Chapter 6 Thermochemistry: Energy Flow and Chemical Change 6-1 Thermochemistry: Energy Flow and Chemical Change 6.1 Forms of Energy and Their Interconversion 6.2 Enthalpy: Chemical Change at Constant Pressure 6.3 Calorimetry: Measuring the Heat of a Chemical or Physical Change 6.4 Stoichiometry of Thermochemical Equations 6.5 Hesss Law: Finding H of Any Reaction 6.6 Standard Enthalpies of Reaction (Hrxn) 6-2 Transfer and Interconversion of Energy Thermodynamics is the study of energy and its transformations.

Thermochemistry is a branch of thermodynamics that deals with the heat involved in chemical and physical changes. When energy is transferred from one object to another, it appears as work and heat. 6-3 Figure 6.1 A chemical system and its surroundings. The system in this case is the contents of the reaction flask. The surroundings comprise everything else, including the flask itself. 6-4 The System and Its Surroundings A meaningful study of any transfer of energy requires that

we first clearly define both the system and its surroundings. System + Surroundings = Universe The internal energy, E, of a system is the sum of the potential and kinetic energies of all the particles present. The total energy of the universe remains constant. A change in the energy of the system must be accompanied by an equal and opposite change in the energy of the surroundings. 6-5 Figure 6.2 Energy diagrams for the transfer of internal energy (E) between a system and its surroundings. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. E = Efinal - Einitial = Eproducts - Ereactants

6-6 Figure 6.3 The two cases where energy is transferred as heat only. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. The system releases heat The system absorbs heat 6-7 Figure 6.4A The two cases where energy is transferred as work only. Zn(s) + 2H+(aq) + 2Cl-(aq)

H2(g) + Zn2+(aq) + 2Cl-(aq) The system does work on the surroundings. 6-8 Figure 6.4B The two cases where energy is transferred as work only. The system has work done on it by the surroundings. 6-9 Table 6.1 The Sign Conventions* for q, w, and E q

+ w = E + + + + - depends on sizes of q and w

- + depends on sizes of q and w - - - * For q: + means system gains heat; - means system releases heat. * For w: + means word done on system; - means work done by system. 6-10 The Law of Energy Conservation The first law of Thermodynamics states that the total energy of the universe is constant.

Energy is conserved, and is neither created nor destroyed. Energy is transferred in the form of heat and/or work. Euniverse = Esystem + Esurroundings = 0 6-11 Units of Energy The SI unit of energy is the joule (J). 1 J = 1 kgm2/s2 The calorie was once defined as the quantity of energy needed to raise the temperature of 1 g of water by 1C. 1 cal = 4.184 J The British Thermal Unit (Btu) is often used to rate appliances. 1 Btu is equivalent to 1055 J. 6-12 Figure 6.5 Some quantities of energy.

6-13 Sample Problem 6.1 Determining the Change in Internal Energy of a System PROBLEM: When gasoline burns in a car engine, the heat released causes the products CO2 and H2O to expand, which pushes the pistons outward. Excess heat is removed by the cars radiator. If the expanding gases do 451 J of work on the pistons and the system releases 325 J to the surroundings as heat, calculate the change in energy (E) in J, kJ, and kcal. PLAN: 6-14 Define the system and surroundings and assign signs to q

and w correctly. Then E = q + w. The answer can then be converted from J to kJ and to kcal. Sample Problem 6.1 SOLUTION: Heat is given out by a chemical reaction, so it makes sense to define the system as the reactants and products involved. The pistons, the radiator and the rest of the car then comprise the surroundings. Heat is given out by the system, so q = - 325 J The gases expand to push the pistons, so the system does work on the surroundings and w = - 451 J E = q + w = -776 J x 6-15 -325 J + (-451 J) = -776 J 1 kJ

= -0.776 kJ 3 10 J 1 kcal -0.776 kJ x = -0.185 kcal 4.184 kJ Enthalpy: Chemical Change at Constant Pressure E = q + w To determine E, both heat and work must be measured. The most common chemical work is PV work the work done when the volume of a system changes in the presence of an external pressure.

Enthalpy (H) is defined as E + PV so H = E + PV If a system remains at constant pressure and its volume does not change much, then H E 6-16 Figure 6.6 Two different paths for the energy change of a system. Even though q and w for the two paths are different, the total E is the same for both. 6-17 Figure 6.7

Pressure-volume work. An expanding gas pushing back the atmosphere does PV work (w = -PV). 6-18 H as a measure of E H is the change in heat for a system at constant pressure. qP = E + PV = H H E for reactions that do not involve gases for reactions in which the total amount (mol) of gas does not change for reactions in which qP is much larger than PV, even if the total mol of gas does change.

