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Chapter 5 Thermochemistry © 2012 Pearson Education, Inc. Why is a pitcher able to throw a baseball faster than he could throw a bowling ball? A. It is harder to hold a baseball ball than a bowling ball. B. For the same input of energy, an object of smaller mass will have the larger speed. C. For the same input of energy, an object of larger mass will have the larger speed. D. The spinning of the bowling ball causes it to drop to the ground faster. Thermochemistry Energy • Energy is the ability to do work or transfer heat. – Energy used to cause an object that has mass to move is called work. – Energy used to cause the temperature of an object to rise is called heat. Thermochemistry © 2012 Pearson Education, Inc. Definitions: Work • Energy used to move an object over some distance is work: • w=Fd where w is work, F is the force, and d is the distance over which the force is exerted. Thermochemistry © 2012 Pearson Education, Inc. Heat • Energy can also be transferred as heat. • Heat flows from warmer objects to cooler objects. Thermochemistry © 2012 Pearson Education, Inc. Kinetic Energy Kinetic energy is energy an object possesses by virtue of its motion: 1 Ek = mv2 2 Thermochemistry © 2012 Pearson Education, Inc. Conversion of Energy • Energy can be converted from one type to another. • For example, the cyclist in the figure has potential energy as she sits on top of the hill. Thermochemistry © 2012 Pearson Education, Inc. Conversion of Energy • As she coasts down the hill, her potential energy is converted to kinetic energy. • At the bottom, all the potential energy she had at the top of the hill is now kinetic energy. Thermochemistry © 2012 Pearson Education, Inc. A. Yes B. No Thermochemistry A. B. C. D. Increases as potential energy is converted to kinetic energy. Increases as the kinetic energy is converted to potential energy. Decreases as potential energy is converted to kinetic energy. Decreases as the kinetic energy is converted to potential energy. Thermochemistry Potential Energy • Potential energy is energy an object possesses by virtue of its position or chemical composition. • The most important form of potential energy in molecules is electrostatic potential energy, Eel: K Q 1Q 2 Eel = d Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Becomes more positive and decreasing in absolute magnitude Becomes more negative and increasing in absolute magnitude. Becomes more positive and increasing in absolute magnitude Becomes more negative and decreasing in absolute magnitude. Thermochemistry Units of Energy • The SI unit of energy is the joule (J): kg m2 1 J = 1 s2 • An older, non-SI unit is still in widespread use: the calorie (cal): 1 cal = 4.184 J Thermochemistry © 2012 Pearson Education, Inc. Definitions: System and Surroundings • The system includes the molecules we want to study (here, the hydrogen and oxygen molecules). • The surroundings are everything else (here, the cylinder and piston). Thermochemistry © 2012 Pearson Education, Inc. A. Yes, because matter is not transferred to the surroundings. B. No, because matter is transferred to the surroundings. Thermochemistry A. Closed system B. Isolated system C. Open system Thermochemistry Sample Exercise 5.1 Describing and Calculating Energy Changes A bowler lifts a 5.4-kg (12-lb) bowling ball from ground level to a height of 1.6 m (5.2 ft) and then drops it. (a) What happens to the potential energy of the ball as it is raised? (b) What quantity of work, in J, is used to raise the ball? (c) After the ball is dropped, it gains kinetic energy. If all the work done in part (b) has been converted to kinetic energy by the time the ball strikes the ground, what is the ball’s speed just before it hits the ground? (Note The force due to gravity is F = m g, where m is the mass of the object and g is the gravitational constant; g = 9.8 m/s2.) Solution Practice Exercise What is the kinetic energy, in J, of (a) an Ar atom moving at a speed of 650 m/s, (b) a mole of Ar atoms moving at 650 m/s? (Hint: 1 amu = 1.66 10–27 kg.) Thermochemistry First Law of Thermodynamics • Energy is neither created nor destroyed. • In other words, the total energy of the universe is a constant; if the system loses energy, it must be gained by the surroundings, and vice versa. Thermochemistry © 2012 Pearson Education, Inc. Internal Energy The internal energy of a system is the sum of all kinetic and potential energies of all components of the system; we call it E. Thermochemistry © 2012 Pearson Education, Inc. Internal Energy By definition, the change in internal energy, E, is the final energy of the system minus the initial energy of the system: E = Efinal − Einitial Thermochemistry © 2012 Pearson Education, Inc. A. Positive number B. Zero C. Negative number Thermochemistry Changes in Internal Energy • If E > 0, Efinal > Einitial – Therefore, the system absorbed energy from the surroundings. – This energy change is called endergonic. Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy • If E < 0, Efinal < Einitial – Therefore, the system released energy to the surroundings. – This energy change is called exergonic. Thermochemistry © 2012 Pearson Education, Inc. Changes in Internal Energy • When energy is exchanged between the system and the surroundings, it is exchanged as either heat (q) or work (w). • That is, E = q + w. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. No, because we are not given specific values for w and q. No, because it an open system. Yes, because we know the signs of w and q. No, because we do not know the temperature of the system. Thermochemistry E, q, w, and Their Signs Thermochemistry © 2012 Pearson Education, Inc. Sample Exercise 5.2 Relating Heat and Work to Changes of Internal Energy Gases A(g) and B(g) are confined in a cylinder-and-piston arrangement like that in Figure 5.4 and react to form a solid product C(s): A(g) + B(g) C(s). As the reaction occurs, the system loses 1150 J of heat to the surroundings. The piston moves downward as the gases react to form a solid. As the volume of the gas decreases under the constant pressure of the atmosphere, the surroundings do 480 J of work on the system. What is the change in the internal energy of the system? Solution Practice Exercise Calculate the change in the internal energy for a process in which a system absorbs 140 J of heat from the surroundings and does 85 J of work on the surroundings. Thermochemistry Exchange of Heat between System and Surroundings • When heat is absorbed by the system from the surroundings, the process is endothermic. Thermochemistry © 2012 Pearson Education, Inc. Exchange of Heat between System and Surroundings • When heat is absorbed by the system from the surroundings, the process is endothermic. • When heat is released by the system into the surroundings, the process is exothermic. Thermochemistry © 2012 Pearson Education, Inc. A. Endothermic, because heat is released to the surroundings. B. Exothermic, because heat is released to the surroundings. C. Endothermic, because heat is absorbed from the surroundings. D. Exothermic, because heat is absorbed from the surroundings. Thermochemistry State Functions Usually we have no way of knowing the internal energy of a system; finding that value is simply too complex a problem. Thermochemistry © 2012 Pearson Education, Inc. State Functions • However, we do know that the internal energy of a system is independent of the path by which the system achieved that state. – In the system depicted in the figure, the water could have reached room temperature from either direction. Thermochemistry © 2012 Pearson Education, Inc. State Functions • Therefore, internal energy is a state function. • It depends only on the present state of the system, not on the path by which the system arrived at that state. • And so, E depends only on Einitial and Efinal. Thermochemistry © 2012 Pearson Education, Inc. A. It is determined by many factors and involves data such as amount of check, date and payee. B. It is a numerical value calculated from known amounts of checks. C. It is calculated from multiple transactions. D It only depends on the net total of all transactions, and not on the ways money is transferred into or out of the account. Thermochemistry State Functions • However, q and w are not state functions. • Whether the battery is shorted out or is discharged by running the fan, its E is the same. – But q and w are different in the two cases. