website MHHE Website QUESTIONS AND PROBLEMS HF(aq) 88n H (aq) F (aq) 6.103 Why are cold, damp air and hot, humid air more uncomfortable than dry air at the same temperatures? (The specific heats of water vapor and air are approximately 1.9 J/g °C and 1.0 J/g °C, respectively.) 6.104 From the enthalpy of formation for CO2 and the following information, calculate the standard enthalpy of formation for carbon monoxide (CO). CO(g) 1 2 O2(g) 88n CO2(g) H° 283.0 kJ 6.109 Acetylene (C2H2) and benzene (C6H6) have the same empirical formula. In fact, benzene can be made from acetylene as follows: 3C2H2(g) 88n C6H6(l) 6.110 Why can’t we obtain it directly by measuring the enthalpy of the following reaction? C(graphite) 1 2 O2(g) 88n CO(g) 6.105 A 46-kg person drinks 500 g of milk, which has a “caloric” value of approximately 3.0 kJ/g. If only 17% of the energy in milk is converted to mechanical work, how high (in meters) can the person climb based on this energy intake? [Hint: The work done in ascending is given by mgh, where m is the mass (in kilograms), g the gravitational acceleration (9.8 m/s2), and h the height (in meters).] 6.106 The height of Niagara Falls on the American side is 51 meters. (a) Calculate the potential energy of 1.0 g of water at the top of the falls relative to the ground level. (b) What is the speed of the falling water if all of the potential energy is converted to kinetic energy? (c) What would be the increase in temperature of the water if all the kinetic energy were converted to heat? (See previous problem for suggestions.) 6.107 In the nineteenth century two scientists named Dulong and Petit noticed that for a solid element, the product of its molar mass and its specific heat is approximately 25 J/°C. This observation, now called Dulong and Petit’s law, was used to estimate the specific heat of metals. Verify the law for the metals listed in Table 6.1. The law does not apply to one of the metals. Which one is it? Why? 6.108 Determine the standard enthalpy of formation of ethanol (C2H5OH) from its standard enthalpy of combustion ( 1367.4 kJ/mol). Back Forward Main Menu TOC 239 6.111 6.112 6.113 6.114 The enthalpies of combustion for C2H2 and C6H6 are 1299.4 kJ/mol and 3267.4 kJ/mol, respectively. Calculate the standard enthalpies of formation of C2H2 and C6H6 and hence the enthalpy change for the formation of C6H6 from C2H2. Ice at 0°C is placed in a Styrofoam cup containing 361 g of a soft drink at 23°C. The specific heat of the drink is about the same as that of water. Some ice remains after the ice and soft drink reach an equilibrium temperature of 0°C. Determine the mass of ice that has melted. Ignore the heat capacity of the cup. (Hint: It takes 334 J to melt 1 g of ice at 0°C.) A gas company in Massachusetts charges $1.30 for 15 ft3 of natural gas (CH4) measured at 20°C and 1.0 atm. Calculate the cost of heating 200 mL of water (enough to make a cup of coffee or tea) from 20°C to 100°C. Assume that only 50 percent of the heat generated by the combustion is used to heat the water: the rest of the heat is lost to the surroundings. Calculate the internal energy of a Goodyear blimp filled with helium gas at 1.2 105 Pa. The volume of the blimp is 5.5 103 m3. If all the energy were used to heat 10.0 tons of copper at 21°C, calculate the final temperature of the metal. (Hint: See Section 5.7 for help in calculating the internal energy of a gas. 1 ton 9.072 105 g.) Decomposition reactions are usually endothermic, whereas combination reactions are usually exothermic. Give a qualitative explanation for these trends. Acetylene (C2H2) can be made by reacting calcium carbide (CaC2) with water. (a) Write an equation for the reaction. (b) What is the maximum amount of heat (in joules) that can be obtained from the combustion of acetylene, starting with 74.6 g of CaC2? 6.1 6.47 103 kJ. 34.3 kJ. 6.3 727 kJ/mol. 6.4 22.49°C. 6.5 86 kJ/mol. 41.83 kJ/g. 6.7 (a) 0, (b) 286 J. 6.8 111.7 kJ. Answers to Practice Exercises: 6.2 6.6 Study Guide TOC Textbook Website MHHE Website C HEMICAL M YSTERY The Exploding Tire† I t was supposed to be a routine job: Fix the flat tire on Harvey Smith’s car. The owner of Tom’s Garage, Tom Lee, gave the tire to Jerry to work on, while he went outside to pump gas. A few minutes later, Tom heard a loud bang. He rushed inside to find the tire blown to pieces, a wall collapsed, equipment damaged, and Jerry lying on the floor, unconscious and bleeding. Luckily Jerry’s injury was not serious. As he lay in the hospital recovering, the mystery of the exploding tire unfolded. The tire had gone flat when Harvey drove over a nail. Being a cautious driver, Harvy carried a can of instant tire repair in the car, so he was able to reinflate the tire and drive safely home. The can of tire repair Harvey used contained latex (natural rubber) dissolved in a liquid propellant, which is a mixture of propane (C3H8) and butane (C4H10). Propane and butane are gases under atmospheric conditions but exist as liquids under compression in the can. When the valve on the top of the can is pressed, it opens, releasing the pressure inside. The mixture boils, forming a latex foam which is propelled by the gases into the tire to seal the puncture while the gas reinflates the tire. The pressure in a flat tire is approximately one atmosphere, or roughly 15 pounds per square inch (psi). Using the aerosol tire repair, Harvey reinflated his damaged tire to a pressure of 35 psi. This is called the gauge pressure, which is the pressure of the tire above the atmospheric pressure. Thus the total pressure in the tire was actually (15 35) psi, or 50 psi. One problem with using natural gases like propane and butane as propellants is that they are highly flammable. In fact, these gases can react explosively when mixed with air at a concentration of 2 percent to 9 percent by volume. Jerry was aware of the hazards of repairing Harvey’s tire and took precautions to avoid an accident. First he let out the excess gas in the tire. Next he reinflated the tire to 35 psi with air. And he repeated the procedure once. Clearly, this is a dilution process intended to gradually decrease the concentrations of propane and butane. The fact that the tire exploded means that Jerry had not diluted the gases enough. But what was the source of ignition? When Jerry found the nail hole in the tire, he used a tire reamer, a metal file-like instrument, to clean dirt and loose rubber from the hole before applying a rubber plug and liquid sealant. The last thing Jerry remembered was pulling the reamer out of the hole. The next thing he knew he was lying in the hospital, hurting all over. To solve this mystery, make use of the following clues. CHEMICAL CLUES 1. Write balanced equations for the combustion of propane and butane. The products are carbon dioxide and water. †Adapted with permission from “The Exploding Tire,” by Jay A. Young, CHEM MATTERS, April, 1988, p. 12. Copyright 1995 American Chemical Society. 240 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website When Harvey inflated his flat tire to 35 psi, the composition by volume of the propane and butane gases is given by (35 psi/50 psi) 100%, or 70%. When Jerry deflated the tire the first time, the pressure fell to 15 psi but the composition remained at 70%. Based on these facts, calculate the percent composition of propane and butane at the end of two deflation-inflation steps. Does it fall within the explosive range? 3. Given that Harvey’s flat tire is a steel-belted tire, explain how the ignition of the gas mixture might have been triggered. (A steel-belted tire has two belts of steel wire for outer reinforcement and two belts of polyester cord for inner reinforcement.) 2. Instant flat tire repair. Nowadays such products are made of nonflammable substances. 241 Back Forward Main Menu TOC Study Guide TOC Textbook Website MHHE Website ...
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