6-19 Figure 6.8 Enthalpy diagrams for exothermic and endothermic processes. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. CH4(g) + 2O2(g) CO2(g) + H2O(g) A 6-20 Exothermic process Heat is given out. H2O(s) H2O(l) B

Endothermic process Heat is taken in. Sample Problem 6.2 Drawing Enthalpy Diagrams and Determining the Sign of H PROBLEM: In each of the following cases, determine the sign of H, state whether the reaction is exothermic or endothermic, and draw and enthalpy diagram. (a) H2(g) + O2(g) H2O(l) + 285.8 kJ (b) 40.7 kJ + H2O(l) H2O(g) PLAN: From each equation, note whether heat is a reactant or a product. If heat is taken in as a reactant, the process is

endothermic. If heat is released as a product, the process is exothermic. For the enthalpy diagram, the arrow always points from reactants to products. For endothermic reactions, the products are at a higher energy than the reactants, since the reactants take in heat to form the products. 6-21 Sample Problem 6.2 SOLUTION: (a) H2(g) + O2(g) H2O(l) + 285.8 kJ Heat is a product for this reaction and is therefore given out, so the reaction is exothermic. The reactants are at a higher energy than the products.

Energy H2(g) + O2(g) (reactants) 6-22 H = -285.8 kJ H2O(l) (products) EXOTHERMIC Sample Problem 6.2 SOLUTION: (b) 40.7 kJ + H2O(l) H2O(g) Heat is a reactant in this reaction and is therefore absorbed, so the reaction is endothermic. The reactants are at a lower energy

than the products. Energy H2O(g) 6-23 (reactants) H = + 40.7 kJ H2O(l) (products) ENDOTHERMIC Calorimetry q = c x m x T

q = heat lost or gained c = specific heat capacity m = mass in g T = Tfinal - Tinitial The specific heat capacity (c) of a substance is the quantity of heat required to change the temperature of 1 gram of the substance by 1 K. 6-24 Table 6.2 Specific Heat Capacities (c) of Some Elements, Compounds, and Materials Substance Specific Heat Capacity (J/gK) Elements

Substance Solid materials aluminum, Al 0.900 wood 1.76 graphite,C 0.711 cement 0.88

iron, Fe 0.450 glass 0.84 copper, Cu 0.387 granite 0.79 gold, Au 0.129

steel 0.45 Compounds 6-25 Specific Heat Capacity (J/gK) water, H2O(l) 4.184 ethyl alcohol, C2H5OH(l) 2.46

ethylene glycol, (CH2OH)2(l) 2.42 carbon tetrachloride, CCl4(l) 0.862 Sample Problem 6.3 Finding the Quantity of Heat from a Temperature Change PROBLEM: A layer of copper welded to the bottom of a skillet weighs 125 g. How much heat is needed to raise the temperature of the copper layer from 25C to 300.C? The specific heat capacity (c) of Cu is 0.387 J/gK. PLAN: We know the mass (125 g) and c (0.387 J/gK) of Cu

and can find T in C, which equals T in K. We can use the equation q = cmT to calculate the heat. SOLUTION: T = Tfinal Tinitial = 300. 25 = 275C = 275 K q = cmT = 0.387 J x 125 g x 275 K gK 6-26 = 1.33x104 J Figure 6.9 Coffee-cup calorimeter. This device measures the heat transferred at constant pressure (qP).

6-27 Sample Problem 6.4 Determining the Specific Heat Capacity of a Solid PROBLEM: A 22.05 g solid is heated in a test-tube to 100.00C and added to 50.00 g of water in a coffee-cup calorimeter. The water temperature changes from 25.10C to 28.49C. Find the specific heat capacity of the solid. PLAN: Since the water and the solid are in contact, heat is transferred from the solid to the water until they reach the same Tfinal. In addition, the heat given out by the solid (-qsolid) is equal to the heat absorbed by the water (qwater). SOLUTION: Twater = Tfinal Tinitial = (28.49C 25.10C) = 3.39C = 3.39 K Tsolid = Tfinal Tinitial = (28.49C 100.00C) = -71.51C = -71.51 K 6-28

Sample Problem 6.4 c csolid = = 6-29 H2O x mass H2O x T H2O masssolid x Tsolid

4.184 J/gK x 50.00 g x 3.39 K 22.05 g x (-71.51 K) = 0.450 J/gK Sample Problem 6.5 Determining the Enthalpy Change of an Aqueous Reaction PROBLEM: 50.0 mL of 0.500 M NaOH is placed in a coffee-cup calorimeter at 25.00oC and 25.0 mL of 0.500 M HCl is carefully added, also at 25.00oC. After stirring, the final temperature is 27.21oC. Calculate qsoln (in J) and the change in enthalpy, H, (in kJ/mol of H2O formed). Assume that the total volume is the sum of the individual volumes, that d = 1.00 g/mL and c = 4.184 J/gK PLAN: Heat flows from the reaction (the system) to its surroundings (the solution). Since qrxn = qsoln, we can find the heat of the reaction by calculating the heat absorbed by the solution.