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. w>0 w=0 w<0 Cannot be determined Thermochemistry Enthalpy • If a process takes place at constant pressure (as the majority of processes we study do) and the only work done is this pressure–volume work, we can account for heat flow during the process by measuring the enthalpy of the system. • Enthalpy is the internal energy plus the product of pressure and volume: H = E + PV © 2012 Pearson Education, Inc. Thermochemistry A. H is italicized in the equation. B. E is a state function in the equation and this is the minimum requirement. C. P and V are state functions in the equation and this is the minimum requirement. D. All other terms in the equation, E, P and V, are state functions. Thermochemistry Enthalpy • When the system changes at constant pressure, the change in enthalpy, H, is H = (E + PV) • This can be written H = E + PV Thermochemistry © 2012 Pearson Education, Inc. Enthalpy • Since E = q + w and w = −PV, we can substitute these into the enthalpy expression: H = E + PV H = (q + w) − w H = q • So, at constant pressure, the change in enthalpy is the heat gained or lost. Thermochemistry © 2012 Pearson Education, Inc. Work Usually in an open container the only work done is by a gas pushing on the surroundings (or by the surroundings pushing on the gas). Thermochemistry © 2012 Pearson Education, Inc. Work We can measure the work done by the gas if the reaction is done in a vessel that has been fitted with a piston: w = −PV Thermochemistry © 2012 Pearson Education, Inc. A. If Zn(s) is the limiting reactant then more gas is formed and more work will be done. B. If Zn(s) is the limiting reactant then less gas is produced and less work will be done. C. If HCl(aq) is the limiting reactant then addition of Zn(s) increases the amount of gas formed and more work is done. D. If HCl(aq) is the limiting reactant then addition of Zn(s) decreases the amount of gas formed and less work is done. Thermochemistry A. Yes, because pressure may change the value of work in w = –PΔV. B. No, because ΔV = 0. Thermochemistry Endothermicity and Exothermicity • A process is endothermic when H is positive. • A process is exothermic when H is negative. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Ruler Graduated cylinder Barometer Thermometer Thermochemistry Sample Exercise 5.3 Determining the Sign of H Indicate the sign of the enthalpy change, H, in these processes carried out under atmospheric pressure and indicate whether each process is endothermic or exothermic: (a) An ice cube melts; (b) 1 g of butane (C4H10) is combusted in sufficient oxygen to give complete combustion to CO2 and H2O. Solution Practice Exercise Molten gold poured into a mold solidifies at atmospheric pressure. With the gold defined as the system, is the solidification an exothermic or endothermic process? Thermochemistry Enthalpy of Reaction The change in enthalpy, H, is the enthalpy of the products minus the enthalpy of the reactants: H = Hproducts − Hreactants Thermochemistry © 2012 Pearson Education, Inc. Enthalpy of Reaction This quantity, H, is called the enthalpy of reaction, or the heat of reaction. Thermochemistry © 2012 Pearson Education, Inc. A. Yes, because the reactants and products are the same. B. No, because only half as much matter is involved. C. Yes, because mass does not affect enthalpy change. D. No, because enthalpy is a state function. Thermochemistry The Truth about Enthalpy 1. Enthalpy is an extensive property. 2. H for a reaction in the forward direction is equal in size, but opposite in sign, to H for the reverse reaction. 3. H for a reaction depends on the state of the products and the state of the reactants. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Exothermic, because the temperature of the system increases Exothermic, because the temperature of the system decreases Endothermic, because the temperature of the system increases Endothermic, because the temperature of the system decreases Thermochemistry Sample Exercise 5.4 Relating H to Quantities of Reactants and Products How much heat is released when 4.50 g of methane gas is burned in a constant-pressure system? (Use the information given in Equation 5.18.) Solution Practice Exercise Hydrogen peroxide can decompose to water and oxygen by the reaction 2 H2O2(l) 2 H2O(l) + O2(g) H = –196 kJ Calculate the quantity of heat released when 5.00 g of H2O2(l) decomposes at constant pressure. Thermochemistry Heat Capacity and Specific Heat We define specific heat capacity (or simply specific heat) as the amount of energy required to raise the temperature of 1 g of a substance by 1 K (or 1 C). Thermochemistry © 2012 Pearson Education, Inc. Heat Capacity and Specific Heat Specific heat, then, is Specific heat = heat transferred mass temperature change s= q m T Thermochemistry © 2012 Pearson Education, Inc. A. Hg(l) C. Al(s) B. Fe(s) D. H2O(l) Thermochemistry Heat Capacity and Specific Heat The amount of energy required to raise the temperature of an object by 1 K (1C) is its heat capacity. heat capacity, then, is heat capacity = Cp = heat transferred temperature change or, heat capacity = mass x specific heat Thermochemistry © 2012 Pearson Education, Inc. Sample Exercise 5.5 Relating Heat, Temperature Change, and Heat Capacity (a) How much heat is needed to warm 250 g of water (about 1 cup) from 22 C (about room temperature) to 98 C (near its boiling point)? (b) What is the molar heat capacity of water? Solution Practice Exercise (a) Large beds of rocks are used in some solar-heated homes to store heat. Assume that the specific heat of the rocks is 0.82 J/g–K. Calculate the quantity of heat absorbed by 50.0 kg of rocks if their temperature increases by 12.0 C. (b) What temperature change would these rocks undergo if they emitted 450 kJ of heat? Thermochemistry Calorimetry Since we cannot know the exact enthalpy of the reactants and products, we measure H through calorimetry, the measurement of heat flow. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Styrofoam cups are very fragile. To create a vacuum between the two cups Gives greater support to the equipment setup Provides more thermal insulation Thermochemistry Constant Pressure Calorimetry By carrying out a reaction in aqueous solution in a simple calorimeter such as this one, one can indirectly measure the heat change for the system by measuring the heat change for the water in the calorimeter. Thermochemistry © 2012 Pearson Education, Inc. Constant Pressure Calorimetry Because the specific heat for water is well known (4.184 J/g-K), we can measure H for the reaction with this equation: q = m s T Thermochemistry © 2012 Pearson Education, Inc. Sample Exercise 5.6 Measuring H Using a Coffee-Cup Calorimeter When a student mixes 50 mL of 1.0 M HCl and 50 mL of 1.0 M NaOH in a coffee-cup calorimeter, the temperature of the resultant solution increases from 21.0 C to 27.5 C. Calculate the enthalpy change for the reaction in kj/mol HCl, assuming that the calorimeter loses only a negligible quantity of heat, that the total volume of the solution is 100 mL, that its density is 1.0 g/mL, and that its specific heat is 4.18 J/g–K. Solution Practice Exercise When 50.0 mL of 0.100 MgNO3 and 50.0 mL of 0.100 M HCl are mixed in a constant-pressure calorimeter, the temperature of the mixture increases from 22.30 C to 23.11 C. The temperature increase is caused by the following reaction: AgNO3(aq) + HCl(aq) AgCl(s) + HNO3(aq) Calculate H for this reaction in AgNO3, assuming that the combined solution has a mass of 100.0 g and a specific heat of 4.18 J/g C. Thermochemistry Bomb Calorimetry • Reactions can be carried out in a sealed “bomb” such as this one. • The heat absorbed (or released) by the water is a very good approximation of the enthalpy change for the reaction. Thermochemistry © 2012 Pearson Education, Inc. Bomb Calorimetry • Because the volume in the bomb calorimeter is constant, what is measured is really the change in internal energy, E, not H. • For most reactions, the difference is very small. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Ensures a heterogeneous mixture Keeps particles slowly moving Ensures solution is at same temperature Permits titration of the solution Thermochemistry Sample Exercise 5.7 Measuring qrxn Using a Bomb Calorimeter The combustion of methylhydrazine (CH6N2), a liquid rocket fuel, produces N2(g), CO2(g), and H2O(l): 2 CH6N2(l) + 5 O2(g) 2 N2(g) + 2 CO2(g) + 6 H2O(l) When 4.00 g of methylhydrazine is combusted in a bomb calorimeter, the temperature of the calorimeter increases from 25.00 C to 39.50 C. In a separate experiment the heat capacity of the calorimeter is measured to be 7.794 kJ/ C. Calculate the heat of reaction for the combustion of a mole of CH6N2. Solution Practice Exercise A 0.5865-g sample of lactic acid (HC3H5O3) is burned in a calorimeter whose heat capacity is 4.812 kJ/C. The temperature increases from 23.10 C to 24.95 C. Calculate the heat of combustion of lactic acid (a) per gram and (b) per mole. Thermochemistry Hess’s Law H is well known for many reactions, and it is inconvenient to measure H for every reaction in which we are interested. • However, we can estimate H using published H values and the properties of enthalpy. Thermochemistry © 2012 Pearson Education, Inc. Hess’s Law Hess’s law states that “if a reaction is carried out in a series of steps, H for the overall reaction will be equal to the sum of the enthalpy changes for the individual steps.” Thermochemistry © 2012 Pearson Education, Inc. Hess’s Law Because H is a state function, the total enthalpy change depends only on the initial state of the reactants and the final state of the products. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. D. Condensation of 2 H2O(g) to 2 H2O(l) Vaporization of 2 H2O(l) to 2 H2O(g) Conversion of 2 H2O(l) to 2 O2(g) Conversion of CO2(g) to CH4(g) Thermochemistry A. B. C. D. No change. Sign of H changes. Value of H doubles. Value of H decreases by half. Thermochemistry Sample Exercise 5.8 Using Hess’s Law to Calculate H The enthalpy of reaction for the combustion of C to CO2 is –393.5 kJ/mol C, and the enthalpy for the combustion of CO to CO2 is –283.0 kJ/mol C: 1. 2. Using these data, calculate the enthalpy for the combustion of C to CO: 3. Solution Practice Exercise Carbon occurs in two forms, graphite and diamond. The enthalpy of the combustion of graphite is –393.5 kJ/mol, and that of diamond is –395.4 kJ/mol: C(graphite) + O2(g) CO2(g) H = –393.5 kJ C(diamond) + O2(g) CO2(g) H = –395.4 kJ Calculate H for the conversion of graphite to diamond: C(graphite) C(diamond) H = ? Answer: +1.9 kJ Thermochemistry Sample Exercise 5.9 Using Three Equations with Hess’s Law to Calculate H Calculate H for the reaction 2 C(s) + H2(g) C2H2(g) given the following chemical equations and their respective enthalpy changes: Solution Practice Exercise Calculate H for the reaction NO(g) + O(g) NO2(g) given the following information Thermochemistry A. B. C. D. All would change. None would change. ΔH3 does not change. ΔH1 does not change. Thermochemistry Enthalpies of Formation An enthalpy of formation, Hf, is defined as the enthalpy change for the reaction in which a compound is made from its constituent elements in their elemental forms. Thermochemistry © 2012 Pearson Education, Inc. Standard Enthalpies of Formation Standard enthalpies of formation, Hf°, are measured under standard conditions (25 °C and 1.00 atm pressure). Thermochemistry © 2012 Pearson Education, Inc. A. Yes, because it is just another elemental form of oxygen. B. No, because it is not the most stable form of the element oxygen at the given conditions. C. Yes, because changing the subscripts of an elemental formula does not change standard heat of formation. D. No, because there is a temperature change when ozone is formed. Thermochemistry Sample Exercise 5.10 Equations Associated with Enthalpies of Formation For which of these reactions at 25 C does the enthalpy change represent a standard enthalpy of formation? For each that does not, what changes are needed to make it an equation whose H is