6-30 Sample Problem 6.5 SOLUTION: (a) To find qsoln: Total mass (g) of the solution = (25.0 mL + 50.0 mL) x 1.00 g/mL = 75.0 g Tsoln = 27.21C 25.00C = 2.21C = 2.21 K qsoln = csoln x masssoln x Tsoln = (4.184 J/gK)(75.0 g)(2.21 K) (b) To find Hrxn we first need a balanced equation: HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l) 6-31 = 693 K Sample Problem 6.5 For HCl: 25.0 mL HCl x 1 L x 0.500 mol 103 mL

1L x 1 mol H2O 1 mol HCl = 0.0125 mol H2O For NaOH: 50.0 mL NaOH x 1 L x 103 mL 0.500 mol x 1 mol H2O = 0.0250 mol H2O 1L 1 mol NaOH HCl is limiting, and the amount of H2O formed is 0.0125 mol. Hrxn =

6-32 qrxn mol H2O = -693 J x 1 kJ 103J 0.0125 mol = -55.4 kJ/mol H2O Figure 6.10 A bomb calorimeter. This device measures the heat released at constant volume (qV).

6-33 Sample Problem 6.6 Calculating the Heat of a Combustion Reaction PROBLEM: A manufacturer claims that its new dietetic dessert has fewer than 10 Calories per serving. To test the claim, a chemist at the Department of Consumer Affairs places one serving in a bomb calorimeter and burns it in O 2. The initial temperature is 21.862C and the temperature rises to 26.799C. If the heat capacity of the calorimeter is 8.151 kJ/K, is the manufacturers claim correct? PLAN: When the dessert (system) burns, the heat released is absorbed by the calorimeter: -qsystem = qcalorimeter To verify the energy provided by the dessert, we calculate qcalorimeter.

6-34 Sample Problem 6.6 SOLUTION: Tcalorimeter = Tfinal Tinitial = 26.799C 21.862C = 4.937C = 4.937 K qcalorimeter = heat capacity x T = 8.151 kJ/K x 4.937 K = 40.24 kJ 40.24 kJ x kcal = 9.63 kcal or Calories 4.184 kJ The manufacturers claim is true, since the heat produced is less than 10 Calories. 6-35

Stoichiometry of Thermochemical Equations A thermochemical equation is a balanced equation that includes Hrxn. The sign of H indicates whether the reaction is exothermic or endothermic. The magnitude of H is proportional to the amount of substance. The value of H can be used in a calculation in the same way as a mole ratio. 6-36 Figure 6.11 The relationship between amount (mol) of substance and the energy (kJ) transferred as heat during a reaction 6-37 Sample Problem 6.7

Using the Enthalpy Change of a Reaction (H) to Find Amounts of Substance PROBLEM: The major source of aluminum in the world is bauxite (mostly aluminum oxide). Its thermal decomposition can be represented by the equation Al2O3(s) 2Al(s) + 3 O2(g) Hrxn = 1676 kJ 2 If aluminum is produced this way, how many grams of aluminum can form when 1.000x103 kJ of heat is transferred? PLAN: From the balanced equation and H, we see that 2 mol of Al is formed when 1676 kJ of heat is absorbed. heat (kJ) 1676 kJ = 2 mol Al mol of Al

mass (g) of Al multiply by M 6-38 Sample Problem 6.7 SOLUTION: 1.000x103 kJ x 2 mol Al 1676 kJ 6-39 x 26.98 g Al 1 mol Al = 32.20 g Al Hess Law Hesss law states that the enthalpy change of an overall process is the sum of the enthalpy changes of its

individual steps. Hoverall = H1 + H2 + . + Hn H for an overall reaction can be calculated if the H values for the individual steps are known. 6-40 Calculating H for an overall process Identify the target equation, the step whose H is unknown. Manipulate each equation with known H values so that the target amount of each substance is on the correct side of the equation.

Change the sign of H when you reverse an equation. Multiply amount (mol) and H by the same factor. Add the manipulated equations and their resulting H values to get the target equation and its H. 6-41 Note the amount of each reactant and product. All substances except those in the target equation must cancel. Sample Problem 6.8

Using Hesss Law to Calculate an Unknown H PROBLEM: Two gaseous pollutants that form in auto exhausts are CO and NO. An environmental chemist is studying ways to convert them to less harmful gases through the following reaction: CO(g) + NO(g) CO2(g) + N2(g) H = ? Given the following information, calculate the unknown H: Equation A: CO(g) + O2(g) CO2(g) HA = -283.0 kJ Equation B: N2(g) + O2(g) 2NO(g) HB = 180.6 kJ PLAN: 6-42 Manipulate Equations A and/or B and their H values to get to

the target equation and its H. All substances except those in the target equation must cancel. Sample Problem 6.8 SOLUTION: Multiply Equation B by and reverse it: NO(g) N2(g) + O2(g); H = - 90.3 kJ Add the manipulated equations together: Equation A: CO(g) + O2(g) CO2(g) Equation B: (reversed) NO(g) CO(g) + NO(g) H = -283.0 kJ N2(g) + O2(g)

H = - 90.3 kJ CO2(g) + N2(g) Hrxn = -373.3 kJ 6-43 Table 6.3 Selected Standard Enthalpies of Formation at 25C (298K) Formula Calcium Ca(s) CaO(s) CaCO3(s) Hf (kJ/mol) Formula 0 -635.1 -1206.9 Carbon

C(graphite) C(diamond) CO(g) CO2(g) CH4(g) CH3OH(l) HCN(g) CSs(l) Chlorine Cl(g) 6-44 0 1.9 -110.5 -393.5 -74.9 -238.6 135

87.9 121.0 Hf (kJ/mol) Formula Hf (kJ/mol) 0 -92.3 Hydrogen H(g) H2(g) Silver Ag(s) AgCl(s) 218 0 Sodium

Nitrogen N2(g) NH3(g) NO(g) 0 -45.9 90.3 Cl2(g) HCl(g) Oxygen O2(g) O3(g) H2O(g) H2O(l) 0 143

-241.8 -285.8 Na(s) Na(g) NaCl(s) 0 -127.0 0 107.8 -411.1 Sulfur S8(rhombic) 0 S8(monoclinic) 0.3 SO2(g) -296.8 SO3(g)

-396.0 Sample Problem 6.9 Writing Formation Equations PROBLEM: Write balanced equations for the formation of 1 mol of the following compounds from their elements in their standard states and include Hf. (a) Silver chloride, AgCl, a solid at standard conditions. (b) Calcium carbonate, CaCO3, a solid at standard conditions. (c) Hydrogen cyanide, HCN, a gas at standard conditions. PLAN: Write the elements as reactants and 1 mol of the compound as the product formed. Make sure all substances are in their standard states. Balance the equations and find the value of Hf in Table 6.3 or Appendix B. 6-45

Sample Problem 6.9 SOLUTION: (a) Silver chloride, AgCl, a solid at standard conditions. Ag(s) + Cl2(g) AgCl(s) Hf = -127.0 kJ (b) Calcium carbonate, CaCO3, a solid at standard conditions. 3 Ca(s) + C(graphite) + 2 O2(g) CaCO3(s) Hf = -1206.9 kJ (c) Hydrogen cyanide, HCN, a gas at standard conditions. H2(g) + C(graphite) + N2(g) HCN(g) 6-46 Hf = 135 kJ

Figure 6.12 The two-step process for determining Hrxn from Hf values. 6-47 Sample Problem 6.10 Calculating Hrxn from Hf Values PROBLEM: Nitric acid, whose worldwide annual production is about 8 billion kilograms, is used to make many products, including fertilizer, dyes, and explosives. The first step in the industrial production process is the oxidation of ammonia: 4NH3(g) + 5O2(g) 4NO(g) + 6H2O(g) Calculate Hrxn from Hf values. PLAN: Use the Hf values from Table 6.3 or Appendix B and apply the equation Hrxn = mHf (products) - nHf (reactants)

6-48 Sample Problem 6.10 SOLUTION: Hrxn = mHf (products) - nHf (reactants) Hrxn = [4(Hof of NO(g) + 6(Hof of H2O(g)] - [4(H of NH3(g) + 5(H of O2(g)] = (4 mol)(90.3 kJ/mol) + (6 mol)(-241.8 kJ/mol) [(4 mol)(-45.9 kJ/mol) + (5 mol)(0 kJ/mol)] = -906 kJ Hrxn = -906 kJ 6-49 Chemical Connections Figure B6.1 6-50

The trapping of heat by the atmosphere. Chemical Connections Figure B6.2 Evidence for the enhanced greenhouse effect. Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display. Since the mid-19th century, atmospheric CO2 has increased. Since the mid-19th century, average global temperature has risen 0.6C. 6-51