an enthalpy of formation? (a) (b) (c) Solution Practice Exercise Write the equation corresponding to the standard enthalpy of formation of liquid carbon tetrachloride (CCl4). Thermochemistry Calculation of H C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) • The sum of these equations is C3H8(g) 3C(graphite) + 4H2(g) 3C(graphite) + 3O2(g) 3CO2(g) 4H2(g) + 2O2(g) 4H2O(l) C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) Thermochemistry © 2012 Pearson Education, Inc. Calculation of H We can use Hess’s law in this way: H = nHf,products – mHf,reactants where n and m are the stoichiometric coefficients. Thermochemistry © 2012 Pearson Education, Inc. Calculation of H C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(l) H = [3(−393.5 kJ) + 4(−285.8 kJ)] – [1(−103.85 kJ) + 5(0 kJ)] = [(−1180.5 kJ) + (−1143.2 kJ)] – [(−103.85 kJ) + (0 kJ)] = (−2323.7 kJ) – (−103.85 kJ) = −2219.9 kJ Thermochemistry © 2012 Pearson Education, Inc. Sample Exercise 5.11 Calculating an Enthalpy of Reaction from Enthalpies of Formation (a) Calculate the standard enthalpy change for the combustion of 1 mol of benzene, C6H6(l), to CO2(g) and H2O(l). (b) Compare the quantity of heat produced by combustion of 1.00 g propane with that produced by 1.00 g benzene. Solution Practice Exercise Comment Both propane and benzene are hydrocarbons. As a rule, the energy obtained from the combustion of a gram of hydrocarbon is between 40 and 50 kJ. Thermochemistry Sample Exercise 5.12 Calculating an Enthalpy of Formation from Enthalpies of Reaction The standard enthalpy change for the reaction CaCO3(s) CaO(s) + CO2(g) is 178.1 kJ. Use Table 5.3 to calculate the standard enthalpy of formation of CaCO3(s). Solution Practice Exercise Given the following standard enthalpy change, use the standard enthalpies of formation in Table 5.3 to calculate the standard enthalpy of formation of CuO(s): CuO(s) + H2(g) Cu(s) + H2O(l) H = –129.7 kJ Thermochemistry Energy in Foods Most of the fuel in the food we eat comes from carbohydrates and fats. Thermochemistry © 2012 Pearson Education, Inc. A. B. C. Carbohydrates Proteins Fats Thermochemistry A. Grams of fat B. Grams of total carbohydrate C. Grams of protein Thermochemistry Sample Exercise 5.13 Comparing Fuel Values Celery contains carbohydrates in the form of starch and cellulose, which have essentially the same fuel values when combusted in a bomb calorimeter. When we eat celery, however, our bodies receive fuel value from the starch only. What can we conclude about the difference between starch and cellulose as foods? Solution Practice Exercise The nutrition label on a bottle of canola oil indicates that 10 g of the oil has a fuel value of 86 kcal. A similar label on a bottle of pancake syrup indicates that 60 mL (about 60 g) has a fuel value of 200 kcal. Account for the difference. Thermochemistry Sample Exercise 5.14 Estimating the Fuel Value of a Food from Its Composition (a) A 28-g (1-oz) serving of a popular breakfast cereal served with 120 mL of skim milk provides 8 g protein, 26 g carbohydrates, and 2 g fat. Using the average fuel values of these substances, estimate the fuel value (caloric content) of this serving. (b) A person of average weight uses about 100 Cal/mi when running or jogging. How many servings of this cereal provide the fuel value requirements to run 3 mi? Solution Practice Exercise Dry red beans contain 62% carbohydrate, 22% protein, and 1.5% fat. Estimate the fuel value of these beans. (b) During a very light activity, such as reading or watching television, the average adult expends about 7 kJ/min. How many minutes of such activity can be sustained by the energy provided by a serving of chicken noodle soup containing 13 g protein, 15 g carbohydrate, and 5 g fat? Thermochemistry Energy in Fuels The vast majority of the energy consumed in this country comes from fossil fuels. Thermochemistry © 2012 Pearson Education, Inc. A. It is easily stored as a gas for a long time. B. Its product of combustion is only H2O(g). C. Safety considerations using hydrogen gas are minimal. D. It is easily found in nature as an element. Thermochemistry